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# Chapter 5 Thomas 9 Ed F 2008

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## Chapter 5 Thomas 9 Ed F 2008Document Transcript

• CrnprpR Applications of Integrals OYERVIEW Many things we want to know can be calculated with integrals: the areasbetweencurves,the volumes and sudaceareasof solids, the lengthsof curves, the amount of work it takes to pump liquids fiom belowground, the forces againsr floodgates,the coordinatesof the points where solid objects will balance.We define all of these as limits of Riemann sums of continuous functions on closed intervals. that is, as integrals,and evaluatethese limits with calculus. There is a pattern to how we define the integrals in applications, a pattern that, once learned,enablesus to define new integralswhen we need them. We look at specific applications 0rst, then examine the pattern and show how it leads to integrals in new situations. AreasBetweenCurves This section shows how to lind the areas of regions in the coordinate plane bv integrating the functions that define the regions, boundaries. The Basic Formula a Limitof Riemann as Sums Supposewe want to find the area of a region that is bounded above by the curve -y =.l(r), below by the curve ) = g(.r), and on the left and right by the lines .r : 4 and r : D (Fig. 5.1). The region might accidentallyhave a shapewhose area w e c o u l d f i n L w i r h g e o m e t r l b u r i l / a n d g a r e i t r b i l r a r ) o n r i n u o ut,u n c r r o n,r l . c e usuaily have to flnd the area with an integral. To see what the integml should be, we lirst approximate the region with,? basedon a parlition p : {.r0, ... , r,,} of [ri, D] (Fig. 5.2, on verticalrectangles rr, the following page). The area of the tth rectangle(Fig. 5.3, on rhe followrng page; ls AA1 : tsjcfi1 x width : Ll (c.t) g(ci)l A4. We then approximatethe area of the region by adding the areasof the ,?rectanqles: A* f ^A, -Itr,,.r grr.,rlA,:,. ' The region between y: f(x) and As ll P | -+ 0 the sums on rhe right approachthe limit y: g ( x ) a n d t h e l i n e sx : a a n d x = b . _ g(,r)l d,r because /j t/(r) 365
• Chapter5: Applicationsof Integrals 356 T - s(ck) f(t) Stctll ::.: We approximate regionwith the perpendicular the x-axis. to AAk = areaof kth rectangle,f(c*) - g(cr) = height, lxk = width rectangles this integral' /and g arecontinuous. takethearea theregionto be thevalueof of We That is, lim )-l/(c,)-gtcrtlAxr - fo rf ,r,- A- Stxt]dx nP o-- J Definition with /(r) > g(x) throughout b], ihenthe area g [4, Ifl ancl arecontinuous the curves : /(*) andy : g(t) ftom a to b is the y of the regionbetween integral [/ - 8 | liom a to D: o[ rb U(x)- g(x)ldx. (1) A= J To apply Eq. (1) we take the following steps. How to Find the Area Between Two Curves Graph the cunes and draw a tepresentative rectangle. Thts rcveals l. which curve is / (upper curve) and which is g (1owercurye). It also helps find the limits of integration if you do not aheady know them. ,s'I ,=sinr 2. Find the limits of integration. I--''- 3. Write a formula for f(x) - g(x). Simplify it if you can. 4. lntegrate [f(x) - gQ, fron a to b Tl1e number you get is the area. H Find the area between lv = sec2 and y : 3i1;5 from 0 to z/4' x EXAMPLE 1 Solution Step t; We sketch the cutves and a vertical rectangle(Fig. 5.4). The upper curve is the gmph of /(;r) = secz;r: the lower is the graph of g(;) : sipI' Sfep 2; The limits of integration are already given: a :0' b = r /4' ,'.]].The regionin Example'lwith a - SteP3: f (x) - 8(x) = sec2;r sin't rectangle. typicalapproximating
• 5.1 AreasBetweenCurves 367 Step 4: {r' *- sinx)r/x : ftan + cos -r x]i/a lo'o f'.+l,to+:+ -t L f, Curves That Intersect When a region is detemined by curves that intersect, the intersection points give the limits of integration. - x2 Find the area of the region enclosedby the parabola ! :2 EXAMPLE 2 : -y. and the line y Solution Sfep t; Sketch the curves and a vertical rectangle (Fig. 5.5). Identifying the upper (-{,,f(-r) -x. The ,{-coordinates and the lower curyes, we take /(x) :2- g(r): xz arLd pointsare the limits of integration. of the intersecdon Step 2; We find the limits of integation by solving | :2 - xz and y : -1 ri- ( 1 ,1 ) multaneously for .r.' -x [ q r l L t/r1 r ) . u r d . q ( f r 2-xz : x2-x-2:0 R.rir.. (x+l)(x -2):0 Fxrr.' (a s(r)) -1, x: x :2. sot. The region runs from x : - 1 to.r : 2. The limits of integrationare a : - l, b : 2. Step 3: 5.5 The regionin Example with a 2 typicalapproximatingrectangle. Step 4 t2 f -2 -t12 - : o: rf,.,>e(x)ldx I t z + x - x z t d x = l z r + z - ,I l' JJ J-t I r / 4 8 / | 1 (o*r-r,l (-'*l*r/ ?qq - -'2 At 3 2 l Technofogy The Intersectionof Two Graphs One of the difficult and some- times frustrating parts of integmtion applications is finding the limits of inte- gration. To do this you often have to find the zeroes of a function or the intenection points of two curues. g(r) using a graphing utility, you enter To solve the equation /(x): yr: f@) and y2: g@)
• Chapter Applications lntegrals of 5: 368 you andusethe grapherroutine to find the Pointsof intersection'Altematively' - g(T) : 0 with a root finder'Try bothprocedures lun solve*tJ /(:r) quation with and g(x):3-x' f (x):lnx hidden When points of intersectionarc not clearly revealedor you suspect further use of calculus behavior,additional work with the graphing utility or may be necessary, ISECT =.79205996845 curves : Inx and lz:3 - x' usinga built-infundion y1 a) The intersecting intersection to find the - : x O) Usinga built-inroot finder to find the zero of f(x) Inx 3 + with ChangingFormulas Boundaries points' we partition If the formula for a bounding curve changesat one or mole th,gionintosubregionsthatconespondtotheformulachangesandapplyEq. (1) to eachsubregion. Find the areaof the region in the fint quadrantthat is bounded EXAMPLE 3 - 2' : aboveby y : .r/i and below by the t-axis and the line y x solution is the graphof Step t.' The sketch(Fig. 5.6) showsthat the region'supperboundary <2to g(x) = fromg(;r):0for0 S r changes f(; : JV. Theloweiboundary -' Z tot Z : t : 4 (thereis agreement r : 2)' We partition the region at r : 2 at rectanglefor eachsubregion' into subregiJns ani f and tketh a representative A 5.6 When the formulafor a boundingcurve changes, area integralchanges match to the (Example3). 'fe limits of integrationfor regionA area : O andb :,2' The lefchand step 2: li#t for region B rs a = 2. To find the right-hand limit, we solve the equations
• 5.1 AreasBetween Curves 369 y : rG andy : x - 2 simultaneously r.' for EqratcI (.r)rnd ^- - z v^ .9(11 a:(x.-2)2:x.2-4x+4 SqLrafeboth x 2- 5 x + 4 : o Re\$'r'ite. (.r-1)(r-4):0 FacL(r'. -_1 -_i Solve. Only the value.x=4 satisfies equation the The valuer:1is an Ji:x-2. extraneous introduced squaring. right-hand root by The limit is b:4. f (x)- S@): -0: S t e p3 . ' F o r 05 x = 2 : - e ( x ) : , 6E - ( x - 2q : ) Ji - x +2 F o r2 a x a 4 : f(x) Step 4.' We add the areaof subregions and B to find the total area: A f)-f4 = T o t a fr e a | J i d x I l t J x - x + 2 t d x a -o - - - - / J Jz areaof B arca of A 'l2 f) f1 -2 14 fil.+l1*'''-v+2,1, /2 - ?ef,,_o_ (1,,0,',, .' 8)_ _8 r r , r{zt'''-z++1 ?,t, : !. -, , 33 lntegrationwith Respect y to If a region's bounding curves are describedby functions of y, the approximating rectanglesare horizontal insteadof vertical and the basic formula has y in place of r. t 1 I L*,,, use tJteformula In Eq. (2),/always denotes right-hand the - stytldy. e: l.' vrrt (2) curve and I the left-hand curve, so fO) - e(y) is nonnegative.
• Chapter 5: Applicationsof lntegrals 370 Find the area of the region in Example 3 by integrating with EXAMPLE 4 rcspect to ). Solution (c()), ).) x:y+2 Step t; We sketchthe region and a typical horiz,ontalrectanglebasedon a panition of an interval ofy values(Fig. 5.7). The region's right-handboundary is the line x : y f 2, so l (y) : y + 2. The left-hand boundary is the curve ir : y2, so g(y) : y2. Step 2: The lower limit of integration is l = 0. We find the upper lirnit by solving I : y *2 and r: )2 simultaneously y; for 5.7 lt takestwo integrations find the l+t:l to areaof this regionif we integrate with 2=o ,t-) respectto x It takes only one if we with respect y (Example integrate to 4). ( ) + 1 ) ( ) 2 ): 0 Iurt,, 1'- _t ) S ,i l . The upperlimit of integmtion D : 2. (The valuey : -1 givesa point of inter- is sectlor belowthe r-axis.) Step3: =)+2 y2=2iy )? /())-s()) Step 4: v,rt- : e: 1,,'' srt,rat 1 2 + y - y ' ) l d y l,' f r,2 rt l' : +; 1l L2Y 3 ln 8 l0 - o.+ t - 4 3 3 This is the result of Example 3, fbund with less work. Integrals with Formulas from Geometry Combining andgeometry. The fastest way to find an areamay be to combinecalculus The Area of the Region in Example 3 Found the Fastesi EXAMPLE5 way Find the area of the region in Example 3. So/ution The area we want is the areabetweenthe curye I : Ji,O = x S 4, and 1:rf the x-axis, minus the area of a triangle with base 2 and height 2 (Fig. 5.8): 2 lr2rr2r / x r u = f or t a , .2 '.2 2 ,11 J .lo 5.8 The areaof the blue regionis the :3ttl-o -'z:+' area underthe parabola = 1& minus , y the areaof the triangle. JJ -
• Exercises 5.1 371 Moral of Examples3-5 It is sometimes easierto lind the areabetween two with respect y instead ,r. Also, it may help to combine curves integrating by to of geometry calculus. and After sketching region,takea moment determine the to the bestwav to oroceed. Exercises 5.1 Find the areasof the shadedregions in Exercises 1-8. l. 2.
• Chapter Applications Integrals of 5: 372 16. y : x 2 - 2 x and Y=r area. In Exercises findthetotalshaded 9-12, and y = -tt2 14x t7. ) : 12 10. 9, ] 18. y = l - l x ' ano ]:x-14 (-3,5) 19. y = x 4 - 4 x 2 + 4 and Y=a2 I and y:0 20, y = r n @ = 7 , a>0, _,_,2 5y = x * 6 (How many intersection points are 21. y = {fl nd there?) 2x3 - (x212)+4 5x 22,y:lx2-41'and ): 2 I Find the areasof the regions enclosedby the lines and curves 23-30. Exercises -3) : r : 0 , a n d) : 3 x:2y2, 23. 24. x = y2 u1t4 a: y 12 and' 4x - Y = 16 25. y2 - 4x:4 26, x - y2 :0 arld x +2Y2 =3 ^nd r+3Y2:2 2 7 ,x * y 2 : O and ,r*)a:2 2 8 .x - y 2 / t : 0 (-2,4 y 29.x =y2-l and x='yJl 3 0 . r = ) l 3- y 2 a n d x = 2 y -2 -l Find the areasof the regions enclosedby the curves in Exercises I 31'34. and .xa-Y:l 31.4x2+y:4 I a n d 3 x 2- y : 4 y :O 3 2 .x 3 I, (3,5) for xZ0 and :r*)a:1, 3 3 .x * 4 y 2 : 4 and 4.t*y2:0 3 4 ,x a y z : 3 12. Find the arcas of the regions enclosedby the lines and curves in ) I+. 35-42. Exercises (3,6) 6 and y = sin2.r, 0:.x57t 3 5 .y = 2 s 1 n x and y:5e92.r, -7t13=x Sr13 v 36.):8cos.r .t2 and 1:1 3 7 .y : c o s ( r x / 2 ) 38. y = sin(nx/2) and 1t:r (3,1) and'x:tt/4 39, l:sec2.r, y=tarfx, x:-tr/4, and .x:-tan2), -7r14=r 3n/4 4O. :ta#y x r and -x=0, 0=y<7t/2 4 1 .r : 3 s i n l / i o s y -' 4 2 .y : s s s 2 1 1 t a J 3 1 r r d y : r 1 / 1 , - l = i r : l 3) a rcgion enclosed the curve by 43. Find the areaof the propeller-shaped Find the areasof the regions enclosedby the lines and curves in - y3 : 0 andthe line.r - Y : 0. t 13-22. Exercises region enclosedby the 44. Find the area of the propeller-shaped 1 3 . y : a 2- 2 clruesx -yll3 : 0 andr - )l/5 = 0. ar'd y:2 14. y =2x - x2 arLd Y: -J 45, Find the areaof the region in the first quadrantboundedby the line I :1, the line,t :2, the curve1 : 1/-r'?, the -t-aris. and 15. y :1+ and ) :8r
• Exercises 5.1 373 Suppose area of the region betweenthe graph of a positive 46. Find the area of the tdangular region in the first quadrant the continuousfunction / and the .x-axisfrom x : 4 to x : b is 4 boundedon the left by the y-axis and on the right by the curves units. Find the areabetween curves)r: /(.r) and the ) = sin.rand): cos,r. square y-2f(x)ftomx:atox-b. belowby the pambola : ,r2 andabove by 47, The regionbounded ) Which of the following integals, if either calculates areaof the : 4 is to be parlitioned into two subsections equal of the line ) region shownhere?Give reasonsfor your answer. areaby cutting acrossit with the horizontal line ), = c. the shaded a) Sketchthe region and draw a line ) : c acrossit that looks fL f) -l-x)d:(= I (x 2xdx al | about right. In terms of c, what are the coordinates the of J J-t I points where the line and parabolaintersect?Add them to ' rot tl -zxax bt / r-x-(r))dx- your figure. | JI J-t b) Find c by integratingwith respectto ),. (This puts c in the limits of integration.) to.x.(Thisputsc into the c) Find c by integntingwith respect integrand well.) as ''2 and Find the areaof the regionbetween curve )):3 the the line y = 1 by integratingwith respectto (a) t, (b) ). 49. Find the areaof the region in the first quadrantboundedon the left by the ),-axis,below by the line y = t /4, aboveleft by the curve y = I * .,[, and aboveright by he (i.jtr..te : 2/ Ji, y 50. Find the area of the region in the first quadrantboundedon the left by the y-axis, below by the curve r : 2/t, aboveleft by right by the line i :3 - ). the curyer : () - l)'?,andabove 54. True, sometimestue, or never tue? The area of the region between graphsof the continuous functions): /(i) and the : g(r) andthe verticallinesr : a andx: b (a < b) rs ) rb I lf @) - s(x)ld)c. J Give reasonsfor your answer. S CASExplorationsand Proiects In Exercises 55-58, you will find the area between curves in the plane when you cannot find their points of intersectionusing simple algebra. Use a CAS to perform the following steps: The figure here shows triangle AOC inscribed in the region cut a) Plot the curves together to see what they look like and how many ftom the parabola ) - .r2 by the line y: a2. Find the limit of (hey hae. points of inrersection the latio of the area of the triangle to the area of the parabolic b) Use the numerical equation solver in your CAS to find all the region as 4 approaches zerc. points of intersection. - g(r)l over consecutive pairs of intersection c) Integate f(r) values. d) Sum together the integrals found in part (c). ^- ^ -2, I . g 1 x .:1 - J x 5 5 . '/3 2 3 : 1x1 t4 1 0 .g ( . r r : 8 - l 2 r 5 6 ./ t x t : i - t r ' - 57. /(r) =x *sin(2tc), g(x) : 13 58. /(.x) : -r2sqsir, g(tc) : x1 - x
• Chapter 5: Applicationsof Integrals 374 Volumes Slicing by Finding From the areasof regions with curved boundaries,we can calculatethe volumes of cylinders with curved basesby multiplying base area by height. From the volumes of such cylinders, we can calculate the volumes of other solids. 5licing Supposewe want to find the volume of a solid like the one shown in Fig. 5.9. At Cross sectionR(jr). Its area is A(ir). each point r in the closed inteNal la, rl the cross section of the solid is a region R(x) whose area is A(r). This makesA a real-valuedfunction of .r' If it is also a continuous lunction of .r, we can use it to define and calculate the volume of the solid as an integral in the following way. We partition the interval [4, ]l along the r-a{is in the usual manner and slice the solid, as we would a loaf of bread, by planes petpendicular to the -r-axis at the partition points. The tth slice, the one between the planes at xk r and tn, has approximatelythe same volume as the cylinder betweenthesetwo planes basedon the region R(,rr) (Fig. 5.10). The volume of this cylinder is Vr : base area x height sedion lf the areaA(x) of the cross : A(rr) x (distancebetween the planes at -rr r and xr) functionof x, we can R(x)is a continuous find the volumeof the solid bY : A(xr)Axr. A(x) from a to b. integrating The volume of the solid is therelbre approximatedby the cylinder volume sum - r,. r,r-. ?-,'^'^ This is a Riemann sum for the function A(r) on [c, Dl. We expectthe approxlmahons from these sums to improve as the norm of the partition of la, bl goes to zero, so we define their limiting integral to be the volume of the solid. Approxrmatlng Plane at.r[ I cllinderbased p l a n en r r , on]((rt] -'i.---------' -, The cylinder's base is the region R(,v*). NOTTO SCAI-E view of the sliceof the solidbetweenthe . Enlarged cylinder' planes xk r and xk and its approximating at
• 5 . 2 F i n d i n g o l u m eb y S l i c i n g 3 7 5 V s Definition The volume of a solid of known integrable cross-sectionarea A(x) from x = a to x: D is the integralofA from a to r. v : fo e61a,. (j) J T o a p p l l E q . { l ) . w e r a k et h e f o l l o w i n g: r e p . . How to FindVolumesby the Method of Slicing 1 Sketch the solid and a typical cross section. 2. Find a formula for A(r). 3. Find the limits of integration. 4. Integrate A(x) to find the volume. EXAMPLE 7 A pyramid 3 m high has a square base that is 3 mon a side. The cross section of the pyramid perpendicularto the altitude r m down from the ve ex is a squarer m on a side. Find the volume of the pyramid. Solution Step 1: A sketcll. We draw the pyramid with its altitude along the r-axis and its vertexat the origin and includea typical crosssection(Fig. 5.11). Typical cfoss section --r\$€t -=_,,., sections the pyramidin Example The cross of 1 are 5quare5. Step 2: A.fonnula for A(x). The cross section at r is a squarex meters on a side, so lts area ls A(r) : 12' Step 3: Tlle lintits of integration. The squaresgo from r :0 to;r :3 Step 4: The volwne. 'l' lb t' - r. -, At^trlt - v=I | t:dr r |=9 J.. J,, .l , The volume is 9 mr.
• Chapter Applications Integrals of 376 5: wedge cut from a cylinderof radius3 by two planes. is 2 A curved EXAMPLE Bonaventura Cavalieri ('l 598-1647) planecrosses One planeis perpendicular the axis of the cylinder.The second to Cavalied, a studentof Galileo's, discovered the nrst plane at a 45' angle at the centerof the cylinder. Find the volume of the that if two plane rcgions can be afiangedto wedge. lie over the sameinterval of the i-axis 1n sucha way that they haveidentical vertical Solution crosssectionsat every point, then the rcgions perpen- Step 1: A sketch. draw the wedgeand sketcha typical crosssection We havethe sarnearea.The theorem(and a letter dicularto the:r-axis(Fig. 5.12). from Galileo) were of recommendation enoughto win Cavalied a chai at the Univelsityof Bolognain 1629.The solid 2lt -7 geometryversionin Example 3, which Cavalieri neverproved,was given his name by later geometen. have Crcsssections the samelongthat everypoint in [d, b] 5-t2 The wedgeof Example slicedperpendicular 2. sections redangles. are to the x-axis. The cross Step 2: Theformula for A(x). The cross section at r is a rectangleof area A(') : (heishtxwidth) (zJe - 'r) : (r) : z x t t: Y - x ' . ^, Step 3: The linxits of integration. The rectangles from r :0 to.r :3. run Step 4: The volune. ?b ^3 Ax)dx: | 2tcJ9x2dx - v:l Ja JO 113 : -;e - xr3/' )o I ,f : o+?e),/' 1! = lf r1r. nncsfalc- J . L n d s u b s t i l U r eb r c k . tr - 19 Cavalieri's Theorerrr EXAMPLE 3 Cavalieri's theoremsaysthat solids with equal altitudesand identical parallel cross- theorem. Thesesolids 5.13 Cavalieri's section areashave the samevolume (Fig. 5.13). We can seethis imrnediately from havethe samevolume.You can illustrate Eq. (1) becauset}Ie cross-sectionarea function A(.r) is the same in each case. f this yourself with stacks coins. of
• Exercises 5.2 377 Exercises 5.2 Areas Cross-Section In Exercises and 2, find a fomula for the arcaA(r) of the cross I sectionsof the solid perpendicular the l-axrs. to 1. The solid lies betweenplanesperpendicular the r-axis at i = to -l and,r:1. In eachcase,the crosssections perpendicular to the r-axis betweentheseplanesrun from the semicircley = -^4 - ,r2 to the semicircley : 4[ - ,2. a) The cross sectionsme circular disks with diametersin the xy-plane. planesperpendicular ther-axis at.r : 0 The solidliesbetween to andl : 4. The crosssectionsperpendicular the-t-axisbetween to these planes run from the parabola y = -aE to the parabola r: Ji. a) The cross sectionsare circular disks with diametersin the r)-plane. b) The crcsssections squares are with bases ther),-plane. in b) The crosssectionsare squareswith basesin the r)-plane. c) The cross sectionsarc squarcswith diagonalsin the xy- plane. (The length of a square'sdiagonalis .,4 times the lengthof its sides.) c) The crosssections squares are with diagonals ther)-plane. in d) The crosssectionsare equilateraltriangleswith basesin the rJ-plane. Volumes Slicing by Find the volumesof the solids in Exercises3-12. 3. The solid lies betweenplanesperpendicular the,-axis at .r : 0 to and r :4. The crosssections perpendicular the axis on the to The crosssectionsarc equilateraltriangleswith bases the in interval 0 5 i :4 are squareswhose diagonalsrun ftom the d) parabola1 : -.vE to the parabolaf : Jtr. ry-plane.
