Chapter 5 Thomas 9 Ed F 2008


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Chapter 5 Thomas 9 Ed F 2008

  1. 1. 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
  2. 2. 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
  3. 3. 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@)
  4. 4. 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. 5. 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.
  6. 6. 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 -
  7. 7. 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.
  8. 8. 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
  9. 9. 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
  10. 10. 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
  11. 11. 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.
  12. 12. 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
  13. 13. 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.
  14. 14. 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.
  15. 15. 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.
  16. 16. 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. -
  17. 17. 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
  18. 18. 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:
  19. 19. 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.
  20. 20. 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.