• 374 chapter5: Applications Integrals of 4, The solid lies between planes perpendicularto the r-axis at -x : - 1 and r : I . The cross sections perpendicular to the ir axis are run from lhe parabolay - x7 to circulardisks whosediamerers the Darabola :2 - t2. t 5. The solid lies betweenplanesperpendicular the r-axis at to Cavalieri's Theorem r : - I andr : 1. The crosssections perpendicular the axis to ll. A twisted solid. A square sidelengths lies in a planeper- of betweentheseplanesare vertical squares whosebase edges pendicular a line L Onevertex the square on t. As this to of lies run from the semicircle - - 1(1- rt lo rhe semicircle - y J squ&emovesa distance alongl, the square l? tums one revo- !7-7. lution aboutZ to generatea corkscrewlike column with square perpendicular ther-axis at r = 6. The solidlies between planes to crosssections. I and n: l. The crosssections perpendicular the ,r-axis to a) Find the volume of the column. planes squares between these are whosediagonals frcm the run b) What will the volumebe if the squaretums twice instead semicircle : -{ -7 to the semicircle y: JT=V. Qn ) Givercasons your answer of once? for lengthof a square's diagonal r/Z timesthelengthof its sides.) is berween curve) : 2 vfi; planes 12. A solid lies between 7. t he base thesolidis thereSion of lhe The qoss sections perpendicular tbe -{-axisat and the interval[0, r] on the.r-axis. perpen- to -r = 0 ard .r : 12.The cross dicularto the r-axis are sections planes perpendicular by to ther-axis arecirculardisks whosediametersrun from the line ) : -r/2 to the line ) : -r. Explain why the solid has the same volumeas a right circular conewith baseradius3 and height12. 13. Cavalieri'soriginal theorem. ProveCavalieri's original theo veftical equilateral triangles with bases running from the a) rem (marginalnote, page 376), assuming that each region is I-axrs to me culve: boundedabove belowby thegraphs continuous and of functions. b) vertical squareswith basesrunning from the r-axis to the t4. The volume of a hemisphere(a classical cutve. application af Cav- alieri's theorem). Derive the formula V : (2/3)zR3 for the 8. The solid lies between planes pe.pendicularto the r-axis at -x = volumeof a hemisphere radiusR by comparing crosssec- of its -r /3 and x : xr/3. The crcss sectionsperpendicularto the r- tionswith the crosssections a solid right circularcylinderof of axis are mdiusR andheightR from which a solidright circularconeof circular disks with diameters running from the curve y : a) baseradius andheishtR hasbeenremoved. R tan,r to the curve ): secr; verlical squares whose base edges run from the curve y : b) tanr to the cuNe ): SeCr. 9. The solid lies betweenplanesperpendicularto the y-axis at y : 0 and ) : 2. The cross sectionsperpendicularto the ]-axis are cir- cular disks with diametersrunning from they-axis to the parabola 10. The base of the solid is the disk j'? + ),2 5 1. The cross sections I and y - I by planesperpendicularto the )-axis between], - are isoscelesdght triangles with one leg in the disk.
• 5.3 Volumesof Solidsof Revolution-Disksand Washers 379 rn::rt sotids Revorution-Disks of E n The most common applicarion of the method of slicing is to solids of revolution. Solids of revolution are solids whose shapescan be generatedby revoJvingplane regions about axes. Thread spools are solids of revolution; so are hand weights and billiard balls. Solids of revolutign sometimeshave volumes we can find with formulas liom geometry, as in the case of a billiard ball. But when we want to find the volume of a blimp or to predict the weight of a part we are going to have turned on a lathe, formulas from geometry are of little help and we tum to calculus lor lhe un:* ers. 6 revolution Axisof revolution s # If we can arange for the region to be the region betweenthe graph of a contin- uous function I = R(r), a 5 x < ,, and the x-axis, and for the axis of revolution to be the x-axis (Fig. 5.14), we can find the solid's volume in the following way. The typical cross section of the solid perpendicularto the axis of revolution is a disk of radius R(x) and area A(.r) : 71136iut1: rlR(x)12. : The solid's volume, being the integral ofA from.r - a to x : b, is the integral of Cross sectionperpendjcularto zfR(x)1'zfrom a to D. the axis at r is a disk ofarea ,4(r) = 7r(radius)2 tt (R(,r))2 = Volume of a Solid of Revolution (Rotation About the .r-axis) The volume of the solid generatedby revolving about the.{-axis the region betweenthe -r-axis and the graph of the continuousfunction ) : R(r), a S :r 5 b, is v = [ , ftua;ur1t : [' r l R 1 x ) l 2 l r . a* (1) z J. J,, EXAMPLE I The region betweenthe curye y :.,rE,O <.{ S 4, and the r-axis and the x-axisfrom a to b about the x-axis. is revolved about the .x axis to generatea solid. Find its volume.
• 380 Chapter Applications lntegrals 5: of :l, Solution We draw ligures showing the region, a typical radius, and the generated solid (Fig. 5.15).The volumeis fb v : I trlR(x)'dx F - qr l r f1 : I rlrtTax /l1ft i Jo- (4t2 f4 - u . ,t 2 l a =tr lo x d x LJo =z-:8n. 1 l J ,, .f How to Find Volumes Using Eq. (1) 1, Draw the regionandidentifythe radiusfunctionR(x). 2. Square R(r) andmultiply by z. 3. Intesrate find the volume. to The axis of revolution the next example not the .x-axis, the rule for in is but calculating volumeis thesame: the Integmtelr(radius)2 betweenappropriate limits. (b) 5.15 The region(a) and solid(b) in EXAMPLE2 Find the volumeof the solid generated revolvingthe region by Exampl1.e bounded y : .rf andthe lines) : l, x:4 about line y : 1. by the Solution We drawfiguresshowing region,a typicalradius,andthe generated the solid (Fig. 5.16).The volumeis F f( lrr V: I r l 'R ( x ) 1 d r z J, r1 : I n l' J x _ t ) - d x Jt f4 : o I 'l , z E+ r ) a , Jt ^ .4 ^ tt tn lx- - + x l : -I :Tl 2':x ^ J o Lz lr -j To find the volume of a solid generated by revolving a region between the )-axis and a curve x : R()), c < l : d, about the y-axis, we use Eq. (i) with r replaced by ). Volume of a Solid of Revolution (Rotation About the y-axis) (b) 5.16 The region(a) and solid(b) in Example2. -
• 5.3 Volumesof Solidsof Revolution_Disksand Washers 391 EXAMPLE 3 Find the volume of the solid generatedby revolving Lnereglon betweenthe y-axis and the curve x:2/y,I S l,:4, aboutthe y-axis. Solution We draw figures showing the region, a typical radius, and the generated solid (Fig. 5.17).The volumeis v: :zln6)1'z ay l_ l (lr r l,o = ;) o, t4 /'r2 J, /11 I r r^ 4 T r'1. l3l I - tlt : qr | -: | : 4n r . t -t Jt v L )lr L4J .l EXAMPLE 4 Find the volume of the solid generatedby revolving the region betweenthe parabola :12 + I and the line I :3 aboutthe line; :3. I so/ut on We draw ligures showing the region, a typical radius, and the generated solid (Fig. 5.18). The volume is ( N1 ,: l.r r: ) J nrlR(y)'?dy pD /11r:1 rf -ll rl2 - y212 : dy I (b) /1 :r 5.17 The region(a) and solid(b) in L4-4y2+y4ldy I Example3. .,,51f : 7 r | 4 y - ; - y '+ a I 4 l J )l^ I 64rJ2 t5 R(])=3-()'?+ 1) ) z 0 -.'t r:t-+r I (a.) (b) 5.78 The region(a) and solid(b) in Example4. l
• Chapter 5: Applicationsof lntegrals 382 The WasherMethod If the region we revolve to generate a solid does not border on or cross the axis of revolution, the solid has a hole in it (Fig. 5.19). The cross sectionsperpendicular to the axis of revolution ate washersinstead of disks. The dimensionsof a typical washer are R(-r) Outerradius: , rx ) Inner radius: areais The washer's A(x) : TtlR(;)lz rVG)12: o (tn{)l' - tt(r)lt). - The Washer Fonnula for Finding Volumes fb .lrtx)f)dx v: I (3) ra ( [ R r x ) | ' : | | tl outer inner radius mdius squarcd squarcd ) = R(-r) Notice that the function integrated in Eq. (3) is t(Rz - r2), not T (R - r)'z. Also notice that Eq. (3) gives the disk method fomula if /(jr) is zero throughout [a, r]. Thus. the disk method is a soecial case of the washer method. The region boundedby the curve y : xz + | and the line y : EXAMPLE 5 -x t 3 is revolved about the -r-axis to generatea solid. Find the volume of the solid. 5.?9 The crosssectionsof the solid of revolution generated here are washers, Solution not disks, the integralf Ak) dx leads so Sfep t; Draw the region and sketcha line segmentacrossit perpendicularto the to a slightlydifferentformula. (the red segment Fig. 5.20). in a;dsof revolution Step 2,' Find the limits of integrationby finding the.r-coordinates the intersection of points. x2+l:-x+3 ,'+*-2:o (x+2)(x-l):0 , -_1 .-l Step 3; Find the outer and inner radii of the washerthat would be swept out by the line segmentif it were revolved about the x-axis along with the region. ftVe drewthe washer Fig. 5.21,but in your own work you neednot do that.)These in radii are the distances the endsof the line sesmentfrom the axis of revolution. of R(x) : -1 -1- 3 Outer radius: r(x):a241 Inner radius:
• 5.3 Volumes Solids Revolution-Disks Washers 383 of of and +-1 (1,2) lntegratlon j . . i T h er e g i o ni n E x a m p l 5 s p a n n e d y e b a line segmentperpendicular the axis to of revolution. when the regionis 5.21 The inner and outer radii of the cmss section Washer about the x-axis, line revolved the Outerradius: = r+3 washersweptout by the line segmentin R(r) will generatea washer. segment r(.r) = 12 + I Innerradius: Fi9.5.20. Step 4; Evaluate the volume integral. rb - ,= ([Rrx)l' lr(x)]') dx F ( t 1 - il J,, lt ^ = ({ x | 3)' tx' I lt'tdt I ltn!1 l /_rtr fl = / z(S 6x x2 xa;dx J-2 r' r'l' tllr -'l*t^ i sl, s LI How to FindVolumesby the WasherMethod Draw the region and sketch a line segmentacross it perpend.icularto l. the q,is of revolution. When the region is revolved, this segmentwill generatea typical washer cross section of the generatedsolid. 2. Find the limils of integration. Find the outer qnd inner rqdii of the washer swept out by the line segment. 4. Integrate to flnd the volume. To find the volume of a solid generatedby revolving a region about the )-axis, we use the stepslisted above but integratewith respectto l instead of ,r. The regionboundedby the parabola : x2 and the line y : 2y EXAMPLE 6 I quadrantis revolved about the )-axis to generatea solid. Find the volume in the first of the solid.
• 384 Chapter Applications Integrals 5: of Solution Step t.' Draw the region and sketch a line segment across it perpendicular to the axis of revolution, in this case the ),-axis (Fig. 5.22). Step 2: The line and parabola intersect at ) : 0 and l : 4, so the limits of inte- grationare c :0 andd :4. Step 3.' The radii of the washer swept out by the line segment are R(y) : a/ry, r(y) : y/2 (Figs.5.22 and 5.23). ? t22 The region,limitsof integration, 523 The washersweptout by the line and radii in Example 6. segmentin Fig.5.22. Step 4: Eq. (-l ) 'irh i - ,: , (tnci)l, tr(y)t2) dy 1 : l,^ -ltrJ')r, (ltl' sleps I .rnd -l :L'(,-';)*:l+-i]:, 8 3 EXAMPLE 7 The region in the first quadrantenclosedby the parabola y :2e2, the y-axis, and the line ) : I is revolved about the line,r : 3/2 to generatea solid. )=*2or*:f t 5ol Find the volume of the solid. 4!.=;-ti Solution Step 1.' Draw the region and sketch a line segment across it perpendicular to the Er axis of revolution, in this case the line ; :3/2 (Fig. 5.24). Step 2: Tbe limits of integration are ) : 0 to l : 1. Step 3.' The radii of the washer swept out by the line segment arc R(y) :3/2, 524 The region,limitsof integration, r(y) : (3/2) - -/1, (Fiss. 5.24 and 5.25). and radii in Example 7.
• Exercises 5.3 385 The washersweptout by the line segmentin 5.-25 Fig.5.24. Step 4: = n(Ln{r)l' - t'(y)1'?)dy J, :1( [ ; ] ' - ft -0)'),, [; 3tr fl =nlzftz+l: (3{y-y)dy :t I 2 l Exercises 5.3 Volumes the Disk Method by Find the volumes of the solids generatedby revolving the regions bounded by the lines and cu es in Exercises5-10 about the r-axis. ln Exercises 1-4, find the volume of the solid generated revolving by y:Q, r:2 5,,y:n2, x:2 6.y:tl, the shadedregion about the given axis. ):0, i.J-Je 7. 8. y:-r-r2, 2. Aboutthe y-axis r:o -v:o l. About the r-axis jr:0 9. y:v€osir, 0Sx=n/2, ):0, 1 0 .I - 5sga. 1=0, r:-n/4, x:r/4 In Exercises and 12, lind the volumeof the solid generated 1l by revolving regionaboutthe givenline. the above the line y : 11. The regionin the iirst quadrant bounded by J1, blo* by the curve ) : secr tanr, and on the left by the aboutthe line y = 1i -r-axis, 4. Aboutthe r-axis above theline y : 2, 12. Theregionin thefirstquadrant bounded by below thecurve by r : r/2, andon theleft by l:2sin.t.0l aboutthe line 1 : 2 the ;y-axis, Find the volumesof the solidsgenerated revolvingthe regions by bounded the linesandcuNesin Exercises 13-18aboutthe)-axis. by -1, t: I 1 3 .x : J 5 y 2 , . v= 0 , y : y:2 1 4 ,x : ! 3 / 2 , a=e,
• Chapter5: App)icatlons )ntegra)s ol 3ab 1 5 .x - D i n T , o = y = r 1 2 , x : 0 = 1, ) :,r'?, below by the i-axis, and on the right by the line, about the line i - I 1 6 ,x = J c o d t y l - \$ , 2=j i0, ,r:o r 36. The region in the second quadrant bounded above by the curve 1 7 .x : 2 l O + 1 ) , x : 0 , ) : 0 , ):3 - 1, 1, : -r3, below by the x-axis, and on the left by the line r : 1 8 .x : J z . y l b , 2 + 1 ) , r c : 0 , 1 : 1 : -2 about the line r Volumes the WasherMethod by Volumes Solids Revolution of of Find the volumes of the solids generated by revolving the shaded 37, Find the volume of the solid generatedby revolving the region bounded by y : ^vE and the lines ) - 2 and r : 0 about regions in Exercises 19 and 20 about the indicated axes. 20. The y-axis 19, The,r-axis the r-axis; a) b) the )-axis; the line ) :21 c) the line i - 4. d) 38. Find the volume of the solid generated by revolving the tdalgular and t - I about region bounded by the lines ) :2x, y :0, the line;r = 1; a) the line;Y :2. b) Find the volume of the solid generatedby revolving the region bounded by the parabola;l :,r2 and the line ]l: I about the line 1, : l; a) Find the volumes of the solids generatedby revolving the regions theliney:2; b) boundedby the lines and curvesin Exercises21 28 about the t-axis. the line ): -1. c) jr, 22. y:);s, 21.y:a, y:1, ;r:0 l: r:1 40. By integration, find the volume of the solid genemtedby re- regionwith vertices 0), (r, 0), (0, r) (0, volving the triangular x-u zJ. J=tJx, l:2, about -2, x -o 24. y = y: Ji, a) the,t-axis; 2 5 .y : x 2 ] 1 , ) : r + 3 b) the )-axis. y:2-r 2 6 .! : 4 - x 2 , R 41. oesigninga wok. You aredesigning wok frying panthatwill a -tr 141x 3n /4 27, ) :5gsa, bowl with handles. bit of expe - y:.15, A be shaped like a spherical mentationat homepe6uadesyou that you can get one that holds ;r : I 28. y : sec,r, l : tant, Y :0, about3 L if you make it 9 cm deepand give the spherea radius 29-34,Iind the volumeof the solidgenerated revolv- by In Exercises of 16 cm. To be sure,you picturethe wok as a solid of revolu- ing eachregionaboutthe )-axis. tion, as shownhere,and calculate volumewith an integral. its To the nearestcubic centimeter,what volume do you really get? 29. The regionenclosed the trianglewith vertices 0), (2, l), (1, by (1L: 1000 m3) c and(1, 1) 30. The regionenclosed the trianglewith vertices l), (1, 0), (0, ) (cm) by and(1, l) above the parabola rc2: 256 31. The regionin the first quadrant by bounded 12, belowby the r-axis, andon the right by the line .I :2 ],: above the curvey: ^,6 and belowby by 32. The regionbounded thelineY:1 33, The regionin the nrst quadrantbounded the left by the circle on x2 + y2 - 3, on the dght by the line .t : r./3, andabove the by line y : .7t 34, Theregionbounded theleft by theline i : 4 andon theright on _' r'+ )' - l) bY theclrcle 9 cmdeep In Exercises and 36, find the volumeof the solid generated by 35 Designinga plumb bob. Havingbeenasked designa brass to about givenaxis. region the revol each ing plumb bob that will weigh in the neighborhood 190 g, you of decideto shape like the solid of rcvolutionshownhere.Find it 35. The region in the frrst quadrantbounded above by the cuNe
• Shells 387 5.4 Cylindrical GRAPHER Graph the solid's volume as a function ofc, lirst f,l d the plunb bob's volume. If you specify a brass that weighs 8'5 for 0: c: I and then on a larger domain What happens g/cml, how much will the plumb bob weigh (to the nearestgram)? to the volume of the solid as . moves away ftom [0, 1]l Does this make sense physically? Give reasons for your r'(cm) answer' irr.r/ trrll. YoLlare designingan auxiliary fuel tank B 44. ln,r.'xri';rf that will 1lt under a helicopter's fuselage to extend its range After some expedmentationat your drawing board, you decide to shapethe tank like the surface generatedby revolving the curve (r2116).-4 = r 5 4. about the it-axis (dimensions in r (cm) ): I feet). a) How many cubic I'eet of fuel will the tank hold (to the : The arch ) - sinn, 0 : r = 'r, is revolved about the line -v ', nearestcubic foot)'i A cubic foot holds 7 481 gal lf the helicopter gets 2 mi to 5 26. 0 5 c = 1, to genemtethe solid in Fig. b) the gallon, how many additional niles will the helicopter be able to fly once the tank is installed (to the nearestmile)? about 45. Tlt e vol u me ctfa.o/ trJ The disk,r2+ l2 : d2 is revolved shapedlike a doughnut 1h 1;ns,1: b (b > d) to generatea solid 'f d:- - and called a loru.t. Find its volume. @irt, I'1,, uQll since it is the area of a semicircle of radius tl ) ra2 f2, A hemisphericalbowl of radius a contains water to a depth 46. a) ,. Find the volume of waier in the bowl' (Relatedrates)Water runs inlo a sunkenconcretehemispher b) ical bowl of radius 5 m at the rate of 0 2 mr/ sec How fast js the water level in the bowl rising when the water is 4 m deeP? '/al!1ni': 41. lt.tin91 ihe .on5i5le/l.J/ i/l e 'al. ritl 'i'iiniiiau al of are all consistent with the The volume lbrmulas in this secfion standardformulas from geometrY. a) As a case jn point, show that if you tevolve the region :.:.',Exercise asksforthevalue f cthat minimizes :r o 43 7 and the r-axis enclosed by the semicircle )'- '/rA the volumeof this solid. solid sphere'the disk formula about the x-axis to generatea for volume (Eq l) will give (4/3)r dr just as it should Find the value ol c that minimizes the volume of the solid' a) b) Use calculus to find the volume of a right circular cone of What is the minimum value? h e i g h t , a n db a ' e r r d i u . , What value of c in [0, 1l maximizesthe volume of the solid? b) .: .. ... . :: .:..: : j : ..::.. .: : Shells Cylindrical When we need to find the volume of a solid of revolution' cylindrical shells some- times work better than washers (Fig. 5 27, on the following page) In part' the reil)on is lhat the lormulil lhe) leudlo does not requiresquaring The ShellFormula Supposewe revolve the tinted region in Fig 5.28 (on the following page) about the of the solid, we can approxlmate )-axis to genemtea solid. To estimatethe volume partition P of the interval fc, Dl over which the basedon a ihe region with rectangles region stands.The typical approximatingrectangleis AJrr units wide by I (c1) units high, where c1 is the midpoint of the rcctangle'sbase.A formula ftom geometrytells
• Chapter Applications Integrals of 5: 388 Rectangle =/(cr) height 5.28 The shellsweptout by the kth 5.27 A, solid of revolution approx- rectangle. imatedby cylindrical shells. us that the volume of the shell swept out by the rectangleis LVy : )v x averageshell radius x shell height x thickness, which in our case is LVL :21T x ck x f(c) x Lxk. We approximatethe volume of the solid by adding the volumes of the shells swept out by the r?rectanglesbased on P. v ,= -,tY r ' J t|ttAx! i: /r )rzJtcl The limit of this sumas llPll + 0 givesthe volumeof the solid: n fb v- l i m f 2 z c k . f l c ^ r l x k : | 2 n xl t x t d x . tp .o?a J The Shell Formula for Revolution About the J-axis between the The volume of the solid generatedby revolving the region betw, and the graph of a continuousfunction ) : /(x) > 0,0 <: d = x = b , a< r-axis the y-axis is about sn:rr .h,,r!/,ta1 t lb v = I z t I r a d r u s l . t l t l ' l a r : I 2 n rf t x t d t . l (1) l J. / he'ght/ The regionbounded the cuwe y: \$, the r-axis, and the by EXAMPLE 1 line x : 4 is revolvedaboutthe,-axis to generate solid.Find the volumeof the a solid. Interval integration of Solution Step t: Sketchthe region and draw a line segmentacrossit parallel to the axis 5.29 The region,shelldimensions,n q a height(shellheight)anddistance (Fig.5.29). from Labelthesegmenfs of revolution intervalof integrationin Example . l
• 5.4 Cylindrical Shells 389 (shellradius). the axisof revolution The width of the segment the shellthickness is One way to remember Eq. (1) is to imagine dr. (We drew the shell in Fig. 5.30, but you neednot do that.) cutting and umolling a cylindrical shell to get a (nearly) flat rectangular solid. ) shell radius = Innercircumference 2rt 4 5.30 The shellsweptout by the line segmentin Fig.5.29. Step 2i Find the limits of integration: r runs from a :0 to D :4. v: l.'z,(;fli) * (,,il!il,) __.L T E q .( l ) d,Y= tbickness Values from stcps : zo1,;1.tri a, Alnosl a rectangularsolid 1and2 fo Y= lenglh x height x thickness : 21tx -f(x . dx f4 f) la 119- ': : 2t I x3t2 = 2n l'= xs/z| : dx J .,/o L) lo E (1) Equation is for verticalaxesof revolution. horizontal For axes,we replacethe .r's with y's. The Shell Formula for Revolution About the r-axis 'ifl) v: [ z, ( raolus (J:l].)o : foznyystay (2) Jr. J. // nelgnr/ (for /(y) > 0 and0 < c < y < d) 4t' The regionbounded the curve y: Ji, the,r-axis,and the EXAMPLE2 by Shell height . line x : 4 is revolvedaboutthe r-axis to senerate solid. Find the volume of the a (4,2) solid. Shell Solution = thickness d) Step t.' Sketchthe region and draw a line segmentacrossit parallel to the axis of (Fig. 5.31).Labelthe segment's length(shellheight)anddistance revolution ftom (shell radius).The width of the segment the shell thickness the axis of revolution is dy. (We drew the shell in Fig. 5.32, shownon the following page,but you neednot 5.31 The region,shelldimensions,and do that.) intervalof integrationin Example 2.
• i 390 Chapter 5: Applicationsof lntegrals Step 2; Identify the limits of integration: runs from c:0 to d:2. lr Step 3.' lntegrateto lind the volume. .hell IJ shell / / v- I 2 n | r a , .t t u s 1 . ' . - , l d v 1 c 7 nergnl// J, l l) : I 2r (g@ _ 12) d)- J,, .'''l' - ) ^ l )t^ nf .l=8r a lo L Shell This agreeswith the disk method of calculation in Section 5.3, Example 1. ,..:.: The shellsweptout by the l r n e s e g m e nitn F i g .5 . 3 1 . How to Usethe ShellMethod Regardlessof the position of the axis of revolution (horizontal or vertical), the stepstbr implementing the shell method are these: l. Draw the region arul sketch a line segment across it parallel to the axis of revolution. Label the segment'sheight or length (shell height), distance from the axis of revolution (shell radius), and width (shell thickness). 2. Find the limits of irtegration. Integrate the product 2z (shell radius) (shell height) with respectto the 3. appropriatevariable (,t or l) to find the volume. ln the next example, the axis of revolution is the vertical line -{ : 2. lnterval of integfation EXAMPLE 3 The region in the first quadrantboundedby the parabolay :.r2, . r+l+2 -v the )-axis, and the line l' : 1 is revolved about the line x :2 to generatea solid. Shell radius Find the volurne of the solid. I : : T h e r e g i o ns h e l ld i m e n s l o na n d , s, Solution intervalof integrationin Example 3. Step 1; Draw a line segmentacrossthe region parallel to the axis of revolution (the line r : 2) (Fig. 5.33). Label rhe segment'sheight (shell height), distancetuom rhe axis of revolution (shell radius), and width (in this case,dx). (We drew the shell in Fig. 5.34, but you need not do that.) Shellheirht Iine,=2 Shell Step 2: The limits of integration:r runs from a :0 to b = 1. radius Step 3: , = ['z, |, heli /.'l.ll ) a,^ ) raorus Jd / nergnt/ l - lo'rrtr-x)(r-x2)ctr .; -2x'+x3)dx =2, InteI.''aL of fo'{z-, rntegratron 13ir i. i.,r The shellsweptout by the line s e g m e nitn F i g . 5 . 3 3 . 6
• 5.4 Cylindricalhells 391 S Table 5.1 summarizesthe washerand shell methodsfor the solid generatedby revolving the region bounded bY ) : x and ) : 12 aboui the coordinateaxes. For this particular region, both methods work well for both axes of revolution. But this is not always the case. When a region is revolved about the y-axis, for example, and washersare used, we must integratewith respectto l. However, it may not be possible to expressthe integrand in terms of ). In such a case, the shell method allows us to integrate with respectto x instead. The washer and shell methods for calculating volumes of solids of revolution always agree.In Section 6.1 (Exercise52), we will be able to prove the equivalence for a broad class of solids. Table 5.1 Washers shells vs. GENERATINGSEGMENT-LTO AXIS: WASHERS. GENERATING SEGMENTll TO AXIS: SHELLS. The region bounded v by ] l or J--shlihishr=fi-n t- .t. ,/ > R;'r ':l': ,---, N:f=.tt'l't.'l'l^ 2z _ ls Fr- I zrlx)[x xt),t^: !
• 392 Chapter Applications lntegrals of 5: 5.4 Exercises x: r, x :4 rt. y:311251, !:0, the volumes of the In Exercises 1-6, use the shell method to find by generated Use the shell methodto find the volumesof the solids revolving the shaded region about the indicated .otiO. gnrutO ty lines in Exercises bounded the curvesald by revolvingthe regions axis. 15-22aboutthe t-axis 1 5 .j { = . / t , a--Y' Y:2 ] ;l2, a=-Y, Y:2 1 6 .x : x:Y 1 8 .x : 2 Y - Y 2 , x:0 lJ.7:)y-y2, Y:2 2 0 'Y : a , Y:2x' y:1 1 9 .y : 1 . x , 2 1 .r - J r , Y = O , Y = x - 2 22. =aE, y:O, !:2-x y ofthe volumes In Exercises and24' usethe shellmethod findthe to 23 regionsaboutthe indicated by revolving the shaded solids generated at(es, The r-axis 23. a) The line Y: t b) The tine )' : 8/5 c) The line Y: -2l5 d) 6. The )-axls 9,I 'V + 9 r' The i-axis 24. a) T h el i n e) : 2 b) The line Y = 5 c) The line 1: -5l8 d) by generated Use the shell methodto find the volumesof the solids and lines in Exercises revolving the regionsboundedby the curves 7-14 aboutthe )-axis. 7. y: x, Y: -x12' x:2 Y= tl2, x: I 8. y:2x, generated re- by In Exercises25-32, find the volumesof the solids .r:0,for'r Z0 t', 9. ,: Y:2-x, the given a'{es If you thin} it would be volving the regions about to do better;o usedisksor washers any giveninstance' free feel r:0 in 1 0 ,) : 2 - x 2 , Y=f, so. y:Q, :r:4 ll. y:1Q, (a) the 25. The trianglewith vertices 1), (1, 2), and(2, 2) about (1, 1 2 .y : 2 x - 1 , Ji, t=O l: (c) the line t - l0/3; (d) the line;v: I x-axis;Oi the )-aris; .x:112, x=2 7 3 .y : l l x , ):0, -
• 5.5 Lengthsof PlaneCurves 393 26. Theregionin thelirst quadrant bounded thecurve.r- I - y3 by 0<r=T/4 36. Lets(x) - and the ]-a{is abour(a) the r-axis; (b) the line l: 1 {;'*l'z'. 27. The regionin the first quadrant bounded .r : y - y3,.r : I, by a) Showthatr8(.x): (tanx)2,0 <x =n14. and)' : I about(a) rhex-axis;(b) they-axis;(c) theline jr : l; b) Find the volumeof the solid generated revolvingthe by (d) the line I - I shaded regionaboutthe)-axis. 28, The triangular regionbounded the lines 2J : x l 4,! : x, by andj = 0 abour ther-axis; (b) the y-axis;(c) the line x : {; (a) (d) the line y : 8 29. The regionin the first quadrant bounded y :;r by and) = 4.r about(a) the i-axis; (b) the line y : 8 30. The regionbounded y: uE and) = x2/8 abour(a) the _r by axis;(b) the ]-axis The region shownhere is to be revolvedabout the,-axis to bounded )-:2x - x2 and 1:r 31. The region by n5eu1 1uy,5 generate solid.Whichofrhe methods a (disk,washer, (b) the line r : 1 shell)could J,-axis; you use to nnd the volumeof the solid?How many integrals 32. The regionbounded y : ^,tr, y :2, t: 0 about(a) rhe .r- by wouldbe required eachcase? in Explatn. axis;(b) the y-axis;(c) the tine r - 4; (d) the line ) :2 ] 33. The regionin the first quadrant that is bounded aboveby the curve), : 1/rrl4, on the left by rheline r : 1/16,andbelowby the line ): l, is revolved aboutthe _r-axis generate solid. to a Find the volumeof the solidby (a) rhe washer (b) method; rhe shellmethod. 34. The regionin the first quadrant that is boundedaboveby the The region shown here is to be revolved about the y_axis to c\$vey : I/rE, on rhe left by the lirle y : y14,andbelowby generatea solid. Which ofthe merhods(disk, washer,shell) could the line ) - I is revolved aboutthe ),axis to generate solid. a you use to find the volume of the solid? How many integrals Find the volumeof the solid by (a) the washer methodl(b) the would be rcquired in each case?Give reasonsfor vour answers. shellmethod. It7* o' 'r^1E 3 5 . L e r/ r . r r - | G i n x:u. lr. a) Showthatf(.r) : sinr,0: r : ri. b) Find the volumeof the solid generated revolvingthe by shaded regionaboutthel-axis. Supposethat the function /(r) is nonnegativeand continuousfor x - 0. Supposealso that, for every positive number ,, revolving the region enclosedby the graph ofJ the coordinate axes, and the line r - , about the )-axis generates solid of volume 2n03. a Find /(.x). Lengths Plane of Curves We approximatethe length of a curved path in the plane the way we use a ruler to estimate the length of a curved road on a map, by measuringfrom point to point with straight-line segmentsand adding the results.There is a limit to the accumcy of such an estimate,howevet imposed in part by how accuratelywe measureand rn part by how many line segmentswe use.
• chapter5: Applications Integrals of 394 uslng With calculus we can usually do a better job becausewe can imagine as we please,each set of segmentsmaking a poly- straighlline segmentsas short way' gona-lpath that fits the curve more tightly than before When we proceedthis of the polygonal paths approach a ljmit we can iuitt u .-oottl cuNe, the lengths calculate with an integral. Formula The Basic x = a to x: b' Supposewe want to find the length of the curve y: /(;) from. points on the We pafition [4, b] in the usual way and connect the conesponding the cuNe (Fig' curve with line segments form a polygonal path that approximates to lf we can find a formula for the length of the path' we will have a formula 5.35). for approximating the length of the curve. 5.35 A typicalsegment of a polygonalpath approximating curveAB' the PQ (Ax1)2* (Ayr The length of a typical line segmentP0 (seethe figure) is The lensth of the curve is therefore approximatedby the sum (1) -.44'-)' + (^r,-t. /- finer' We expect the approximation to improve as the partition of [4, b] -becomes in (l) approacha calculable limit as the and we would liie to show that the sums in (1) in a norm of the partition goes to zero. To show this, we rewrite the sum Theorem from Chapter 4' Our form to which we can apply the Integral Existence Tangent starting point is the Mean Value Theorem for derivatives' parallel to chord Gt, fGr)) Definition A function with a continuous first derivative is said to be smooth and its graph is called a smooth curve' (ct' /(cr)) on the If/is smooth, by the Mean Value Theorem there is a point tangentis parallel to the segmentP0 (Fig 5 36) curve betweenP and O where the rltctt* At this point Aju+l [-- Ayr. Ay1 : /'(c1)A,r1. or 5.36 Enlargement of the arc PQ in Fig. f'@n: ^x,,
• 5.5 Lengths Plain of Curves 395 With this substitution Alr, the sumsin (l) takethe form for -:------------- ' - l((rrr'rrrl - ,^ t r - ,,/ ' ,.-J , /. . . ^,1-r )' - | | (J (ci ))'Axa. '-.., L L/ Because a 1y1x112 continuous [4, D], the limit of the sumson the right is on ./1 as the norm of the partitiongoesto zero it I: Jl+ UiYrh. We definethe lengthof the curveto be the valueof this integral. Definition Iflis smooth on [a, D], the length of the culve ] : /(.r) from a to b is ffi,,:1' L: Jr + c,ly ay. (2) J EXAMPLE I Find the length of the curve 4a x ) ' ' - 1 . 'v : 0<x11. -- .t Solution We useEq. (2) with a :0, b: 1, and 4J2 .,,,, '_3 dy 4A 3,, : z ^ z ^ ,' ) r/x - . E= I Z.* ., ,2 ( o r : ( z , f z x t , z: s, ^ . / dx/ The length of the curve from x : 0 to .r : I is ,r I Eq. (l) wilh u a ar: I_ I 4+ lr 11 . lntcgrirlc- itnd r-,,,,'] ] ],t* t o lrsf. D Deaf with Discontinuities dyldx ing in At a point on a curve where dy / d x fails to exist, dx f dy may exist and we may be able to find the curve's length by expressingr as a function of y and applying the following analogueof Eq. (2): Formula for the Length of a SmoothCurve x = g(J), c 1y I d Jr + rcoy ay. (3)
• Chapter Applications Integrals of 5: 395 (xl2)2/3 fromx:0tox:2' Find the length of the curve y: EXAMPLE 2 The derivative Solution |/2' av -2-(^-'/l a-= iz) ;/ z/:: (2)' is not defined .r :0, so we cannotfind the cuwe'slengthwith Eq at of )' We thereforerewrite the equationto express in terms x / 2/3 }:l;, 5 . 3 7 T h eg r a p ho l y = ( x l 2 l ' 1 f r o m - x = 0 tox=2 is;lsothe qaph oI x =2Y11' -2 ./ lo lhe po\$er 3/1. fromy:0toY:1. -- ),,312 Sole lo r. Fromthisweseethatthecuwewhoselengthwewantisalsothegraphoft:21312 f r o my : g t o y : 1 ( F i g . 5 . 3 7 ) . The derivative , /'l dx - 3Y'/z ;:'i)Yz is continuous [0, l]. Wemaytherefore Eq.(3) to findthecuve'slength: use on ,FW at: ': l' [q. (])illr l'1..wa:' Letll:l+9.r. - - ! . ? , r +9- u ) 3 / ' l ' 9l ln substitL(c rck. b - : -_:(10J10 r) x 2.27. 27' The Short DifferentialFormula The equations L'l'.(n' = L: J. (4) dy dx (*)' by are often written with differentials instead of derivatives. This is done formally the dx ar.,ddy thinking of the derivatives as quotients of differentials and bringing insicle the radicals to cancel the denominators. In the first integral we have In the second integral we have '.(E)' ,ta,, +a Thus the integrals in (4) reduce to the same differential formula: fb - L: Jdx?+drz. J
• 5.5 Lengthsof PlainCurves 397 Of course, dx and d;t must be expressedin terms of a common variable, and appropriatelimits of integration must be found before the integration in Eq. (5) is performed. We can shortenEq. (5) still further Think of dr and dy as two sidesof a small triangle whose hypotenuse is ds : Jdx2 -t cly2 (Frg.5.38). The differential ds is then regarded as a differential of arc length that can be integrated between appropdate limits to give the length of the cuNe. With /dx, + dy2 set equal to ds, the integral in Eq. (5) simply becomesthe integral of ds. 5.38 Diagram remembering for the Definition eguation ds = Jdx2 + dyz. TheArc Length Differential the Differential and Formula for Arc Length ds -- /n4ayz L:lds arc lengdr diferential formula differential for arc lenglh :i<Curveswith Infinite Length As you may recall from Section 2.6, Helga von Koch's snowflake curue K is the limit curve of an infinite sequence C2,...,C^,... of triangularpolygonal C1, cuwes. Figure 5.39 shows the first four curves in the sequence.Each time we r.- j introduce a new vertex in the construction process,it remains as a veltex in all Curve I Curve 2 Cur,re 3 Curve 4 subsequent curves and becomesa point on the limit curve K. This means that each of the C's is itself a polygonal approximation of K-the endpoints of its sides all 5.39 The first four polygonal belonging to K. The length of K should therefore be the limit of the lengths of the approximations the construction in of Helgavon Koch's snowflake. curves C,. At least, that is what it should be if we apply the definition of length ue developed smoothcurves. for What, then, is the limit of the lengthsof the curvesC,,?If the original equilateral triangle C1 has sides of length 1, the total length of C1 is 3. To make C2 from C1, we replace each side of C1 by four segments,each of which is one-third as long as the original side. The total length of C2 is therefore 3(413).To get the length of C3, we multiply by 4/3 again.We do so again to get the length of Ca. By the time we get out to C,, we have a curve of length 3(4/3)' I. CurveNumber 12 3 n ,(i)' 1 / 1 ' ' (i) Length 3 3/ The length of C1e is nearly 40 and the length of C1s0 is greater than 7,000,000,000,000. The lengths grow too rapidly to have a finite limit. Therefore the snowflake curve has no lenglh.or. if you prefer.infinite length. What went wrong? Nothing. The formulas we derived for length are for the graphs of smooth functions, curves that are smooth enough to have a continuously turning tangent at every point. Helga von Koch's snowflake curve is too rough for that, and our derivative-based formulas do not apply. Benoit Mandelbrot's theory of fractals has proved to be a rich sourceof curves with infinite length, curves that when magnified prove to be as rough and varied as they looked before magnification. Like coastlineson an ocean, such curves cannot be smoothedout by rnagnilication (Fig. 5.40, on the following page).
• Chapter Applications Integrals of 5: 398 magnifications a of 5-40 Repeated fractal coastline.Like HelgaVon Koch's snowflake curve,coastslike these are too length. rough to havea measurable 5.5 Exercises from t=0to,r=4 1 0 .y : a t l z FindingIntegrals Lengths Curves of for ll. x=(y3/3)+ l/(4),) from )=lto):3 1-8: In Exercises (Hint: 14 (dttld.l)2 is a perfectsquarc.) a, Set up an inte\$al for the lengthof tle curve 1 2 .x : O 3 / 2 p ) - y r l 2 f r o m ) : 1 t o ) : 9 ar/ rta St t Crapl thecurue seewhalit lookslike. to (Hint: | + (di(ldy)2 is a perfectsquare.) ar/ tta S c; Use your grapher'sor computer'sintegml evaluatorto find the 1 3 . r : ( y 4 4 ) + 1 / ( 8 y ' ? )f t o m y : l t o v : 2 curve'slengthnumericallY. / (Hi t: | + (da/dy)z is a perfectsquare.) -l=x=2 l. y:x2, l ! . a : ( y 36 ) + l / ( 2 y ) f r o m y : 2 t o Y = 3 / -1r/3=a=0 2, J:tun, (Hint: 1 + (dx ldy)z is a perfect square ) 3 . ; 1= s i n y , O : . t < 1 r -t 1( 3 / 8 ) 1 2 +3 , 1 3 x = 8 /5 1 5 .y : ( 3 1 4 ) x a -ll23y =112 a. r:r4-yz, 0=x =2 ( x 3 1 3 )x 2 + x + L / ( 4 x+ 4 ) , 1 6 .y : + ftom (-1,-l) to(7,3) 5. y2+2y:bc+1 tt_- -n/4a)) 1ft/4 / s e c ' r- 1 t u . 17. a : | 6, ) =sinx-.rcos.r, 0=t 5ti tt_ I J3t^ tdr. -2=x s-l tan dt, t 0<xSn/6 7. v= I 18. y : J2 ?| - Find a curve through the point (1, l) whoselength integral 19. a) nll 1y=n14 8. x- I Jsedr-ldr, (Eq. 2) is raF FindingLengths Curves of t : | ,Yl r - 4 x a ' )' Jr 9-18. If you havea grapher' Find the lengthsof the curvesin Exercises you may want to graph thesecurvesto seewhat they look like' How many such cuNes are there? Give reasonsfor your b) answer. g . y : ( 1 / 3 ) ( x+ 2 ) 3 / 2 f r c m x : 0 t o * : 3 2
• Exercises 5.5 399 Find a curvethrough point (0, 1) whoselengthintegral a) the Original sheet (Eq. 3) is t! I I L: I tlt+ ^dy. Jl v , How many such curves are there? Give reasonsfor your b) answer. Find the length of the curve t' - y: I t/cos2rd, 24, Your engineeringflrm is bidding for the contact to construct JN the tunnel shownhere.The tunnel is 300 ft long and 50 ft wide fromi=0tox=n/4. at the base. The cross section is shapedlike one arch of the curvey : 25cos(zr/50). Uponcompletion, tunnel's the inside The length of an astroid. The graphof the equation .r2l3+ surface(excludingthe roadway)will be treatedwith a waterproof )2/3 : 1 is one of a family of curvescalledastroids(not as- that costs\$1.75per squaie foot to apply.How muchwill sealer (see the accom- teroids) becauseof their starlike appearance it cost to apply the sealer?(Hint: Vse numericalintegration to panying figure). Find the length of this particular astroid by find the lengthof the cosine curve.) portion, ): finding the length of hall the f,rst-quadrant e , x2/3/2,f2/4 < r : l, andmultiplyingby 8. ] : 25 cos(7tl50) ) .r (fu Theoryand Examples 25, Is there a smoothcurve y : /(r) whoselength over the interyal eD,a'!Givereasons your alswer. 0 = r : a is always for 26. Using tangent fins to detive the length formula for curves. Assume/is smooth [a, r] andpartitionthe interval[a,,] in on theusualway.In eachsubinterval [rk-r, rr] consmrct targent the l'? at the point (xr-r, f(xr-r)), shownin the flgure. a) Showthat the length of the ,tth tangentfin over the interval Integration Numerical [.rr-r..rr]equals v4d,'t+Tfillrl;;;t. You may have wondercdwhy so rnany of the curveswe have been b) Showthat working yl{!311unusual fomulas. The reasonis that the squarc rL' - -i- .oot Jl + (dy ld.x)2 that appears the integals for length and sur- in t l e n g r o f k t ht a n g e n tn r : lim ) n /l + tI t^)t2dx. h / face area almost never leads to a function whose antiderivativewe J can find, In fact, the squareroot itself is a well-known souce of which is the length I of the curve y : f(x) fuoma to b. integrals.Most integrals for length and surfacearea Donelementary haveto be evaluated numerically,as in Exercises23 and 24. 23. Your metal fabdcationcompanyis bidding for a contractto make sheetsof corrugatedircn roofing like the one shown here. The crosssectiotrs the cofiugatedsheets to conformto the curve of are ad 'y : 0 i n 1 . x . 2s 0<r<20in. If the roofing is to be stampedfrom flat sheetsby a process that does not stretchthe material,how wide should the original materialbe?To find out, usenumericalintegrationto approximate the length of the sine cufle to 2 decimal places.
• 400 chapter5: Applications Integrals of S CASExplorations Projects integal. How doesthe actuallength comparewith the approxi- and Explain your answer mationsas n increases? 27-32, usea CAS to perform the following stepsfor the In Exercises 27. f (x) = {l-7, given curve over the closedinterval I: ' s I 28, f ( x ) : r t / 3 + t 2 / 3 , 0 . t 5 2 a) Plot the cuNe togetherwith the polygonal path approximations for n :2, pointsovertheinteNal.(See 4, 8 partition Fig.5.35.) 29, / ( r ) : s i n ( z i ' ? ) , O = t = J 1 approximationto the length of the cuNe b) Find the corresponding 30. f ( r ) : t 2 c o s x , o < x = 1 t by summingthe lengthsof the line segments. your Y-l l c) Evaluate lengthof thecurveusingan integral. the Compare 3 1 . f(xt: -::------. -; S,r S t approximations /, : 2,4, 8 to the actuallengthgivenby the 4.r'tr z for -l:ir51 1., f(x):13-x2, Areasof Surfaces Revolution of When you jump rope, the rope sweeps out a surface in the space around you, a surface called a surface of revolution. As you can imagine, the area of this surface dependson the rope's length and on how far away each segmentof the rope swings. This section explores the relation between the area of a surface of revolution and the length and reach of the curve that generates i1. The areas of more complicated surfaces will be treated in Chapter 14. The Basic Formula Supposewe want to find the areaof the sufac€ swept out by revolving the graph of a nonnegativefunction y : f (x), a : .x : ,, about the r-axis. We partition [a, b] in the usual way and use the points in the partition to partition the graph into short arcs. Figure 5.41 shows a typical arc PQ and the band it sweeps out as part of the graph of I As the arc PQ revolves about the r-axis, the line segmentjoining P and Q generated revolving 5.41 The surface by sweepsout part of a cone whose axis lies along the;r-axis (magnified view in Fig. the graph of a nonnegative function 5.42). A piece of a cone like this is called a frustum of the cone, J?rslrrm being y : f ( x l , a S X 3 b , a b o u tt h e x - a x i sT h e . Latin for piece. The surface area of the frustum approximates the surface area of surface a union of bandslike the one is swept out by the arc PQ. the band swept out by the arc P0. The surfaceareaof the frustum of a cone (seeFig. 5.43) is 2z times the average of the base radii times the slant height: 'a+ Frustumsurfacearca : 2r L : r (rt + r)L. For the frustum swept out by the segment (Fig. 5.aa), this works out to be P0 F r u s t u mu r f a c a r e a- r t f ( x 1 , t + s e /t,tl./ta*,), +tay,l. The areaof the original surface, being the sum of the areasof the bandsswept out by arcslike arc P_Q, approximated the frustum areasum is by D, ff 6r) + /(xr)).u{A+f+ (AvrP. (1) k=1 We expect the approximation to improve as the partition of [a, l] becomes fine1 and we would like to show that the sums in (l) approacha calculable limit as the joining P and Q 542 The line segment sweeps out a frustumof a cone. norm of the Dartition soes to zero.
• 5.6 Areasof Surfaces Revolution of 401 The importantdimensions the frustumin Fig. of 5.42. To show this, we try to rewrite the sum in (l) as the Riemann sum of some function over the interval from ri to b. As in the calculation of arc length, we begin Segmentlengthl z =.,(.t4-l{i by appealingto the Mean Value Theorem for derivatives. -^[/ P . r- ,/ Ifl is smooth, then by the Mean Value Theorem, there is a point (c1, 1 /(c1)) on r'=r1r) :- - the curye between P and Q where the tangent is parallel to the segmentpO (Fig. A I i-r >1 5.45).At this point, f'(cr): 4l(. Av1 : /,(c1)A.r1. rir f._.u.i_l With this substitution tbr A).r, the sums in (l) take the form Dimensions associated v? t)'+ f1.J^'^). with the drc and segmentPQ. -L,1r.r, 1 r - / r . r r , l u / t. , 7 ' c r r : A r , . ( 2 ) l-l At this point thereis both good newsand bad news. The bad news is that the sums in (2) are not the Riemann sums of any lunction becausethe points r1 1,,ti, and c1 are not the same and there is no way to make them the same.The good news is that this does not matter. A theorem called Bliss,s theorem, fiom advancedcalculus, assuresus that as the norm of the partition of [zr.b] goes to zero, the sums in Eq. (2) convergeto Gi.lG;)) Just the way we want them to. We therefore deline this integral to be the area of 1 Tangent t the sudace swepr our by the graph of/ fiom .? ro D. parallel Definition TheSurface AreaFormula the Revolution for Aboutthex-axis 'r-r 'r If the function ./(-r) > 0 is smooth on fa, bl, the area of the surface gen_ l'-J-rr-l erated by revolving the curye l :.f(x) about the x axis is ,F6 lf f is smooth,the MeanValue rb s : f u2 , , Theorem guarantees existence a d:(: 2 7ft ( x ) J t + ( f , ( x ) ) 2 d x . the (3) of J. point on arc PQ where the tangent is parallel segmentPQ. to
• Chapter Applications Integrals of 5: 402 The square root in Eq. (3) is the same one that appearsin the formula for the length r =2[; of the generating curve. Find the area of the surface generated by revolving the curve EXAMPLE I x =2, aboutthe x-axis (Fig. 5.a6). y:zJi,1 = We evaluate the formula Solution o. r, ,: 1,,fr(fj ,q with dv1 y :2tr, dx FW:'.'(+)' E' -- r ; t-----:--7 /------i---i t vx+r l, lx+L 546 Example calculates areaof 1 the Vx r Y /x this surface, With thesesubstitutions, a*:+n ['uiardx s= ['zn.rJig! Jt Jr Jx 't 12 Rr : +n.'rtx+ l)i/'?Jr:;(3.,/3 2J2. D About the y-axis Revolution For revolution aboutthe y-axis, we interchange and / in Eq. (3). -r Surface Area Formula for Revolution About the '-axis 5,47 Revolving segment about line AB If -r : g(y) > 0 is smoothon [c, d], the areaof the surfacegenerated the y-axisgeneratesa cone whose lateral by surfacearea we can now calculatein two revolving the curve r : g(y) aboutthe y-axis is different ways (Example2). fr19',:1' (4) z g1y1,/1 ({ (yD,ay. + T:heline segment :1- y,0: l : 1, is revolvedaboutthe EXAMPLE2 x :t-axis to generate conein Fig. 5.47. Find its lateral surfacearca. the Solution Herewe havea calculationwe can checkwith a formula from geometry: basecircumference ^ /^ ------- X Stant nelgnl : Laterat surace area : z{z. -
• 5.6 Areasof Surfaces Revolution 403 of To seehow Eq. (4) givesthe sameresult,we take dx x.: | _ J, _t, c:0, d:1,, dy: '.(#)' r+ -'z: rt and calculate s: 1 zo' T -,ztl :zrtlt-i), : r J . ( 1 - ] ' ) : J1. D The results agree, as they should. The Short DifferentialForm The equations ,Fe,J', ': 1.'rl'.(H)' zo' s: 1 are often written in terms of the arc length differential ds : Jdx,lyz as lb ld ht ds and zzx ds. S: s: J. J In the first of these,y is the distance from the x-axis to an element of arc length dr. In the second, .r is the distance from the y-axis to an element of arc length ds. Both integmls have the form tf width): s:/ 2z(radiug(band Jzrods, where p is the radius from the axis of revolution to an element of arc length ds (Fig. 5.48). 5.48 The area of the surfaceswept out by revolving AB about the axisshown arc Axis of . revolution nele ts l2b ^ zltpos, Ine exacr expresston deoends on the formulas for p and ds.
• 404 Chapter Applications Integrals 5: of If you wish to rememberonly one fomula Short Differential Form for surfacearea,you might make it the short s: zoo a, differential form. I In any particularproblem,you would then express radiusfunction p andthe arc the length differential ds in termsof a ctimmonvariableand supplylimits of integration for that variable. EXAMPLE 3 Find the area of the surface generated revolving the curve by y : .r3,0 < x .< | 12, aboutthe .{-axis(Fig. 5.49). so/ution We staft with the short differential form: s: I zooa, : zta' For revolutior aboul the r-nxi. lhc radius I f_ : ,i.l : .,?- +./r' 2ryJdxz+ dy?. I 549 The surfacegenerated by revolving t h e c u r v e y = x 3 , 6 3 x 3 1 1 2 ,a b o u t t h e We then decide whether to expressdy in terms of dx or dx in terms of dy. The x-axis could be the designfor a original form of the equation,y : .r3, makesit easierto express in termsof dr, dy glass (Example champagne 3). so we continuethe calculationwith ! : x3, dy :3x2 d;c, Jdr, + dy, : I dx, + Ax, dil, and :uEaXnar. With thesesubstitutions, becomes variableof integrationand x the rx=l/2 S:, h t y / d x z+ d y 2 Jr=o rt/2 :I znx3^,/t+gxadx Jo s , . L . r j r | l e , ,| _ e r . . , / r rr 0 _ / | /2 . ...1,/, : zttl) . ' , r . ' n t c t ' . r c .J , , b , er , , n i . ltt +gtot'r'| lli J6l J/ bi.k. ln -fr o 3/2 I -1.] ;l'.;6) r Lc/ -',1: l* -'j n f lz5frz -l r( / r25 u 61n 17?a' As with arc length calculations, even the simplest curves can provide e workout. tt
• 405 Exercises 5.6 Exercises 5.6 Area FindingIntegrals Surface for -Y xis a 1 5 .y : J z r - x t , 0.5:r:1.5; I 5 r.:< 5; .r-axis 16. r:.,,6T l, In Exercises 1-8: a) Set up an integral for the area of the surface generatedby re- 17.x-.,-1fi, 0:]11; _,-axis volving the given cuwe about the indicated axis. 113: '-eris I 1 3 .1 = r l , 3 r y ' : I Graph the curve to see what it looks like. If you can, graph the 1 9 .r - 2 / - ' + - v . 15.4: f axi. 0a| surface,too. Use your grapher's or computer's integral evaluator to find the area numerically. sur-face's 0<-r <r7/4: r-axis L v:tanr. ) ]-12, 'r-axis o5r:2; .r_r,:1, l:l':2; ) axis 4. _ t : s i n ) , ) axis 0 = )r: z; r'ai jlI-Ifromt4. lrtorl.4r: r 5. 20. r - tzt=1, <21 -r. 5/8 : t': l; ),axis o. r + 2 - f 1<r r-axis .t -I y-2,3: 7. r t c nr d t , 0 1-axi. It - xaxrs 8. j : l ^ / t 2 ldt, l5r:V5; Areas FindingSurface 9. Find the lateral (side) surface atea of the cone Seneratedby revolving the line segment) : -r/2. 0 = t : '1, about the i-axis Check your answer with the Seometry formula I r-axis (Hlrr.' Express 1/(8r''?),11y12; 21, x:01lq+ : x base circumferencex slant height Lateral surfacearea : of d), and evaluatethe integral S = ,ts : l7tz + dt in terms 10. Find the lateral surface area of the cone generatedby revolving / 2zl ds with appropriatelimits.) the line segment y': xl2.O: t :4 about the )]-axis.Check y axis (lli,rt. Express 22. -,-= Ol3)(x2 +2)tit, 0 <.r i 'D; your answer with the geometry formula ay'? in terms of dx, and evaluatethe integral S : : Jth ,lt t ds wirh appropriatelimits.) J 2nt Lateral surface aJea- x base circumferencex slant height. i 23, Testing the new definition. Show that the surface area of a 11. Find the surfacearea of the cone irustum generatedby revolving sphereof radius a is still 4zar by using Eq. (3) to find rhe area rhe line se8ment) : (x 12)+ (l l2), 1 : r 5 3, about the x axis. of the surface generatedby revolving the curve r : 161lll, Check your result with the geometry formula a = x = a, about the ,r-axis. Frustum surface area - n(rr + /r) x slant height. 24. Testing the new definition. The lateral (side) surface area of a cone of height l? and base radius r should be rr'G h2, 12, Fjnd the sudace areaof the cone frustum generated revolving by the semipeimeter of the base times the slanl height. Show that rhe line segment : (xl2) + (112),I : r :3, aboutthel'-axis. ) this is still the case by linding the area of the surface generated Check your result with the geomefty formula by revolvingthe line segment = (r lh)x,0 = x:,t, aboutthe r Frustum surface area= r(.rt + r:) x slant height. r-axis. Find the areas of the surfacesgeneratedby revolving the curves in W.ite an integral lbr the area of the surface generatedby 25. a) Exercises13 22 about the indicated axes. If you have a grapher,you revolving the cuNe ) - cosr. -1t 12 : x = 1/2, about the may want to graph these curves to see what they look like. j-axis. In Section 7..1 we will see how to evaluate such 13. r -, x'19, integrals. 0 : .r : 2; r-axis Find the surface area numerically. l,q _ r ' l5 4: rxi. E r; calcuraroR l.l. _r G.
• chapter5: Applications Integrals 406 of 26. thc tu[faG af an astroid Find the area of the surface gen- 29. The shadedband shown here is cut ftom a sphereof radius R by erated by revolving about the r axis the poition of the astoid parallel planes ft units apafl. Show that the sudace area of the r2l3 + )2/:r : I shown here. (Hirlt: Revolve the filst-quadrant band is 2z Rft. portion ) : (l ir2lr)3/2,0 < ). = I, about the r-axis and double your result.) a , 1 -r4 r h 'I @ 27, in;tmelincy vroks. Your company decided to put out a deluxe version of the successfulwok you designed in Section 5.3, Ex- ercise 41. The plan is to coat it inside with white enamel and 30, Here is a schematicdrawing of the 90 ft dome used by the U.S. outside wjth blue enamel. Each enamel will be sprayed on 0.5 National Weather Service to house radar in Bozeman, Mont. mm thick belore baking. (Seediagram here.) Your manufacturing How much outside surface is there to paint (not counting a) department wants to know how much enamel to have on hand the bottom)? for a production run of 5000 woks. What do you tell them? (Ne- ffi u) calcuraroR Exprcss answer rhenearesr rhe ro square glect waste and unused mateial and give your answer jn liters. root, Remember that I cmj : I mL, so I L - 1000 cml.) ) (cm) ?l Ti + s r t l^ I I Lenler Radius .. i,/ ----l 45ft i 225 ft | €'rt. t: *2 +1'2 i62 = 256 = 31. Surfacesgenerated by curves that (rcss the axis of revolu Slt.tng bread. Did you know that if you cut a spherical loaf of bread into slices of equal width, each slice will have the tiot. The surface area fbrmula in Eq. (3) was developedunder same amount of crust? To see why, supposethe semicircle y = the assumption that the function / whose graph generatedthe shown here is revolved about the ,r axis to generate .n42 sudace was nonnegativeover the interval [a, b]. For cuNes that 2 a sphere.Let AB be an arc of the semicircle that lies above an cross the axis of revolution. we replace Eq. (3) with the absolute interval ot' length I on the r-axjs. Show that the area swept out value formula by AB does not depend on the location of the interval. (It does tl 5 = | 2ttptls: (5) | 2n f(x) ds. depend on the length of the inteNal.) .t J Use Eq. (5) to find the surfaceareaof the double cone generated by revolving the line segment 'y : x, I : * : 2, about the n- axis. (Exercise 31, co tinue.l.) Find the area of the surface generated by revolving the curye )' - ,'19. J5, about the -r Ji = :! axis. What do you think will happen if you drop the absolute value bars f.om Eq. (5) and attempt to find the surfacearea with rhe formula S : .f 2ft f (x) ds insread?Try it.
• 5.7 Moments and Centers Mass 407 of Numerical Integration Find, to 2 decimalplaces,the areasof the surfaces generated by revolving curyes Exercises the in 33-36 aboutthe;r-axis. 3 3 .y : s i 1 y , 0 = x = n 3 4 .y = a 2 1 4 , 9 . r . 2 35. y=yasi1 2x, -2jt/3 Sr.32n/3 (the curvein Section 3.4,Exercise5) x /-:- ^ Jo. y: . x z . l ) x : . 6 ( r h es u r f a co f r h ep l u m b o bi n e 1 b l2VJ6- tkr Section frk 5.3,Exercise rk 44) ' -^t 37. An alternative derivation of the surfa.e area formula. As_ sume/is smoothon [a, r] andparrition[d, D] in rhe usualway. In the *th subinterval t,tr] construct tangent the [.rr line to the c) Showthat the lateral surfaceareaof the frustumof the cone curye at the midpoint mk : (ak 1+ ri/2, as in the figure here. swept out by the tangentline segmentas it revolvesabout the.r-a.r,is 2nItmttrll - t f'tmrt txt. is S h o w | n a rr t : I t m t , l - 1 t . r t - a, a n d 1 2 : f t n p )- d) Show that the area of the surfacegenerated revolving by Lx* J ,t.n k ) ) : /(.,r) abourthe r-axis over [a, ]l is ). b) Show that the lengrh It oj !!9_jg!gqll4gj!g!q!-nt = fu2,1,,t,1t ,(r,G)id,. in the ,-- I(lfill.':*::-*) / fir i=i or /(rhrrusrum Ith subinterval Lk : lra{xi, + U,@k;6;..1:. is L Momentsand Centers Mass of Many structures and mechanicalsystemsbehaveas if their masses were concentrated at a single point, called the center of mass (Fig. 5.50, on the following page). It is lmportant to know how to locate this point, and doing so is basically a mathematical enterprise.For the moment we deal with one- and two_dimensional obiects. Ihree- dimensional objects are best done wirh rhe multiple integrals of Chapier 13. Masses Along a Line We develop our mathematicalmodel in stages.The first stageis to imagrnemasses m1, m2, ad m 3 on a rigid x-axis supportedby a fulcrum at the origin. Fulcrum at ongrn The resulting system might balance,or it might not. It dependson how large the Mass vs. weight masses are and how they are arranged. Weight is the force that resultsfrom graviry Each mass mk exerts a downward force mpg equal to the magnitude of the pullingon a mass. an objectof massm ls If rnass times the acceleration of gravity. Each of these forces has a tendency ro tum placed a location in wherethe acceleration of the axis about the origin, the way you turn a seesaw.This turning effect, called gravityis & the object'sweighrthereis torque, is measured by multiplying the force mtg by the signed distance xk 1 F:mg from the point of application to the origin. Masses to the left of the orisin exert (asin Newton'ssecond law). negative(counterclockwise)torque. Massesto the dght of the origin exen positive (clockwise) torque.
• Chapter 5: Applicationsof Integrals 408 until we notice 5-50 (a) The motion of this wrenchglidingon ice seemshaphazard that the wrenchis simplyturning about its centerof mass the centerglidesin a as revolve straightline.(b) The planets,asteroids, comets our solarsystem of and (lt centerof mass. liesinsidethe sun.) about their collective The sum of the torquesmeasures tendencyof a systemto lotate about the the This sum is called the system torque. origin. (1) torque: m8xt + m28x2 m384 System + will balance and only if its torqueis zero. if The system If we factorout the I in Eq. (1), we seethat the systemtorqueis g(mfit + mzxz+ m3h). rut{. ot tt the system Thus the torque is the product of the gravitional acceleration g, which is a fea- ture of the environment in which the system happens to reside, and the number (mrxt I mzxz + m3r3), which is a feature of the systemitself, a constantthat stays the same no matter where the system is placed. The number (m 1.r1 m.zxz mzxz) is called the moment of the system about I I the origin.It is the sum of the moments mt)ct, m2x2,m3x3of the individual masses Mo : Moment of system about origin : l*t xr (We shift to sigma notation here to allow for sums with more terms. For lmpxp, read summation mt times xr.) We usually want to know where to place the fulcrum to make the system (4, balance,that is, at what point t to place it to make the torque zero. Special locatiofl for balance The torque of eachmassabout the fulcrum in this speciallocation is rorquez1abour (lr# r;::,r: ) (*il:.-.) ': of (x7 -i)mpg. : When we write the equation that says that the sum of these torques is zero, we get
• 5.7 Moments and Centersof Mass 409 an equation we can solve for t: ilrxi I : o I,to i)mp:6 sf(rr ina't:0 Itr^r^ D*rrr-lrll1:g f*rrr: l*r This last equation tells us to lind -r by dividing the system's moment about the odgin by the system's total mass: system moment about origin L,r*r - Lro system mass The point t is called the system'scenter of mass. Wiresand Thin Rods In many applications,we want to know the center of mass of a rod or a thin stdp of metal. In caseslike these where we can model the distribution of mass with a continuous function, the summation signs in our fonnulas become integrals in a manner we now describe. Imagine a long, thin stdp lying along the r-axis fiom I : a to.{ : D and cut into small pieces of mass Lmp by a partition of the interval [a, r]. The,tth piece is Arr units long and lies approximately .r1 units from the origin. Now observe three things. First, the strip's center of mass r is nearly the same as that of the system of polnt masseswe would get by attaching each mass Am r to the point t(: system moment try systemmass Second,the moment of eachpiece of the strip about rhe origin is approximately rnAmr, so the system moment is approximately the sum of the rrrAl?1r: S l s t e mm o m e n t r r-Arzt. f Third, if the density of the strip at;r1 is d(ra), expressedin tems of mass Density per unit length, and 6 is continuous,then Az1 is approximately equal to d(rp)A4 A material's densityis its massper unit (mass per unit length times length): volume. practice, In however, tendto use we L m k x 8 ( x k )L x k ' we Ltnits canconvenientl) measure.For wires,rods,andnairowstripswe usemass Combining these three observationsgives per unit length.For flat sheets plateswe and system oment m x,Am, _ f ) - x ^ a r , r ,) 4 . r , usemassper unit area. - - :----i----:---: z) system mass EL/|l T6(t-)Ar-
• Chapter Applications Integrals of 410 5: The sum in the last numeratorin Eq. (2) is a Riemann sum for the continuous function;6(x) over the closedintervalla, bl. The sum in the denominator a is Riemannsumfor the function 6(;) over this interval.We expectthe approximations in (2) to improveasthe stdp is partitionedmorefinely, andwe areled to the equation ,: r;:::,,)!, l oxtqx This is the formulawe useto find i. Moment, Mass, and Center of Mass of a Thin Rod or Strip Along the r-axis with Density Function 6(r) xt{ia* Mo : (3a) Momentaboutthe origin: f.b I,u o, u<*t M: (3b) Mass: Mo (3c) Centerof mass: To find a centerof mass,divide momentby M mass. Strips and rods of constant density EXAMPLE1 Show that the centerof massof a straight,thin strip or rod of constantdensity lies halfway betweenits two ends. from J : a to x: b (FiE. So/ution We modelthe stripas a portionof the.x-axis goalis to showthatt: (a + b)/2, the point halfwaybefween andb. d 5.51).Our The key is the density'shaving a constantvalue.This enables to regardthe us function6(.r) in the integrals Eqs. (3) as a constant (call it d), with the result in that ,'),,: urru' u*o':u - o, 5,51 The center of massof a straight, lu l,',0,:ul) thin rod or strip of constantdensity lies halfwav between its ends. = l,'u : u o*:,['], o, l' :6(b-a) d .., - q-) -Mo ,(b- 6(b - ,) alb The ,t s cancel in rhe tr 2 formula fof r. ', / l l0l EXAMPLE2 Avariable density 5-52 We can treat a rod of variable The 1o-m-long in Fig. 5.52thickens from left to dght sothatits density, instead rod see thickness a rod of variabledensity. as of beingconstant, 6(-r) : t a (x/10) kg/m. Find the rod's centerof mass. is Example 2.
• and Centers Mass 411 of 5.7 Moments solution The rod's momentaboutthe origin (Eq. 3a) is tO r 12I r'n rl0 / ( x + , t^ulll a x * (' r + - - ) a ' - t xtdr-l M-o : l lu' Jn Jo Jo 250. loo f x ' r - l* ' l ' - r r ' i L n , i r r r fr t I nf Lfrl r =l lg.t. 301,, L2 -l : The rod's mass(Eq. 3b) is - I * t t - l ' o: x, r)u d r x t d x : r ' n ,(.' - M- l0+5:15kg. i)dx L, ,olo /., Jo The cenlerof masstEq. 3c) is located lhe point al 250 1 50-... M2 _ - I 5 . ) Om . '-M3159 j over a PlaneRegion Distributed Masses Supposewe have a finite collection of masseslocated in the plane' with mass flk at the point (rr, )r) (see Fig. 5.53). The mass of the system is U -L^^. mas.: System Each massn ' has a moment about each axis. Its moment about the;r-axis is nt4y1' and its moment about the 1-axis is zrrr. The moments of the entire system about the two axes are 5.53 Eachmass hasa momentabout mk eachaxt5. u:lln1,vt, Moment about Jr-axis: u':l^tr' Moment about y-axis: The r-coordinate of the system'scenter of mass is deflned to be M, L^,*, M Lmo balances about case,the system With this choiceof .x, as in the one-dimensional the line,r : t (Fie.5.54). centerof massis delinedto be The y-coordinate the system's of - _ M^ -L:r*vr . (s) 'ML^r With this choice of t, the systembalancesabout the line l' : t as well The torques exertedby the massesabout the line y : y cancel out. Thus, as far as balance is concemed,the systembehavesas if a1lits mass were at the single point (t' l') We arrayof masses 5.54 A two-dimensional call this point the system'scenter of mass. balances its centerof mass. on Thin, Flat Plates In many applications, we need to find the center of mass of a thin, flat plate: a disk of aluminum, say, or a triangular sheet of steel ln such caseswe assumethe distribution of mass to be continuous, and the formulas we use to calculatet and t contain integrals insteadof finite sums.The integrals arise in the following way,
• 412 Chapter 5: Applicationsof Integrals Imagine plateoccupying regionin the4)-plane, into thin sffipsparallel the a cut .l srrin ^f m,.s AD to one of the axes(in Fig. 5.55,the y-axis).The centerof massof a typical strip is (i, !). We tleat the strip'smassAru as if it wereconcentrated (;, i). The at momentof the strip aboutthe /-axis is theniAn. The momentof the strip about Equations and (5) then become (4) the r-axis is iA.,??. Li t* _M, M,, )-i Alrr '- r- Dr _^*' M' LL- As in the one-dimensionalcase,the.sumsare Riemannsumsfor integralsand approach theseintegrals limiting valuesas the stripsinto which the plateis cut as become nanowerandnanower. write these We integralssymbolically as lidm t - --v--- and r.:ij A plate cut into thin stripsparallel .l am to the y-axis.The moment exerted by a typicalstrip about eachaxisis the moment its massAm would exert if concentrated the strip's at centerof mass 0,,i1. Mornents,Mass,and Centerof Massof a Thin PlateCovering Region a in thexy-plane U, : I y a* Moment about r-axis: the y (crn) a, : about y-axis: (6) Moment the i a, I Mass: u: a^ I ] l M' M' - -: center of masr: , x )= M, M theseintegrals, picturethe platein the coordinate To evaluate we planeand i 5j The plate in Example 3. sketcha stdp of massparallel to one of the coordinatesaxes.We then expressthe (;, strip's massdm and the coordinates i) of the stxip'scenterof massin terms j of x or y. Finally,we integrate dm, i dm, anddm between limits of integration determined the plate'slocationin the plane. by (.r,2) EXAMPLE3 The triangular plateshownin Fig. 5.56hasa constant densityof :3 glcm2. (a) theplate'smomentM) aboutthe 1,-axis, the plate'smass (b) 6 Find M, and(c) the r-coordinate the plate'scenterof mass(c.m.). of Stripc.m. is halfway. G, i) : (r, r) Solution Method 1: Verticalstrips(Fig. 5.57). a) The momentMr: The typical vertical strip has centerof mass(c.m.): (i, y) : (x, x), Unitsin centimeters nea'. dA :2x dx, length: 2x., .2x dx :6x dx, mass: dtn : 6 dA :3 width: dr, 5.5.2Modelingthe plate in Example 3 distanceof c.m. from y-axis: i :,r. with vertical strips.
• 5.7 Moments and Centersof Mass 413 The moment of the strip about the y-axis is x . . 6 x .d x : 6 x 2 d x . idm: The moment of the plate about the y-axis is therefore f 11 - -11 I 6 x 2d x : 2 x 3 1 : 2 g , c m . M 't : lidm: J Jo lo The plate's mass: b) 'll f ft ox x: ix,lo 3 g. : ^t: a Jdm: Jo c) The r-coordinate of the plate's center of mass: Mt 2g'cm 2 _ : tt' 1- t M By a similar computation we could find Mx andt: Mr/M. Method 2: Horizontal smTs (Fig. 5.58). ) (cmJ a) The moment M/: The y-coordinateof the centerof massof a typical horizontal strip is y (see the figure), so ^2 ! Stripc.m. j, : v. is halfway. /- !, G.t=ff.y) The .x-coordinate the .r-coordinate the point halfwayacross triangle. is of the _L Thismakes theaverage y/2 (thestrip'slefrhand.r-value) I (thestrip,s it of and righrhandx-value): ]F- 1 (Y + | 1 - /2) Y v+2 ,: 2 4' 2- 4 We also have 5.58 Modelingthe plate in Example 3 l e n 2 2h : - a : - - ! . ' st with horizontal strips. width: dy, '---! dA : area; d- 2' '-' oy' mass: dm:6dA:3.2-! z distance c.m.to )-axis: I : of +. Howto Finda Plate's Center of Mass The moment of the strip about the y-axis is Jl - - - - - : . - . - - ' , lt ) 1 1. Picturethe platein the,xy-plane. . idn: ay: gt+_y-tay. 2. Sketcha stdp ofmass parallel to one 4 2 of the coordinateaxesand find its The moment of the plate about the y-axis is dimensions. 3, Find the saip's massdm and center r r2r ,y., l ' : . 3f. l/16 M = I i d m : I ; t + y , ) d y : : l 4 y l :2 s.cm. (: l of mass(i, i). Jo d 6L Jlo 6J,i J 4. Inte\$ate t dm, i dm, altiddm to fid M,, My aD:d, M. The plate's mass: b) 5. Divide thg momentsby the massto f f2 l ?r ,,212 ? calculatet and t. M: ldm-- I :t2-ytdy:;lzy-';l =ia-2J:3s. Jo z 2L z)o Z J
• Chapter 5: Applicationsof Integrals 414 The x-coordinate the plate'scenterof mass: of c) 2 Mr Zg.cm - M 39 3 f By a similarcomputation, couldfind M, andt. we <-r.E l--l If the distribution massin a thin, flat platehas an axis of symmetly, of the centerof masswill lie on this axis.If thereare two axesof symnetry,the center These factsoftenhelp to simplify our work. of masswill lie at their intersection. Find the center of mass of a thin plate of constant density 6 EXAMPLE 4 covering the region bounded above by the parabola y : 4 - x2 and below by the ;r-axis (Fig. 5.59). so/ufion Since the plate is symmetric about the y-axis and its density is constant, the distribution of mass is synmetric about the y-axis and the center of mass lies on the y-axis. This means that .r:0. It remains to fitrd t : Mr/M. A trial calculation with horizontal strips (Fig. 5.59a) leads to an inconvenient integration f 23 yr/a - y dy. M, : (b) Jo 5.59 Modelingthe plate in Example 4 (Fig.5.59b). with vefticalstripsinstead We therefore modelthedistribution mass of with (a) horizontal stripsleads an to The typical vertical strip has integration, we model inconvenient so r L-.2t with (b) verticalstripsinstead. c e n t eo f m a s s c . m ) : { i . i ) - { r . - ; : r { } z/ 4 - x2, length: width: dx, (4 area: dA: x2)dx, mass: dm : 6 dA : 6(4 - x2)dx, distance c.mto r-axis: t:+. from The momentof the strip about the x-axis is 4-x2 3 .6t4 x'tdx= -14- s'1'ix. ia^ = i The momentof the plate aboutlhe x-ais is ru: lra*: l.uuro-,t0 :XI'l.u-2+a':2fa' (7) The mass of the plate is -.a* :3] a. u : I a*: l,';r+ (8)
• and Centers Mass 4'15 of 5.7 Moments Therefore, (256/1s)6 8 M, _ - t- (32/3)6 5 M- The plate'scenterof massis the point ( t . t ) : 1 0/ '-; l '8 tr J./ EXAMPLE5 Variable density Find the centerof massof the plate in Exarnple4 if the density at the point (.r, y) from the point to the y-a,rs. i.s6 : 2x2,twice the square the distance of so/ut on The massdishibution is still symmetricaboutthe y-axis, so r = 0. With 6 : 2x2,Eqs.(7) and (8) become f I t28 x2t4-x22dx M,: lidm: I it+-*'t'd*: I l-z J J-t2 ro48 12 (7't / 1 t 6 x '-: 8 * n + x 6 t d , = ; . : I, lUJ r127) u -- ldm= 6t4-x2rdx: | 2x2r4-x2tdx | J-z J Jt 12 )A (s) A typical small segment of : I G x 2 - z x a ^d - - l - 15' wiehasdm=6ds=6ad0. J-2 y=\$- (,,: Therefore, ( d c o sd , , s i n d ) M, 2048 15 8 _ 'M1052567 The plate'snew centerof massis G,t: (0,;). E Find the centerof massof a wire of constant densityd shaped EXAMPLE5 like a semicircle radiusc. of y: Gig. 5.60).The solution We modelthe wire with the semicircle aboutthe y-axis,soi :@=7 find y, we imagine 0. To distribution mass symmetric of is (Fig. 5.60a)has the wire dividedinto shortsegments. typicalsegment The (b) length: ds : a d0, wire in Example 6. 5,60 The semicircular (a) lhe dimensions variables usedin and 6 ds : dtn : Massper unit lengthtimeslength 6 a d0, mass: (b) findingthe centerof mass. The center i : a sir 0. distance c.m. to x-axis: of doesnot lie on the wire. of mass
• I I Chapter Applications lntegrals of 5: 416 Hence, ffr 2 6a2f- cos9lf, | qm t^ -- !;. : !- t 6att 7( I am The center of mass lies on the axis of symmetry at the point (O,2a/T), abo''Jt two-thirds of the way up from the origin (Fig. 5.60b). tr Centroids When the density function is constant, it cancels out of the numelator and denom- inator of the formulas for 7 and y. This happened in nearly every example in this section. As far as -r and y were concerned,6 might as well have been l. Thus, when the density is consta.nt,the location of the center of mass is a feature of the geometry of the object and not of the material from which it is made. In such cases engineers may call the center of mass the centroid of the shape, as in Find the centroid of a triangle or a solid cone. To do so, just set d equal to 1 and proceed to flnd - and y as before, by dividing moments by masses. 5.7 Exercises 6(.r)=l+(r/3), 0:-t=3 Thin Rods 8. 6 ( x , ) : 2 - ( t / 4 ) , O = r < 4 1. An 80lb child and a l00lb child are balancingon a seesaw. The 80lb child is 5 ft from the fulcrum. How far from the fulcrum rsx34 9. 6 ( x ) : t + ( t / J i ) , is the l00lb child? 10. d ( r l : 3 ( r - l l 2 l r 5 / 2 ) . 0 , 2 5 : . r < I 2. The endsof a log are placedon two scales.One scalereads100 2-x, 0:r<l 11. kg and the other 200 kg. Where is the log's centerof mass? x, |.x =2 3. The endsof two thin steelrodsof equallengthareweldedtogether )t+1, 0:.r<l to make a dght-angledframe. Locatethe frarne'scenterof mass. 12. 6(.r.): l=r=2 2, (Ilizt Wherc is the centerof massof eachrod?) Density with Constant Thin Plates In Exercises13-24, find the centerof massof a thin plate of constant density 6 coveringthe given region. 13. The region boundedby the parabolay: x2 and the line y = 4 14. The regionbounded the parabota : 25 -.r2 andthe x-axis by ), parabola ): t - tc2and the line 15. The rcgion boundedby the 4. You weld the ends of two steel rods into a right-angledframe. 16. The region enclosed the parabolasy : x2 - 3 andy = -2x2 One rod is twice the length of the other. Where is the frame's by (llirt Whereis the centerof massof eachrod?) centerof mass? 17. The region boundedby the J-axis and the curve x : 1,- )3, o<Yst 5-12 give densityfunctionsof thin rods lying alongvarious Exercises intervals of the .r-aris. Use Eqs. (3a--c)to find each rcd's moment 18. The region boundedby the parabola, = y' - y and the line aboutthe odgin, mass,and centet of mass. O=x=2 5.6(a):4, 19. The rcgion boundedby the r-axis and the curve y:cosr, -t /23x =n/2 ls{:3 6.6(:t)=4,
• 417 Exercises 5.7 20. The region betweenthe x axis and the curve ) : sec2 -z/4 < 't, x =ft14 4x and )' : 21. The region bounded by the parabolas ) = 1r2 The region cut from the lirst quadrant by the circle x2 1 22, a) The region bounded by the i-axis and the semicircle ): b) J'-v Compare your answer with the answer in (a). (a) 7 1 The triangular region in the first quadrant between the circle jr2 + )2 = 9 and the lines :r = 3 and y = 3. (llinl. Use geometry to find the area.) l/x3, below by the 24, The region bounded above by the curye ): -1/n3, and on the left and right by the lines r: I curve ): and x : zz> 1 Also, lind lim--;r. with VaryingDensity Thin Plates _.L Find the center of mass of a thin plate covering the region be- tween the r-axis and the curve 1 : 27,12, : i 5 2, if the plate's I density at the point (r, ),) is 6(.r) - 1'?. 26. Find the centerofmass ofa thin plate coveringthe region bounded below by the parabola) : t2 and aboveby the line y : x if the plate's density at the point (x, r) is d(x) : 12r' (b) The region bounded by the cufr'es ): +4/vA and the lines ',1 ' :: The trianglein Exercise (a) The centroid. 29. x : I and r : 4 is revolved about the y-axis to generatea solid (b) The dimensions variables usein locating to and Find the volume ol the solid. a) the centero{ mass. Find the center of mass of a thin plate covering the region b) if the plate's density at the point (r, y) is 6(,r) = l/x. Sketchthe plate and show the centerof massin your sketch c) 28. The region betweenthe curve ) : 2/.t and the r-axis from r - I Show that t: l/3. 3. to.r - 4 is revolved about the r-axis to generatea solid. Extend the argument lo the other sides. 4. Find the volume of the solid. a) Use the result in Exercise 29 to nnd the cenfroids of the triangles Find the center of mass of a thin plate coveling the region b) whose vertices appearin Exercises30-34. (Hint: Draw each triangle il the plate's density at the point (.r, y) is 6(r) : r/x. first.) Sketch the plate and show the centerof mass in your sketch. c) 3 1 . ( 0 , 0 ) ,( 1 . 0 ) ,( 0 , l ) 3 0 . ( - 1 , 0 ) , ( 1 , 0 ) ,( 0 ,3 ) 33. (0,0), (.r,0), (0, r) 3 2 . ( 0 , 0 ) ,( a , 0 ) ,( 0 ,a ) 34. (0,0),(a,0).(a/2, b) ThinWires toward the opposite vertex is the point where the triangle's three 35. Find the moment about the x-axis of a wire of constant density medians intersect.Show that the centroid lies at the inteNection that lies along the curvs ,v: .v& from jr = 0 to:t :2. of the medians by showing thai it too lies one-third of the way 36. Find the moment about the i-axis of a wire of constant density from each side toward the opposite vertex. To do so, take the that lies along the cuNe ) : xr from,t = 0 to r : 1. following steps. 37. Supposethe density of the wire in Example 6 is d:ftsind Standone side ofthe triangle on the t-axis as in Fig. 5 61(b) 1. (l constant).Find the center of mass. Express dm in terms of I and d). 38. Supposethe densityof the wire in Example 6 is , - 1 + k cosrl Use similar triangles to show that L : (blh)(h - y) Sub- 2, (f constant).Find the center of mass. stitute tlis expressionfor l, in your formula for dm.
• Chapter 5: Applicationsof lntegrals 4'18 41. For wires and thin rods of constant density shapedlike circular Formulas Engineering arcs centeredat the origin and symrnetric about the ) axis' the and formulas in Exercises39-42' i ) -coordincle lhe centerof mas' Verify the statements ol 39. The coordinatesof the centroid of a differentiable plane cuNe ac a sina ds are J)as Jx4s length length z i; Y,' N 42, (ContinLratiollof Etercise 4l) Show that when d is small, the distanceI lrom the centroid a) to chord AB is about 2i/3 (in the notation of the figure herc) Whatever the value of p > 0 in the equation )-: xz l(4p), the by taking the following steps. parabolic segmentshown here )-coordinate ol the cenffoid of the : (3/5)a. is 1 Show that l. s l n d - a vc o s d (e) h ZZz. cnneHeRcraptr sina - crcoscv ct-ctcos(l and use TRACE to show that Ilm-a- f @) x 213 equality (You will be able to confiIm the suggested 74 6.6,Exercise ) in Section between atd2hl3) d CnfcufnfoR The enor (difference ffi l) than45' See{or yourself greater is smallevenfor angles side of Eq (9) for a :0 2' by evaluating right-hand the a,n d1 . 0r a d . 0.4,0.6,0.8 :::: E'*:*.,;:--:. * InScience.thetelmlefe6specificallytoaforceactingonabodyandthebody's displacament This section shows how to calculate work The appltca- subsequent tanks tions run frorJcompressing railroad car springs and emptying subteranean lifting satellitesinto orbit' to forcing electronstogether and Work Done bY a ConstantForce When a body moves a distanca d along a straight line as a result of being acted the on by a force of constantmagnitudeF in the direction of motion, we calculate
• 5.8 Work 419 work W done by the force on the body with the formula Joules W: (Constant-forcaformula for work). Fd (1) The joule, abbreviatedJ and pronounced 'jeuel. i namedafter rhe En!li5h phy.icisl Right away we can see a considerabledifference between what we are used to Joule (1818-1889). James Prescott The calling work and what this formula sayswork is. If you push a car down the sbeet, defining equation is you will be doing work on the car, both by your own reckoning and by Eq. (l). I joule - (l newton)(l meter). But if you push against the car and the car does not moveJEq. (l) says you will do no work on the car, even if you push for an hour. t n s y m b o l s ,J - 1 N . m . 1 From Eq. (1) we see that the unit of work in any system is the unit of force It takes a force of about I N to lift an apple from a table. If you lift it I ln you multiplied by the unit of distance.In SI units (SI standsfor Systime International, have done about 1 J of work on the apple. If or lntemational System), the unit of force is a newton, the unit of distance is a you then eat the apple you will have meter, and the unit of work is a newton-meter(N , m). This combination appears consumedabout 80 food calories, the heat so often it has a special name, the joule. In the British sysrem,the unit of work is equivalentof nearly 335,000joules. If this the foot-pound, a unit frequently used by engineers. energy were dircctly useful for mechanical wo[k, it would enable you to lift 335,000 more applesup I m. If you jack up the side of a 2000-lb car 1.25 ft to change a nre EXAMPLE 1 (you have to apply a constant vetical force of about 1000 lb) you will perform 1000 x 1.25 : 1250 ft-lb of work on the car. In SI units, you have applied a force of 4448 N througha distance 0.381 m to do 4448 x 0.381 :: 1695J of work. of D Work Doneby a VariableForce If the force you apply varies along the way, as it will if you are lifting a leaking bucket or compressinga sprirg, the fomula lll: Fd has to be replaced by an integral formula that takes the variation in F into account. Supposethat the force performing the work acts along a line that we can model with the r-axis and that its magnitudeF is a continuousfunction of the position. We want to find the work done over the interval ftom _r : a to r : b. We partition La,D] in the usual way and choosean arbitrary point cr in each subinteryal [.:rp 1,r1]. If the subinteryalis short enough,f being continuous,will not vary much from;r1 1 to .rt. The amount of work done acrossthe interval will be about F(ca) times the distance A;1, the same as it would be if F were constantand we could apply Eq. (l). The total work done from a to b is therefore approximated bv the Rienann sum F(ci)Ar1. (2) ! t:l We expect the approximation to improve as the norm of the pafiition goes ro zero, so we define the work done by the force from 4 to , to be the integral of f. from ato b. Definition The work done by a variable force F(,r) directed along the.r-axis from x:qtox:bis l'F (x) dx. (3)
• 4ZO Chapter Applications Integrals of 5: meters' and The units of the integral are joules if F is in newtons and x is in foot-pounds F is in pounds t in feet' and if I /';r2N alongthe x-axis The work doneby a force of F(r): E{AMPLE 2 from.t:lmtox:10mis 10 . r lr L o ,: J, r l i -u1 :0.9 J. - -l = 10! dx=--l I lr A leaky 5lb bucket is lifted from the ground into the air by EXAMPLE 3 (Fig 5'62)' The ropeweighs0'08 lbift' speed pulling in 20 ft of ropeat a constant It finishes ih uitt ,tu.tt wiih 2 gal of water (16 lb) andleaks at a constantrate' 3 5.62 The leakybucketin ExamPle &ainingjust as it rcaches top' How much work wasspent the a) lifting the wateralone; b) lifting the water and bucket together; c) lifting the watel bucket,and rope? Solution the water's The water elone. The force required to lift the water is equal.to ^) When the weight, which variessteadilyfrom 16 to 0 lb over the 20-ft lift bucket is r ft off the ground,the water weighs 4x '{ / /20-x = 16(l -.)A) : 6-T rb. 1 6 1 ' 1: 1 6 { = - '' ) /' ./ original'weight ProPirrtionleft at elevation .r The work done is I F( x ) x Use L.q. (l) fof .tfi:tblc lol.ces d w: f z ot 4r f : I ( ' o - T ) a : l t o ' - ? l rrzl ol 2 0: 3 2 0 - 1 6 0 = 1 6 0 r r ' l b ' - t/ L Jo (l), it takes5 x 20: 100 b) The water and bucketngether Accotding Eq to 20 ft' Therefore ft.lb to lift a 5Jb weight 160+ i00 : 260 ft ' lb of work were spentlifting the water and bucket together' The water,bucket,and rope.Now the total weight at level r is c) ft lb/ft ll / 4Y (0.08)(20 -L5+ F(x)=l16-;l ),/ _'--v-' -v- weight of rope conslanr variable paid out at wergnt welgnt elevationx of bucket of watel
• 421 5.8 Work The work lifting the rope is f20 (0.08)(20 - x)d,r Work on rope : lr.u-o.orr)0, 1,,'o ;l [ : I l . 6 r - 0 . 0 4 x '1 2 0 1 2 - l o - 1 6f t . I b . | - L Jo The total work for the water, bucket, and rope combined is D 1 6 0 +l o b + f t - 2 7 6 f t . t b . Law for Springs: = kx F Hooke's Hooke's law says that the force it takes to stretch or compressa spring I length units from its natural (unstressed) length is proportional to .r. In symbols, (4) The constant i, measuredin force units per unit length, is a characteristicof the spring, called the force constant (or spring constant) of the spring. Hooke's law (Eq. 4) gives good resultsas long as the force doesn't distort the metal in the spdng. We assumethat the forces in this section are too small to do that. Find the work required to compress a spring from its natuml EXAMPLE 4 length of I ft to a length of 0.75 ft if the force constantis ft: 16 lb/ft. Sorution We picture the uncompressedspring laid out along the r-axis with its movable end at the origin and its fixed end at x : I ft (Fig. 5.63). This enablesus to describethe force required to compressthe spring from 0 to,r with the formula F: 16,r. To compressthe spring from 0 to 0.25 ft, the force must increasefrom 1 6. 0 . 2 5 : 4 l b . F(0) : 16.0 : 0lb : F(0.25) to The work done by F over this interual is Eq.(3) wirha : 0, f0.25 10.25 b:0.25,r(jr): :05ft.lb. l.6xd.x:8x2 w:l | lo.I JO _10 Work doneby F tr fromr=0 to-I=0.25 A spring has a natural length of I m. A force of 24 N stretches EXAMPLE 5 Amount compressed the spring to a length of 1.8 m. (b) Find the force constant/t. a) How much work will it take to stretchthe spring 2 m beyond its natural length? 5.63 The force F neededto hold a spring b) linearlyas increases undercompression How far will a 45-N force stretch the spring? c) the springis compressed. Solution Theforce constant.We find the force constantfrom Eq. (a). A force of 24 N a) shetchesthe spring 0.8 m, so 24 : k(0.8) E q .( 4 )w i t hF : 2 4 . r : 0 . 8 k : 24/0.8: 30 N/m.
• 422 Chapter 5: Applicationsof lntegrals b) The work to stretcll the spring 2 m. We imagine the unstressed spdng hanging along the r-axis with its free end at -{ - 0 (Fig. 5.64). The force required to stretch the spdng.r m beyond its natural length is the force required to pull the free end of the spring -{ units from the origin. Hooke's law with I :30 saysthat this force is F(n) : 30x' The work done by F on the springfrom r :0 m to r :2 m is l2 lt I 3 0 x x : ls;r'ln= 60J. w: d Jo c) How far will a 15-N Jbrce stretch the spring? We substituteF:45 in the equationF: 30.r to find 45 : 30:r. r : 1.5m. or A 45-Nforcewill stretch spdng1.5m. No calculus required find this. the is to l -r(m) Pumping Liquids rom Containers f 5.64 A 24-Nweight stretches spring this 0.8 m beyondits unstressed length. How much work does it take to pump all or part of the liquid from a container? -t----:-: To find out, we imagine lifting the liquid out one thin hodzontal slab at a time and applying the equation W : F d to each slab. We then evaluatethe integral this leads to as the slabsbecome thinner and more numerous.The integral we get each time dependson the weight of the liquid and the dimensionsof the cortainer, but ,*l'',,# the way we nnd the integral is always the same. The next examplesshow what to do. IF EXAMPLE 6 How much work does it take to pump the water from a full updght circuiar cylindrical tank of radius 5 m and height l0 m to a level of 4 m 1{'rr above the top of the tank? So/ution We draw the tank (Fig. 5.65), add coordinateaxes,and imagine the water divided into thin horizontal slabsby planesperpendicularto the }-axis at the points of a partitionP of the inteNal 10, 101. The typical slab between the planes at ) and l + A,v has a volume of 5.65 Tofind the work it takes pump to the waterfroma tank,thinkof Iifting Ay : r (radius):(rhickness) r(5)':Ay : 25r Ly m3. : the wateronethin slabat a time. The folce F required to liti the slab is equal to its weighr, How to FindWork Done During F:9800Ay ( ) S 0 0i , r r Pumping = 9800(25zAy) 245,0002Ar = N. 1. Draw a figure with a coordinate which F mustact is about(14 - )) m, so the work done The distance through system. lifiing the slabis about 2. Find the weight F of a thin horizontal slab ot' liquid. AW : force x distance 245,000r(14 : r)A) J. 3. Find the wolk AW it takes to lift the slab to its destination. The work it takes to lift all the water is approximately 4. Integrate the work expressionfron] l0 it) w ''LLw : I z+s,ooo y)a) J. _ r.t+ the base to the surface of the liquid.
• 5.8 Work 42i This is a Riemannsum for the functio y=zxor'=)t n 245,00M _ y) over the interval < (14 ,;;;;, 0 T t :; 10.The work of pumpingthe tankdry is th. ,; ji;l--' O: fi,it,f -v lro w: _ ytdy= 245.0002 24s,0oor(14 (5, 10) ['u l+ _ r) at Jo Jo , = , y1 = 245,00M1uy r,2t'0 L -71=245'ooon[eol x 69,272,118 69.3 x 106 ^, J. A l-horsep_ower output motor rated, 746 J/seccould at - --'- empty the tank in a little v'rPLJ ure Iess than 26 h 5.66'The oliveoil in Example 7. o EXAMPLE 7 The conical tank in Fig. 5.66 is filled to within 2 ft of the top ng 57tblrt3 . Howmuch ;; ;;,; ;.T,pump theol to ;:l*; fi : i,ff Solution We imagine the oil di perpendrcurar ro *. r;I':1 *: points a,*;i':r'il:,t#;'fo;.0d.0,n.' of rne ryplcatstabbetween planes y the ut anay i Ay nasa volumeof about A l z = n ( r a d i u s t 2 ( tiness, tr / l 2 hicl = o, __ tt n,. r, ) it, The force F(y) required lift this to slabis equalto its weight, F(y) : 57 Av : 5!r, o, tt. w;-jshr w€1cnr unrl : per thyuch__1hich F()) mustact ro tift rhis slabro the nm .T:lt:l-: is about(10 _ y) ft, ,'- ot rhe cone level of the so the work a fif,irg ,fr ljib ,. uOout --::119 5in AW : - y)y2Ay fr . tb. The work done lifting all the slabs from y : 0 to ),-: g to the rim is approximately w- f r r o -y t f a yf t . t b . f This is a Riemannsum for the funcnong7r /4)(10 _ y)y2 onthe intervaltrom ) = 0 to l, _- 8. The work of pumoingthe oil a G ,irn,rr,,rriirniio,*r. ,urn, as the norm of the partition go, to ,.o. f8 's'ti-- l o _ - W: I v t y ld t Jo+--' a7- rE =; :'oy2_y3;dy | 57T f loyr va'l8 = -7' - 3 0 . 5 6 1f r ' l b . 4 | .- _ tr J0
• 424 Chapter 5: Applicationsof lntegrals Exercises 5.8 Work Done by a VariableForce to V2 : 32 in3 if pr : 50 1b/in3and p and y obey the gas law pV'a : constant (for adiabatic processes). l The workers Example changed a largerbucketthat held in 3 to 5 gal (40 lb) of watet but the new bucket had an even larger Springs leak so that it, too, was emptyby the time it reached top. the 9, It took 1800J ofwork to stretch springftom its naturallength Assuming that the waterleakedout at a steady a rate,how much work was donelifting rhe water?(Do not includethe ropeand of 2 m to a lenBthof 5 m. Find the spring'sforceconstant. bucket.) 10. A springhasa natural lengrhof 10 in. An 8001bforcestretches the spdngto 14 in. (a) Find the force constant. How much 2. The bucketin Example is hauled twice asfastso thatthere (b) 3 up is still I gal (8 lb) of waterleft when the bucketreaches work is done in stretching spring from 10 in. to 12 in.? the the (c) How far beyondits naturallength will a 1600-lbforce stretch top. How muchwork is donelifring the waterthis time?(Do not includethe ropeandbucket.) the spring? 11. A forceof2 N will stetcha rubber 3. A mountainclimberis about haulup a 50-mlengthof hanging band2 cm (0.02m). Assuming to rope.How muchwork will it takeif theropeweighs0.624N/m? Hooke'slaw applies, how far will a 4-N forceshetch rubber the band?How much work does it take to stretch the rubber band 4. A bagof sand originallyweighing144lb waslifted at a constant this far? rate.As it rose,sandalsoleaked at a constant out late.The sand was half goneby the rime rhe bag had beenlifted 18 fr. How 12. If a forceof 90 N stetches springI m beyond natural a its length, muchwork wasdonelifting thesand far?(Neglect weight how muchwork doesit taketo stretch spdng5 m beyond this the the its of the bag andlilling equipment.) natumllength? 5. An electricelevator 13. Subwaycar springs. It rakes forceof 21,'114Ib compress with a motor at the top has a multishand a to cableweighing lb/ft. Whenthe car is at the lirst floor, 180ft a coil spring assemblyon a New York City Transit Authority 4.5 of cablearepaidout, andeffectively ft areout whenthe car is subway from irs free heightof 8 in. to its fully compressed car 0 at the top floor.How muchwork doesthe motor do just lifting heightof 5 in. the cable when it takesthe car from the first floor to the top? a) What is the assembly's forceconstant? b) How much work doesit take to compress assembly 6. Whena paticle of masszr is at (_r, it is attracted the the 0), towardthe first half inch?the second inch?Answerto the nearest originwith a forcewhose half magnitude &/r2. If theparticle is starts in . lb. from rest at r - D and is actedon by no otherforces,find the work doneon it by the time it reaches - a,0 < a < b. r (Datacourtesy Bombardier, of Inc., MassTransitDivision,for spring assemblies subwaycarsdeliveredto the New york City in 7. Suppose thegasin a circularcylinder that ofcross-section areaA TransitAuthorityfrom 1985to 1987.) is beingcompressed a piston. by Ifp is thepressure ofthe gasin inchandyis thevolumein cubicinches, pounds square per show 14. A bathrcom scaleis compressed in. whena l50lb person 1/16 thatthe work donein compressing gasfrom state(pr, %) to the stands it. Assuming scale on the behaves a springthatobeys like state(p2, V2)is givenby the equation Hooke's law,how muchdoessomeone compresses scale who the f( P' v') 1/8 in. weigh?How much work is donecompressing scale the Work : pdv. / 1/8in.? J( t1,.vn PumpingLiquidsfrom Containers (I1irr. In the coordinates : A dt. suggested the figvehete,dV in The force against the piston is pA.) The Weight of Water Because variations the earth'sgavitationalfield, the of in weight of a cubic foot of water at sea level can vary from about62.26lb at the equator asmuchas62.59lb nearthe to poles,a variationof about0.57o. cubic foot rhat weighs A about62.4 lb in Melboumeand New York Ciry will weigh 62.5 lb in Juneauand Stockholm. While 62.4 is a typical ligure and a commontextbookvalue.thereis considerable vanat1on. (Continuation oJ Exercise Z) Use the integral in Exercise 7 to nnd the work done in compressingthe gas from Vt:243 in3
• Exercises 5.8 425 15. The rectangular shownhere,with its top at groundlevel,is tank 19. A vertical right circular cylindrical tank measures ft high and 30 usedto catchmnoff water.Assumethat the waterweighs62.4 20 fr in diameter It is full of keroseneweighing 51.2 lb/ft3. How 1b/ft3. much work does it take to pump the keroseneto the level of the a) How muchwork does taketo emptythe tankby pumping top of the tank? it the waterbackto groundleveloncethe tankis full? 20. The cylindrical rank shown here can be filled by pumping water b) If the water is pumpedto groundlevel with a (5/11)-hp from a lake 15 ft below the bottom of the tank. There are two motor (work output250 ft . lb/sec), how long will it take ways to go about it. One is to pump the water through a hose to emptythe full tank(to the nearest minute)? attached to a valve in the bottom of the tank. The other is to c) Showthat the pumpin part (b) will lower the waterlevel attach the hese to the rim of the tank and let the water pour in. 10 ft (halfway) duringthe first 25 min of pumping. Which way will be faster?Give reasonsfor your answer d) Whataretheanswen parts(a) and(b) in a location ro where warerweighs62.26lblft32 62.59lbfif.l ,w @ 21. CALCULATOR truncated The conicalcontainershownhere is full of strawberry milkshake weighs4/9 ozlin3.As you can that 16. The rectangularcistem (storagetank for rainwater) shown here see,the container 7 in. deep,2.5 in. across the base. is at and has its top 10 ft below ground level. The cistem, cunently full, is 3.5 in. across the top (a standard at Brigham,s Boston). at size in to be emptied for inspection by pumping its contents to ground The straw sticksup an inch abovethe top. About how much level. work doesit taketo suck up the milkshal<e throughthe straw (neglecting friction)?Answerin inch-ounces. a) How much work will it take to empty the cistem? b) How long will it take a ( 1/2)-hppump, raredat 275 ft . lb/sec, to pump the tank dry? c) How long will it rake the pump in part (b) ro empry rhe - 17.5 tank halfway? (It will be less rhan half rhe time required to empty the rank completely.) d) What are the answers to parts (a) (c) in a location where (r.15,',7t water weighs 62.261b1ft31 62.59 tbfit3'l Cround level ':.14 1 0t r Dimensions inches in t7, How much work would it take to pump tie water from the tank in Example 6 ro rhe level of the rop of the rank (instead of 4 m 22, a) Supposethe conical container in Example 7 contains milk higher)? (weighing 64.5 lb/ft3) insteadof olive oil. How much work 18. Supposethat, insteadof being full, the tank in Example 6 is only will it take to pump the contents to the rim? half full. How much work does it take to pump the remaining b) How much work will it take to pump the oil in Example 7 water to a level 4 m above the top of the tank? to a level 3 ft above the cone's rim?
• Chapter Applications lntegrals of 5: 426 long will it take to fill the tank the fiIst time? (Includethe time steel tank, you To designthe interior surfaceof a huge stainless waterweiShs 4 lblfi3. 62 it rakes fill the pipe.,Assume rcvolve the curve ) : t2, 0 s t : 4, aboutthe y-axis. The con- to tainer, with dimensionsin meters,is to be filled with seawater, N/m3.How muchworkwill it taketo empty whichweighs10,000 the tant by pumping the water to the tank's top? containers way we do from the 24. We modelpumpingfrom spherical other containe\$, with the axis of integration along the vertical axis of the sphere.Use the f,gure here to find how much work water reservoir of radius it takes to empty a full hemispherical 5 m by pumping the water to a height of 4 m abovethe top of the reservoir.Waterweighs 9800 N/m3. Subme\$iblepump OtherApplications a satellite in orbit. Te stretrgth of the earth's grav- R Zl, erttirg field varies v/ith the distance r from the earth's center, itational magnitude of the \$avitational folce experienced by a and the of mass rrt dudng and after launch is satellite AMG , ,v , _ _ .t , 12 tank 25. You arein chargeof the evacuation repairof the storage and Here,M :5,975 x 102 kg is the eatth'smass,G :6.6'720 x showtrher€.The tank is a hemisphere radius 10 ft and is full of 10 lr N . m2kg-2is the universalgravitationalconstant,and r is weighing 56 lb/ft3. A firm you co[tacted saysit can of benzene measwedin meters.The work it takesto lift a 1000-kgsatellite empty the tank for 1/20 per foot-poundof work. Find the work ftom the earth'ssurfaceto a circular orbit 35,780km abovethe requircd to empty the tank by pumping the benzene an outlet to earth'scenteris thereforegiven by the integral 2 ft abovethe top of the tank. If you have \$5000 budgetedfor ?80 f 15 000l000MG _ the job, can you afford to hirc the fiIm? Work : ,r Jou,.t. / ... ^^^ I J6.3?0.000 Evaluatethe integral.The lower limit of integrationis the earth's mdiusin metersat the launchsite. Cfhis calculationdoesnot take into accountenergy spent lifting the launch vehicle or energy spentbringing the satelliteto o\$it velocity.) E 28. Forcing electrons together Two electrons metersapartrcpel r eachother with a force of 23x1O2e F : -------;- newtons' Suppose electron held fixed at the point (1, 0) on is one a) the r-axis (units in meters).How much work does it take to move a secondelectrol along the t-axis ftom the point 26. Your town hasdecidedto drill a well to inqeaseits water supply. (-1,0) to the origin? As the town engineer,you have determinedthat a water tower Supposean electron is held fixed at each of the points b) will be necessary provide the pressureneededfor distribu- to (-1,0) and(1, 0). How muchwork doesit taketo movea tion, and you have designedthe systemshownhere. The water third electronalong the r-axis from (5, 0) to (3, 0)? is to be pumpedfrom a 300-ft well througha vertical 4-in. pipe into the base of a cylirdrical tart 20 ft in diameteratrd 25 ft Work and KineticEnergy high. The baseof the tank will be 60 ft aboveground. pump The moves a body of mass is a 3-hp pump,mted at 1650ft . lb/sec.To the nearest 29. If a variable force of magnitude F(t) hour,how
• 5.9 FluidPressures Forces 427 and m along the ir-axis frcm -rr to 12, the body's velocity u can be written as di/d/ (where / represents time). Use Newton's Second Weight Ys.Mass Law of Motion F : m(dD/dt) and the Chain Rule Weightis the forcethat resultsfrom gravity pulling on a mass. du du tlx du The two arerelated theequation Newton's by in second law, -dt dt dx dt Weight: massx acceleration. to show that the net work done by the force in moving the body Thus, from ,r1 to 12 is : Newtons kilograms m/sec2, x l.t. l'. W: I F ( r ) d x= . n u i - : Pounds slugsx ft/sec2. inui. J, To convertmassto weight,multiply by the acceleration of whereu1andu2arethebody'svelocities xr and12.In physics at gravity. convert To weightto mass,divideby theacceleration (1/2)mv2is calledtte kineticenergy thebody the expression of of gravity. moving with velocity v. Therefore,the work done by the force equalsthe changein the body's kinetic energ),, and we can find the wo* by calculating change. this In Exercises 30-36, usethe resultof Exercise 29. 30. Tennis.A 2-oz tennisball was served 160frlsec(about109 at was clockedat a phenomenal mph. How much work did 124 mph). How much work was done on the ball to make it go this Sampras haveto do on the 2-oz ball to get it to that speed? fast?(To find the ball'smass from its weight,express weight R 34. football. A quarrerback the ioorbatl88 lvsec (60 threw a 14.5-oz in pounds anddivideby 32 ft/sec2, acceleration gravity.) the of mph).How manyfoot-pounds work weredoneon the ball to of 31, Baseball.How manyfoot-pounds work doesit taketo throw gerir to this speed? of weighs5 oz :0.3125 lb. a baseball mph?A baseball 90 Ei 3S. SoftOti. How muchwork hasto be perlbrmed a 6.5-02 on soft- golf ball is drivenoff the tee at a speed 280 32. Golf. A 1.6-02 ball to pitch it 132fr./sec mph)? (90 of ft/sec(about191mph).How manyfoot-pounds work aredone of 36. A ball bearing. A 2-oz steelball bearingis placedon a ver- gettingthe ball into the air? tical springwhoseforce constant ,t: l8 lb/ft. The springis is E 33. lennis. Dudng the matchin which PeteSampras won the 1990 compressedinches released. 3 and Abouthow high does ballthe U.S. Openmen'stennischampionship, go? Sampras a serve hit bearing that FluidPressures Forces and We make dams thicker at the bottom than at the top (Fig. 5.67) because pressure the against them increaseswith depth. It is a remarkable fact that the pressureat any point on a dam dependsonly on how far below the surfacethe point is and not on how much the surface of the dam happensto be tilted at that point. The pressure, in pounds per squarefoot at a point & feet below the surface,is always 62.4h. The number 62.4 is the weight-density of water in pounds per cubic foot. The formula, pressure : 62.4h, makes sense when you think of the units involved: lb lb x ft. ff ft3 5.67 To withstandthe increasing As you can see,this equationdependsonly on units and not on the fluid involved.The pressure, damsare built thickerasthey go down. pressure& feet below the surfaceof any fluid is the fluid's weight-densitytimes lz.
• chapter5: Applications Integrals of 428 Weight-density Equation The Pressure-Depth A fluid's weighfdensity is its weight per unit In a fluid that is standing still, the pressurep at depth ft is the fluid's volume. Typical values (lb/ft3) are weight-density l, times lr.' 42 Gasoline (1) Mercury 849 64.5 Milk 100 Molasses Olive oil 57 Seawater In this section we use the equation P : u,h to derive a formula for the total b4 Water 62.4 force exerted by a fluid against all or part of a vertical or horizontal containing wall. Formula FluidForce for The Constant-Depth In a container of fluid with a flat horizontal base, the total force exerted by the FF{ *T fluid againstthe base can be calculatedby multiplying the area of the base by the pressureat the base.We can do this becausetotal force equals force per unit area (pressure)times area. (See Fig. 5.68.) If E p, and A are the total fbrce, pressure, and area. then F : total force : force per unit area x area : preSSUIe arca : PA x containers filled with are 5.59 These water to the samedepth and havethe samebasearea.The total force is thereforethe sameon the bottom of do eachcontainer. containers'shapes The not matter here. Fluid Force on a Constant-Depth Surface (2) pA: whA :':. '..: rl're Great i'/lcl.::t:.. EXAMPLE 1 At 1:00 pM. on January 15, 1919, an unusually warm day, a 9O-ft-high, 90-ft- diametercylindrical metal tank in which the Pudtan Distilling Companywas stodng molassesat the comer of Foster and Commercial streetsin Boston's North End exploded. The molassesflooded into the streets,30 ft deep, tapping pedesftians and horses, knocking down buildings, and oozing into homes. It was eventually tracked all over town and even made its way into the suburbs(on trolley cars and people's shoes).It took weeks to clean up. Given that the molassesweighed 100 lb/ft3, what was the total force exerted by the molassesagainstthe bottom of the tank at the time it blew? Assuming the tank was full, we can find out from Eq. (2): SHADEDBAND NOT'IO SCALE Total force : whA - (100)(S0)(n lb. (45)'?) 57.255.526 drawingof the molasses 5.1a.,! schematic How about the force againstthe walls of the tank? For example,what was the t a n k i n E x a m p l 1 . H o w m u c hf o r c ed i d e total force againstthe bottom foot-wide band of tank wall (Fig. 5.69)? The area of the lowestfoot of the vertical wall have to withstandwhen the tank was full? lt the band was takesan integralto find out. Noticethat A : 2Trh - 2r (45)() : gott fe . the proportions the tank were ideal. of
• 5.9 Fluid Pressures and Forces 429 The tank was 90 ft deep, so the pressurenear the bottom was approximately p - u)h : (100)(90)= 9000 lb/ftr. Therefore the total force against the band was approximately (9000)(901r^, 2, 544.690 lb. F - uhA: ) But this is not exactly right. The top of the band was 89 ft below the surface,not 90. and the pressurethere was less. To find out exactly what tbe force on the band was, we need to take into account the variation of the pressureacrossthe band. Formula TheVariable-Depth Strip length ai level l, Supposewe want to know the force exertedby a fluid againstone side of a vertical plate submergedin a fluid of weight-density }y. To llnd it, we model the plate as The force exertedby a fluid against strip is one sideof a thin horizontal a region extending from y : a to | = b in the .r-v-plane (Fig. 5.70). We pa ition a b o u tA F : p r e s s u rx a r e a= w x ( s t r i P e imagine the region to be cut into thin horizontal st ps ftr, Dl in the usual way and depth) x r0)Ay. The plate here is flat, by planesperpendicularto the )-axis at the partition points. The typical stdp tiom _' but it might havebeencurvedinstead, unitswide by a()') unitslong. We assume L(1) to be a continuous l i k et h e v e r t i c aw a l l o f a c y l i n d r i c a l t a n k . to ) + A} is A_- l functionof 1 the case, strip length is the Whatever alongthe surface the plate. of The pressurevaries across the strip from top to bottom, just as it did in the measured molassestank. But if the strip is narrow enough, the pressurewill remain close to its bottorn edge value of u x (strip depth). The fbrce exertedby the fluid against one side of the strip will be about AF = (pressure along bottom edge) x (area) = rrl x (strip depth) x Z(1)A_r. The force against the entire plate will be about bh : -ru (3) x (stripdepth)x L(),)4,-). )- -lr ' /2 1-' The sum in (3) is a Riemannsum for a continuous functionon [4, b]. and we erpect the approximationsto improve as the nolm of the partition goes to zero. We define the force against the plate to be the limit of these sums. Definition Thelntegral Fluid for Force Supposethat a plate submergedvertically in fluid of weight-density w runs from r, : a to ) : b on the )-axis. Let Z()) be the length of the horizontal strip measuredfrom left to right along the surface of the plate at level -r,. Then the force exefied by the fluid against one side of the plate is fh I w . (stripdepth) L(y) dy. . (4) F= J. A flat isoscelesright triangular plate with base 6 1t and height EXAMPLE 2 3 ft is submergedvertically, base up. 2 ft below the surface of a swimming pool. Find the fbrce exertedby the water againstone side of the plate.
• Chapter 5: Applicationsof Integrals 430 Solution We establish a coordinate system to work in by placing the origin at the plate's bottom vertex and running the ]-a-is upward along the plate's axis of symmetry (Fig.5.71). (We will look at other coordinatesystemsin Exercises3 and 4.) The surfaceof the pool lies along the line 1 : 5 and the plate's top edge along the line y : 3. The plate's right-hand edge lies along the line 1' : x, with the upper right vertex at (3, 3). The length of a thin sftip at level ) is L(1') = )'a : 2Y' .r (f0 The depth of the strip beneaththe surtaceis (5 - ;v). The force exertedby the water against one side of the plate is therefore To find the force on one sideof the Lrytcty r=[0,'f )'jl) plate in ExamPle we can 2, submerged d e p r h/ J. system the one here. like usea coordinate f3 : I 62.4(s Y)2!dY Jn f3 124.8 (5) - )') d) I y2+],: 1684.8 lb. 124.8; I LL How to Find Fluid Force Whatever coordinate system you use, you can find the fluid force against one side of a submergedvertical plate or wall by taking these steps. for the length and depth of a typical thin horizontal Find expressions l. strip. 2. Multiply their product by thefluid's weight-densityw and integrate over the interval of depths occupied by the plate or wall We can now calculate exactly the force exertedby the molasses EXAMPLE 3 againstthe bottom l-ft band of the Puritan Distilling Company's storagetank when the tank was full. The tank was a right circular cylindrical tank 90 ft high and 90 ft in diameter' Using a coordinate system with the origin at the bottom of the tank and the )-axis pointing up (Fig. 5.72), we find that the typical hodzontal strip at level ) has 90 - f, Strip depth: zx tank diameter:902. Strip length: The force againstlhe band is therelore tt I Force: / ru(depth)(length) = , 100(90 r)(902)dr dy OTTOSCALE Bottom level Jo Jo tank with the I The molasses fl : 90002I (90- r) dr - 2's30,553 lb origin at the bottom coordinate ( E x a m p l3 ) . e Jt)
• 5.9 FluidPressures Forces 43'l and Surfacelevel of fluid As expected, forceis slightlylessthantheconstant-depth the estimate following Example1. FluidForces and Centroids If we know the location of the centroid of a submergedflat vertical plate (Fig. 5.73), we can take a shortcut to find the force against one side of the plate. From Eq.(4), 7b F: I w> tsrripdepth),4yt(ly Lt J,, :, roivdepth) z(y)r/l x 5.73 The force againstone sideof the 1. p l a t ei s w . h . p l a t ea r e a . : u x (moment about surface level line of region occupied by plate) : u x (depth of plate's centroid) x (area of plate). Fluid Forces and Centroids The force of a fluid of weighrdensity w against one side of a submerged flat vertical plate is the product of 4 rhe distancet from the plare's centroid to the fluid sudace. and the Dlate'sarea: (s) F=whA. EXAMPLE 4 Use Eq. (5) to find the force in Example 2. Sorution The centroid of the triangle (Fig. 5.71) lies on the y-axis, one-third of the way from the baseto the verler,. h -- 3. The rriangle's .o areais 1(base)(height) 4: I :e 1(6)(3) Hence, F : wh .q : 62.4)(3)(9) : 1684.8Ib. D Equation (5) says that the fluid force on one side of a submergedflat veftical plate is the same as it would be if rhe plate's entire area lay ir units beneaththe surface.For many shapes,the location of the centuoidcan be found in a table, and Eq. (5) gives a practical way to find F: Of course,the centroid's location was found by someonewho performed an integration equivalent to evaluating the integral in Eq. (4). We recommendfor now that you practice your mathematicalmodeling by drawing pictures and thinking things through the way we did when we developed Eq. (4). Then check your results, when you conveniently can, with Eq. (5).
• 432 Chapter 5: Applicationsof Integrals Exercises 5.9 The weight-densitiesof the fluids in the following exercisescan be Surface level ,i lbund in the table on page 428. 1fr 1. What was the total fluid force against the cylindrical inside wall of the molassestank in Example I when the tank was full? half l fu11? 4fr 2, What was the total fluid lbrce against the bottom l-ft band of I the irside wall of the molassestank in Example I when the tank was half full? J 3. Calculate the fluid force on one side of the plate in Example 2 using the coordinate system shown here. 8. The plate in Exercise 7 is revolved 180' about line AB so lhat part of the plate sticks out of the lake, as shown here. What force rG0 does the water exert on one face of the plate now? 9, The vertical ends of a watering trough are isoscelestriangles like the one shown here (dimensionsin feet). 4. Calculate the fluid force on one side of the plate in Example 2 Find the flujd lorce againstthe ends when the hough is full. a) usinBthe coordilatesy'lem sho n here. g b) CALCULATORHow many inche. do you hae lo louer rhc water level in the trough to cut the fluid force on the ends J (IT] in half? (Answer to the nearesthalf inch.) c) Does it matter how long the trough is? Give reasons for yout answer. r (fi) ) (fi) The plate in Example 2 is lowered another 2 ft into the water. 5. ' (fr) What is the fluid force on one side of the plate now? The plate in Example 2 is raised to put its top edge at the surface of the pool. What is the fluid force on one side of the plate now? 10.The vertical ends of a watering trough are squares3 ft on a side. 7. The isoscelesffiangular plate shown here is submergedvertically a) Find the fluid force againstthe ends when the trough is full. I ft below the surface ol a freshwater lake. CALCULATOR How many inches do you have to lower the b) Find the fluid force against one face of the plate. a) water level in the trough to reduce the fluid lbrce by 25%? What would be the fluid force on one side of the plate if b) c) Does it matter how long the trough is? Give reasons for the water werc seawaterinstead of freshwater? your answer,
• Exercises 5.9 433 glasswiDdowin a typical The viewing poftion of the rectangular fish tank at the New EnglandAquarium in Boston is 63 in. wide andruns from 0.5 in. below the water's surfaceto 33.5 in. below Find the fluid force against portion ofthe window. this the surface. The weighFdensity seawater 64 lb/ft3.{b caseyou were is of wondering,the glassis 3/4 in. thick and the tank walls extend4 in. abovethe water to keep the ish from jumping out.) 12. A horizontalrectangularfteshwaterf,sh tank with base2 x 4 ft and height 2 ft (interior dimensions)is filled to within 2 in. of the top. a) Find the fluid force againsteachside and end of the tank. b) ffthe tank is sealedand stoodon end (without spilling), so E 19. CALCULATOR The end plates of the tough shown herc were that one of the squareends is the base,what does that do designed withstanda fluid force of 6667 lb. How many cubic to to the fluid forces on the rectangularsides? feet of watercanthe tank hold without exceeding limitation? this I rr. Round down to the nearestcubic foot. measures x 3.75 carton 3.75 CALCULATOR rectangular A milk in. at the baseand is 7.75 in. tall. Find the force of the milk on ) (ft) one side when the cartonis full. (-4, l0) Iu. 5.75 by 3.5 CALCULATOR standardolive oil can measures A in. at the baseand is 10 in. tall. Find the fluid force againstthe baseand eachside when the can is full 15. A semicircularplate 2 ft in diameter sticks straight down into t' fresh water with the diameteralong the surface.Find the folce exertedby the waier on one side of the plate. r (ft) horizontalright circular 16. A tanktruck haulsmilk in a 6-ft-diameter cylin&ical tant. How much force does the milk exert on each end of the tank when the tank is half fuU? 17. The cubical metal tank shown here has a parabolic gate, held swimmingpool shownhere 20. Wateris running into the rectangular in placeby bolts and designed withstanda fluid force of 160 to at the Iate of 1000ft3lh. lb without rupturilg. The liquid you plan to storehas a weight- a) Find the fluid force againstthe triangular drain plate after density of 50 1b/ff. t h of filling. a) What is the fluid force on the gate when the liquid is 2 ft b) The dmin plate is designed withstanda fluid force of 520 to deep? lb. How high can you fill the pool without exceedingthis b) What is the maximumheight to which the containercanbe limitatiol? filled without exceedingits desi\$ limitation? J (ft) 10fr , (f0 Triangulardlain plate Enlarged of view Parabolic Parabolicgate sate - ] (ft) gate Parabolic window tank shownherehasa I ft x I ft square The rectangular 1 ft abovethe base.The window is desimedto withstanda fluid force of 312 lb without cracking. a) What fluid force witl the window have to withstandif the tank is filled with water to a depthof 3 ft? b) To what level can the tank be filled wilh water without Enlargedview of drain plate exceedingthe window's designlimitation?
• Chapter 5: Applicationsof Integrals 434 Waterin Movable end A vertical rcctangular plate 4 units long by , units wide is sub- merged in a fluid of weight density n with its long edgesparallel to the fluid's sudace.Find the averagevalue of the pressurealong the vertical dimension of the plate. Explain your answer. (Continuation of Exercis?21.) Show that the force exertedby the fluid on one side of the plate is the averagevalue of the pressure Sideview (found in Exercise 21) times the area of the plate. warer pours inlo rhe tan-khere at lhe rale of 4 ftl/min. The lank's cross sectionsare 4-ft-diameter semicircles.One end of the tank is movable, but moving it to increasethe volume compressesa spring. The spdng constantis ,t - 100lb/ft. If the end ofthe tank moves 5 ft againstthe spdng, the water will drain out ol a safeiy hole in the bottom at the rate of 5 ft3/min. Will the movable end reach the hole before the tank overflows? Patternand Other The Basic ModelingApplications There is a pattern to what we did in the preceding sections.In each section we wanted to measure something that was modeled or described by one or more continuous functions. In Section 5.1 it was the area between the graphs of two continuousfunctions. In Section 5.2 it was the volume of a solid. In Section 5.8 it was the work done by a force whose magnitude was a continuousfunction, and so on. In each case we respondedby partitioning the interyal on which the function or functions were defined and approximating what we wanted to measure with Riemann sums over the interval. We used the integral defined by the limit of the Riemann sumsto define and calculatewhat we wanted to measure. Table 5.2 shows the pattem. Literally thousandsof things in biology, chemistry, economics, engineering. finance, geology, medicine, and other fields (the list would fill pages)are modeled and calculatedby exactly this process. This section reviews the processand looks at a few more of the integrals it leads to. Traveled Displacement Distance vs. If a body with position function s(t) movesalong a coordinateline without changing direction, we can calculate the total distance it travels from t :.t to t - b by inte\$ating its velocity functior u() from r : a to t - b, as we did in Chapter4. If the body changesdirection one or more times during the trip, we need to integrate the body's speed lu(t)l to find the total distance traveled. Integrating the velocity will give only the body's displacement, s(D)-s(a), the difference between its initial and final positions. To see why, partition the time interval a A t < b into subintervals in the usual way and let Atr denote the length of the lrh interval. If At1 is small enough, the body's velocity u(t) will not change much from /1 I to ft and the right-hand
• 5.10 The BasicPatternand Other Modeling Applications 435 Table 5.2 The phasesof developing an integral to calculate something Phase1 Phase2 Phase3 We partition [a, ,] into subinterr'alsof We describe or model something we want The approximations improve as the lengthA.rr andchoose point ct in each a norm of thg partition goes to zero. to measurein tems of one or more corr- subinterval. tinuous functions def,ned on a closed The Riemann sums apprcach a limit- interval [a, r]. We approximate what we wantto m€asure ing integral. with a finite sum. We use the integral to define and cal- We identifythe sum as a Riemann sum culate what we originally wanted to of a cortinuousfunctionover [d, ,]. measule. I [/(cr) - g(cr)]Arr The area between the curves ) = f(-I). B ( c) l A . r & r ,,lipI[ft.rl y = g(r) on [a, ,] when f(x) Z S(r) tfG - eG)) dx J. The volumeof the solid def,ned revolv- by zlR(cr)]'?A'r z[R(c*)]'?A-rr I lpot ing thecurve): R(-x), 5 r 5 r, about a ther'axis. ttlR(x))'dx J, r(cr)lr The orl done by a conlinuouslariable t a(c*)a.rk ]1poI force of magnitude F(I) directed along the r-axis from a to b l.1ro'
• Chapter Applications Integrals of 436 5: endpoint value u(t1) will give a good approximation of the velocity throughout the interval. Accordingly, the change in the body's position coordinate during the tth time interval will be about u(tk)Ltk. The change will be positive if rr(4) is positive and negativeif u(te) is negative. In either case,the distancetraveled during the ftth interval will be about ,u(r4)la/r. will be approximatell The tolal trip distance i. l u l t t ) (1) Ltt. - t=l The sumin Eq. (1) is a Riemann lu(t)l on theintervallri, b]. sumfor the speed We expect approximations improve thenom of thepartition [4, b] goes of as to the looks as if we shouldbe able to calculate total distance the to zero.It therefore from d to b. In practice, this naveled the body by integrating body'sspeed the by modelpredicts distance the tumsout to be the right thing to do. The mathematical correctlyeverytime. :t. r(t)ldt If we wish to predict how far up or down the line from its initial position a body will end up when a trip is over, we integrate u instead of its absolutevalue. To seewhy, let s (/) be the body's position at time t and let F be an antiderivative r. Then of s(t):F(r)+c for some constant C The displacementcausedby the trip from t : q to t : b ls (F (a)+ C) s(D) s(a) : (F(b) { c) F (b) - F (.a) = ['uftat. : : Displacement ,<l a, l.' The velocity of a body moving along a line from / : 0 to / = EXAMPLE1 hr f2 secwas u(t) = 5cost r/sec. fiaveledand the body'sdisplacement. Find the total distance
• 5.'10 The BasicPattern and Other Modeling Applications 437 Solution f3n /2 Distancehaveled : D i s t a n c ei t h e i n l e g f a lo i s p c c L l . 5costld/ r(r):5cost / tn /2 r3n/2 :l 5cosrdr / t-5corrdr Jo Jr/2 E 13a/2 T/2 -5sin/l :5sinrl 6 JO )nl2 : 5 ( l - 0 )- 5 ( - 1 - l ) : s + 1 0 : 1 5m '' Difhccmc t is ihc intcgrLrl : Displacement s a, o, lo' : r r*,]' : s(-1)- 5(o) -s m : s(0) +5 During the trip, the body traveled5 m forward and 10 m backward a total for tr distance 15 m. This displaced body 5 m to the left (Fig.5.74). s(t):5sinl+r(0) of the fll Delesse's Rule Fo As you may knoq the sugar in an apple starts tuming into starch as soon as the apple is picked, and the longer the apple sits around, the starchierit becomes.You can tell fresh apples from stale by both flavor and consistency. To find out how much starch is in a given apple, we can look at a thin slice under a microscope.The cross sectionsof the starch granuleswill show up clearly, 5.74 The velocity and displacement of and it is easy to estimate the proportion of the viewing area they occupy. This t h e b o d yi n E x a m p l e . 1 two-dimensional propofiion will be the same as the three-dimensionalproportion of uncut starchgranulesin the apple itself. The apparentlymagical equality of these propofiions was f,rst discoveredby a French geologist, Achille Emest Delesse,in the 1840s.Its explanationlies in the notion of averagevalue. Supposewe want to find the proportion of some granular material in a solid and that the sample we haye chosento analyze is a cube whose edgeshave length 575 The stepsleadingto Delesse's rule: l,. We picture the cube with an r-axis along one edge and imagine slicing the (a)a slicethrough a samplecube;(b) the cube with planes perpendicularto points of the interval [0, f] (Fig. 5.75). Call the (c) granularmaterialin the slice; the slab proportion of the area of the slice at x occupied by the granular matedal of interest betweenconsecutive slices determined bv a partitionof [0,L]. (starch,in our apple example) r(x) and assumer is a continuous function of x.
• chapter 5: Applicationsof Integrals 438 Now pafiition the interval 10,tl into subinteNals in the usual way lmagine the cube sliced into thin slices by planes at the subdivision points. The length At1 of the frth subinteNal is the distance between the planes at rt r and nk lf the planes are close enough together, the sections cut from the grains by the planes will resemble cylinders with basesin the plane at xi. The proportion of granular material between the planes will be about the same as the proportion of cylinder base area in the plane at -rr, which in tum will be about r(.:r1).Thus the amount ol granular material in the slab between the two planes will be about l P r o p o r l i o nY l s l r b v o l u m e ) = r l r / ) L 2 A r t . ) The amount of granular material in the entire sample cube will be about r ( x i ) L ' z A T1 . I This sum is a Riemannsum for the function r1r)Lr o'er the interval [0,4]. We expect the approximationsby sums like these to improve as the norm of the sub- division of [0, L] goes to zero and thercfore expect the integral t: r (x) L2 dx to give the amount of glanular material in the sample cube We can obtain the proportion of granular material in the sample by dividing this amount by the cube's volume, L3. If we have chosenour samplewell, this will also be the proportion of granular material in the solid from which the sample was taken. Putting it all together,we get Propoftion of granular Proportion of granular material in the sample cube material in solid rQ)L2dx r x )a x L- | rule I Delesse's 1 rL -- i J 'ato* Achille Ernest Delessewas a mid nineteenth L3 L3 century mining engineer interestedin : averagevalue of r(r) over [0, Z] determining the composition of rocks. To find out how much of a particular mineral a : propo ion of area occupied by granular rock contained,he cut it through, polished an material in a typical cross section. exposediace, and covered the tace with This is Delesse'srule. Once we have found -, the averageof r(,v) over [0, U' we trrnsparent waxed paper. trimmed to size. He then traced on the paper the exposedpoftions have found the proportions of granular material in the solid of the rnineral that interestedhim. Aftef In practice, i is found by averaging over a number of cross sections-There weighing the paper, he cut out the mineral are several things to watch out for in the procass. In addition to the possibility tracesand weighed them. The ratio of the samplesdifficult to find, that the granulescluster in ways that make representative weights gave not only the proportion of thc there is the possibility that we might not recognize a granule's tace for what it surfaceoccupied by the mineral but, more is. Some cross sectionsof normal red blood cells look like disks and ovals, while importanl, the proportion of the entire tock others look suryrisingly like durnbbells.We do not want to dismiss the dumbbells occupied the mineral.This rule is still by as experimentalerror the way one researchgroup did a few years ago used by petroleum geologists today. A two dimensional analogueof it is used to - determine the porosities of the ceramic lllters Integrals BadModels Useless l h , r le t r r c l o r g J n i c o l e c u l ein c h e m i . t r y m Some of the integrals we get from forming Riemann sums do what we want, but laboratoies rnd screenout micrcbes in water othen do not. It all dependson how we chooseto model the problems we want to puriliers. solve. Some choices are good: others are not. Here is an example.
• 5 . 1 0 T h e B a s i c a t t e r na n d O t h e r M o d e l i n gA p p l i c a t i o n s 4 3 9 P . A s o l i do f r e v o l u t i o n Modeled by revoh,ing lrn rpproprialc curvc whose surfacearea r.:J(_rl,d<r<r. we waDl to know Compare Cone frustlrln rpproximatioll (ltirJ) 'lhc suface afea ol Lhe The rrea of the gcometdc model is solid's surface ., f ,,i should be S. . s = 2 7 r l r .ltr+ i ; ] . i , : .! 5.76 The modelingcyclefor surface area. We use the sudace area fbrmula F(!J,' s= l.''z, 1r,t 2) valueand alwaysgivesresults consistent with inlormalion because haspredictive it tiom other sources.In other words, the model we used to derive the tbmrula (Fig. 5.76) was a good one. Why not find the surfhcearea by approximating with cylindrical bands instead of conicalbands,as suggestcd Fig. 5.77?The Riemannsums we get this way in convergejust as nicely as the ones basedon conical bands,and the resulting integml is simpler Insteadof Eq. (2), we get s:I' .l (x) dx. (3) 2tr Atier all, we migllt argue, we used cylinders to derive good volume tbrmulas. so why not use them again to derive sudace area fonnulas? The answeris that the lbrmula in Eq. (3) has no predictivevalue and alnost The comparison step in the with oLhercalculations. nevergives resultsconsistent modelingprocess fails tbr this formula. we end up with a nice looking integral There is a molal here: Jusl because does not mean it will do what we want. Constructingan integral is not enough-we haveto test it too (Exercises and 16). 15 The Theoremsof Pappus In the third century, an Alexandrian Greek named Pappusdiscoveredtrvo formulas that relate centroids to surfaces and solids of revolutiitn. The fomrulas provide lengthycalculations. shortcuts a numberof otherwise to Theorem 'l Theorem Volumes for Pappus's If a plane region is revolved once about a line in the plane that does nol cut through the rcgion's interior, then the volume of the solid it generatesis equal to the region's areatimes the distancetmveled by the regiot]'s centroid during the revolution. If p is the distanceliom the axis of revolution to the (b) centrcid, then 5 r: Why not use(a) cylindrical bands (4) l/:2rpA. instead (b) conicalbandsto of area? surface approximate
• 440 Chapter 5: Applicationsof Integrals Proof We draw the axis of revolution as the r-axis with the region R in the fir:r quadrant (Fig. 5.78). We let ,()) denote the length of the cross section of R perpendicularto the }-axis at )'. We assumel(y) to be continuous. By the method of cylindrical shells, the volume of the solid generatedbr revolvingthe regionaboutthe x-ais is ld ld t : 2r (shellradius)(shell height)dy :2r y L(y) dy. (5) | I J, J, The y-coordinare R s centroidis , of fdP I yL(y)dy I idA J- J,' - 'AA 578 The regionR is to be revolved so that (once) about the x-axis generate to a t, solid.A 1700-year-oldtheoremsays that yL(y)dy:Ay. the solid's volumecan be calculated by multiplying the region'sarea by the distancetraveledby its centroidduring Substituting for the last integralin Eq. (5) gives y:2ryA.With At p equalto the revolution. we have V :2trpA. tr t, EXAMPLE 2 The volume of the torus (doughnut) generatedby revolving a circular disk of radius a about an axis in its plane at a distance D Zt,? from its center (Fig. 5.79) is v : 2ir (b)(r q2) :21r2ba2. -.t 5.79 With Pappuslfirst theorem,we can Distance from axis of find the volumeof a torus without revolution to centroid havingto integrate(Example 2). EXAMPLE 3 Locate the centroid of a semicircularregion. So/ution We model the region as the region betweenthe sernicircley : !/A=7 (Fig. 5.80) and the x-axis and imagine revolving the region about the x-axis ro generatea solid sphere.By symmety, the x-coordinate of the centroid is i:0. 5.80 With Pappus's theorem,we can first With t : p in Eq. (4), we have locatethe centroidof a semicircular (4/3)ra3 V 4 regionwithout havingto integrate '! : - (Example 3). LI 2nA 2n(1/21na2 3n-
• Exercises 5.10 441 Theorem 2 Pappus's Theorem Surfac€ for Areas If an arc of a smooth plane curve is revolved once about a line in the plane that does not cut through the arc's interior, then the area of the sur{ace generated by the arc equals the length of the arc times the distance traveled by the arc's centoid during the revolution.If p is the distancefrom the axrs of revolution to the centroid, then (6) S:2ttpL. The proof we give assumes that we can model the anis of revolution as the r-axis and the arc as the graph of a smooth function of r. Proof V'/edmw the axis of revolution as the r-axis with the arc extending from x = a to x = b in the fint quadrant(Fig. 5.81). The area of the surfacegenerated by the arc is ':1.',:=r l.':' (7) 2try d s l ds. The )-coordinate of the arc's cenboid is x=h ia' J , o' I I ,:,, r_ L r,=b ldt Figurefor Pappus's areatheorem. JF. Hence f'- I Yds:YL. J, -- Substituting for the last inte\$al in Eq. (7) givesS : 2rjL. With p equalto tl, 1', we haveS : 2ttp L. EXAMPLE4 The surface area of the torus in Example 2 S:2r(b)(2ra):4trzba. 5.10 Exercises and Displacement Distance 1. u ( r ) : 5 c o s r , 0 = t = 2 n o:r:2 ln Exercises1 8, the iunction u(1) is the velocity in metersper second = 1_'la) srn t, T .,i a body moving along a coordinateline. (a) Graph lr to seewhere it u(r):6sin3r, O=t=n/2 :. positiveand negative.Then find (b) the total distancetraveledby the 4. , ( / ) = 4 c o s 2 t . rdy during the given time interval and (c) the body's displacement. !=t =1t
• M2 Chapter Applications Integrals 5: of v(t)-49 9.8t, 0:l:10 5- Time Velocity Time Velocity 6. u ( r ) : 8 - 1 . 6 / , 0 : 1 5 1 0 (ir./sec) (in-/sec) (sec, (sec) 1 v(t) : 6t2 - 18t+ 12: 6(t - l(t - 2), 0 s t 3 2 -t l 0 0 6 8. u(t) : 6t2- 18t+ 12: 6(t - I)(t - 2), 0 = t 33 '7 -6 t2 I 9, The function.r : (l /3)t3 - 3t2+ 8t gives position a body the of 22 2 8 2 moving on the horizontal .r-axisat time r > 0 (s in meters,r in l0 3 9 6 seconds). 4 10 0 ,13 Show that the body is moving to the dght at time I = 0. 5 a) b) When doesthe body move to the left? What is the body'spositionat time I = 3? c) Whent : 3, whatis thetotaldistance bodyhastraveled? d) the DelessetRule tt e) GRAPHER Graph.r asa functionof t andcomment the on na 13. The photograph here showsa grid superimposed the polished on relationshipof the graph to the body's motion. face of a piece of granite. Use the grid and Delesse'srule to 10. The functions : -t3 + 6t2 - 9t givesthe positionof a body estimatethe proportion of shrimp-coloredgnnular material in moving on the horizontal .r-axisat time , > 0 (.t in meters,t in the rcck. seconds). a) Show that the body is moving to the left at r = 0. b) When doesthe body move to the right? c) Does the body ever move to the right of the origin? Give reasons your answer. fot d) What is the body'spositionat time r :3? e) What is the total distancethe particle has taveled by the time t : 3? aa ra f) GRAPHER Gruph.ras a functionof t andcomment the on relationshipof the graph to the body's motion. Herearethe velocity graphsof two bodiesmovingon a coordinate line. Find the total distancehaveledand the body's displacement 14, The photograph hereshowsa grid superimposed a micrcscopic on for the given time interval. view of a stainedsectionof humanlung tissue.The clear spaces (called between cellsarecross the sections ofthe lung'sair sacks syllable)-Usethe grid andDelesse's ah)eoli,accett on the second rule to estimatethe proportion of air spacein the lung. v (m/sec) v (m/sec) t (sec) ModelingSurface Area 15. Modeling sLirfacearea. The lateralsurfaceareaof the cone w 12. CALCULATORThe tableat the top of the next column showsthe swepr by revolving linesegment : xlJl.O 1r = J3. out lhe ) aboutthe1-axisshouldbe ( U2xbasecircumferencexslant velociry of a modeltrain enginemoving back andfofth on a hack height) : (112)(2tr)(2): 22. What do you get if you useEq. (3) with for 10 sec.UseSimpson's to find the resulting rule displacement andtotal distance traveled. f(*):,/Jil
• 5.10 M3 Exercises 64 lb/Itr . How many pounds of water does P ipedream displace'! r':*,0<r<rE (Displacementis given in pounds for small craft, and long tons 1:l u long ton : 2240 lbl for larger vessels..i (18, I ) Elementsof yacht DesiStr,Francis S- Kinney' (Data from Sftene's Inc., 1962) Mead & ComPanY, Dodd, (Cotltinuatio'1 of Exercise /7/ A boat's 18. pdsmatic coefficient is the ratio of the displacementvolume to the volume of a prism whose height equals the boat's waterline length and whose base equals the area of the boat's largest sub merged cross section. The best sailboats have prismatic coefli- cients between0.51 and0 54.Find Pipedredn's prismatic coefll- The onlY surface for which Eq (3) 16. cient, given a waterline length of 25.4 ft and a largestsubmerged gives the areawe want is a cylinder. Show that Eq (3) gives S : cross section area of 16.14 ft2 (at Station 6). 2rrh fot the cylinder swept out by revolving the line segment y : r,0 1 ,r < i, about the .ir-axis. TheTheorems PaPPus of 4, To find the volume of water dis- (2,4) 19. The square region with vertices (0, 2), (2' 0)' (1' 2). and placed by a sailboat, the common practice is to partjtion the axis to generatea solid Find the volume is revolved about the,r waterline into l0 subinteNals of equal length' neasurc the cross and sutface area of the solid section arca A(r) of the submergedpoftion of the hull at each 20, Use a theorem of Pappus to lind the volume generatedby re- parlition point, and then use Simpson's rule to estimatethe inte- volving about the line .I :5 the tdangulal region bounded by gral of A(,r) from one end of the waterljne to the other' The table the coordinate axes and the line 2ir + ) :6 (As you saw in here lists the area measurements stations 0 through 10' as at 5.7, the centroid of a ffiangle lies at the Exercise 31 of Section the partition points are called, for the cruisrng sloop Pipedream' intenection of the medians, one-third of the way from the mid- shown here. The common subinterval length (distance between point of each side toward the opposite vertex .) consecutivestations) is i :2 54 ft (about 2' 6 1/2, chosenfor 21. Find the volume of the torus generatedby revolving the circle the convenienceof the builder). 1 x- 2 . t ' - 1 : : I c b o u l h e r a i : t 22. Use the theoremsof Pappusto find the lateral surface area and the volumeof c riShtcircularcone 23. Use the secondfheorem of Pappusand the fact that the suface area of a sphere of radius a is 4t tr2 to find the centroid of the semicircle 1' - './eF =. - 24. As found in Exercise 23, the centroid ol dre semicircle y ;ri lies at the point (0,2d/1). Find the alea of the surface ^,62 swept out by revolving the semicircle about the line I - a' 25. The area of the region R enclosed by the semiellipse -v: anl ttte;;-axis is ( I /2) z a ll and the volume of the Estinate Pipedreon's displacementvolume to the nearest (bI n@-'V a) ) generated revolving R about the.lr-axisis ('1/3)z aDl' by eliipsoid cubic foot. the cenffoid ofR. Notice the remarkablefact tbat Lhelocation Find is independentof d Submerged area (ft2) Station 26. As found in Example 3, the centroid ot' the region enclosed lies at the point by the i-axis and the semicircle .,-- 'JE = 0 (0,4al3r). Find the volume of the solid generatedby revolving 1.0? I this regionaboul the lrne Y = ,-1 3.84 2 '7.82 3 :i d 27. The region of Exercise 26 is revolved abouf the line ) t2.20 4 of the solid' to generatea solid. Find the volume 15.18 5 - 28. As found in Exercise 23. the centroid of the semicircle I 16.1,1 6 - x'1 lies at the point (0, 2a/t). Find the areaof the surface ,[2 14.00 7 - a' generatedby revolving the semicircle about the line t =:t 9.21 8 29. Find the moment about the i-axis of the semicilcular region 1n 3.21 Example 3. If you use resuks already known' you will not need 0 l0 to integrate. which weighs The figures in the table are fbr seawater, b)
• 4M Chapter 5: Applicationsof Integrals ro YouR Quesrroxs Guroe Rpvrew 1. How do you deflneandcalculate areaof the regionbetween the 9, How do you locatethe centerof massof a thin llat plate of the graphs two continuous of functions? Give an example. material? Give an example. 2. How do you defineand calculate volumes solidsby the the 10. How do you define and calculateth€ work done by a variable of methodof slicing? Give an example alonga portionofthe r-axis?How do you calculate forcedirected the work it takesto pumpa liquid from a tank?Giveexamples. 3, How are the disk and washermethods calculating for volumes derivedfrom the methodof slicing?Give examples volume 11. How do you calculate force exerted a liquid againsta of the by calculations these by methods. pofiion of a verticalwall? Give an example. 4. Describe method cylindricalshells. the ol you Give an example. 12. Suppose know the velocityfunctionu(r) of a body thatwill bemovingbackandforth alonga coordinate from time , : a line 5. How do you defineand calculate lengthof the graphof a the to time /: r. How can you predicthow much the motionwill smoothfunctionover a closed interval? Give an €xample. What shift the body'sposition? How canyou predictthe totaldistance aboutfunctions that do not have continuousfirst derivatives? the body will travel? 6. How do you define and calculatethe area of the surfaceswept 13. What doesDelesse's say?Give an example. rule out by revolving graphof a smoothfunction) : /(.r), d : the r : b, aboutthejr-axis?Give an example. 14. Whatdo Pappus's theorems two say?Giveexamples how they of are used to calculate surface areasand volumes and to locate 7. What is a centerof mass? centroids. 8. How do you locate center mass a straight, the of of narow rod or 15. Thereis a basicpatternto the way we constuctedintegrals in strip of material?Give an example.If the density of the matedal this chapter. What is it? Give examples. is constant, cantell right awaywherethe center massis. you of Where is it? CneprBR PRecrrcBExERCTSES Areas 4. x3+u\$= l, :r:0, for 0S.t:1 ):0, Find the areas of the regions enclosed by the curves and lines in ) Exercises 1-12. ) : l/r2, 1, ):x, x:2 2.y: x, y:r/tr. x=z 3. Ji+.,/r- 1, r:0, y:Q y ] 5. jr -l)_, ,r:u. ):J 6. x:4-y2, t:0 7, y2:4x. y=4x-2 8. ,y2 4.r*4, = y:4x-16 9. ) : sinr, y:x, 0=a:n14
• Exercises 445 Practice y2 :4x and the line ,r = I in thc ,l-)-plane Each cross section -n12:x=ft/2 sin-rl, r:1. 10.l: perpendicularto the-r-axisis an equilateraltriangle with one edge 0=iJz 11.):2sintc, ):sin2r, in the plane. (The triangles all lie on the same side of the plane.) I t 1 3< x = 7 t / 3 sec2x. 12.y:8cos,r, ): 25, Find the volume of the solid generatedby revolving the region bounded by the r-axis, the cutve r':3:t4, and the lines i : I 13, Find the area of the triangular region bounded on the letl by r + y - 2.on the right by ) = rr, and above by ) - 2. and r - - I about (a) the 'I-axis; (b) the )-axis: (c) the line i - 1; (d) the line ) = 3. 14. Find the area of the triangulai' region bounded on the left by andbelow y 1: I b 26. Find the volume of the solid generatedby revolving the tri- y-f,onthcrightby):6-n, angular region bounded by the cuNe -! : '1/x3 and the lines 15. Find the exheme values of /(r) : .rl 3lv2and nnd the areaoI r - 1 and ) - 1/2 about (a) the:r-axis;(b) the ] axis; (c) the the region enclosedby the graph of/ and the ,t-axis line 'I : 2; (d) the line 1 - 4. 16, Find the area of the region cut from the first quadrant by the 27, Fjnd the volume of the solid generatedby revolving the region ctyre xt /2 + ))t/2 - at /2. bounded on the left by the parabolar : )2 + I and on the right 17. Find the total area of the region enclosedby the curve r : 12/l by the line r : 5 about (a) the i axis; (b) the )'-axis: (c) the line l andthelinest=]and): sinr 18, Find the total area of the region between the curves ): 28. Find the volume of the solid generatedby revolving the region a n dJ : c o s t f o r 0 = t < 3 r 1 2 . bounded by the parabola )2 :4x and the line lr - t about (a) the r-axis; (b) the ]-axis; (c) the line li :4: (d) the line -1:'1. Volumes 29, Find the volume of the solid generatedby revolving the trian- Find the volumes of the solids in Exercises 19-24. gular region bounded by the,r-axis, the line jr :1/3, and the 19. The solid lies betweenplanesperpendicularto the jr-axis at t : 0 curve ): tan ir in the first quadrantabout tle -t-axis perpendicularto the jl-axis between and.r : l. The cross sections 30. Find the volume of the solid generatedby revolving the region these planes are circular disks whose diameters run ftom the boLrnded the curve ) - sin:r and the lines -r - 0, 'r : rr, and by parabolay = r2 to the parabola ) : .r'/; : 2 a b o u t h e l i n el : 2 . I 20, The base of the solid is the region in the first quadrant between 31. Find tbe volume of the solid generatedby revolving the region the line 1 : x and the parabola ) - 2./4. The cross sections 2l about (a) the r- between the x-axis and the curve i :,!2 of the solid perpendicularto the ).-axis are equilateral t angles (b) the line ] - -l; (c) the line jr - 2; (d) the line )' = 2 axis; whose basesstretch from the line to the curve. 32, Find the volume of the solid generatedby revolving about the 21, The solid lies between planes perpendicular to the t-axis at .r-axis regionbounded r-.= 2tanr, -r':0,x: r /4, ancl by the a : r 14 andx: 5ir/4. The cross sectionsbetweenthesephnes x : n 14. (The region lies in the first and third quadnnts and are circular disks whose diametelsrun lrom the curve ) : 2cosr resemblesa skewedbow tie.) t o t h e c u r v e) : 2 s i n n 33. A round hole of radius .'6 ft is bored through the center of a 22. The solid lies betweenplanesperpendicularto the,r axis at ,t : 0 solid sphereof radius 2 ft. Find the volume of material removed and r :6. The cross sectionsbetween theseplanes :rre squares trom the sphere. whosebasesrun from the i-axis up to the cuNe xt /2 + !1/2 : J6. The profile of a football resemblesthc ellipse ffi:a. carcufnfOn shown here. Find the football's volume to the nearestcubic inch. The solid lies betweenplanesperpendicularto the ir-axis at r. - 0 Lengths Curves of The cross sectionsot' the solid perpendicularto the and,r:4. Find the lengths of the curves in Exercises35 38. ir-a,is between these planes arc circular disks whose diameters 3 5 , y : 1 r / :- ( l l 3 ) x 3 / 2 , l=x=1 run from the curve x2 :4) to the curve -12:'h. 1:):8 36, r:)':/r, 24, The base of the solid is the region bounded by the parabola