This document provides information about multiple integrals and examples of evaluating them. It begins by defining multiple integrals and iterated integrals. It then gives 4 examples of evaluating double and triple integrals over different regions. These regions include rectangles, triangles, and solids. The document also discusses Fubini's theorem, which allows reversing the order of integration in certain cases. It concludes by providing an example of converting an integral from Cartesian to polar coordinates.

1. MULTIPLE INTEGRALS
rahimahj@ump.edu.my

2.  
xddyyxfdxdyyxf
dydxyxfdxdyyxf
dycbxayxR
x,yf
b
a
d
c
b
a
d
c
d
c
b
a
d
c
b
a
  
  















),(),(
),(),(
then
,:),(
regionrrectangulaaincontinuousis)(If
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Iterated Integral

3.  


1
0
2
1
1
1
2
2
30(ii)
)23((i)
x
x
ydydx
dydxxyx
Example1
Evaluate the iterated integrals.
rahimahj@ump.edu.my

4.  
 
 
14
)1(2)2(226
])1()1(3[])1()1(3[
3)23(
obtainweintegrals,iteratedofdefinitiontheUsing(i)
332
1
3
2
1
2
2
1
2222
2
1
1
1
22
2
1
1
1
2






 



xdxx
dxxxxx
dxxyyxdydxxyx
y
y
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Solution :

5. rahimahj@ump.edu.my
 
 
2
3535
)1515(
1530
obtainweintegrals,iteratedofdefinitiontheUsing(ii)
1
0
53
1
0
42
1
0
2
1
0
2
2





 


xx
dxxx
dxyydydx
xy
xy
x
x

6. rahimahj@ump.edu.my
theorem.sFubini’–integralsiteratedanas
calculatedbecanfunctioncontinuousanyofintegrals
doublethe1943),-(1879FubiniGaudiotoAccording
 dycbxayxR  ,:),(
ifregionrrectangulaaontheoremsFubini’
  
b
a
d
c
d
c
b
a
dydxyxfdxdyyxfdAyxf ),(),(),(
R
then
Fubini’s Theorem

7. rahimahj@ump.edu.my
  
b
a
y
yR
dydxyxfdAyxf
2
1
),(),(   
d
c
x
xR
dxdyyxfdAyxf
2
1
),(),(

8.  
(-2,1)and(3,1)(0,0),erticesv
h theregion witrtriangulaclosedtheis;),((iii)
sinand0
,,0boundedregiontheis;),((ii)
20,2:),(;4),((i)
),(Evaluate
2
2
Rxyyxf
xyy
xxRyyxf
yyxyyxRyxyxf
dAyxf
R






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Example2

9. rahimahj@ump.edu.my
Solution :

10.  
5
36
5
2
4
2
)26(
]})(2[]2)2(2{[
2)4()4(
2
0
54
3
2
0
432
2
0
32222
2
0
22
2
0
2
2
2












 






yy
y
dyyyy
dyyyyy
dyxyxdxdyyxdAyx
yx
yx
y
y
yx
yxR
rahimahj@ump.edu.my

11. rahimahj@ump.edu.my
asdillustrateisRregionThe(ii)
Solution :

12. rahimahj@ump.edu.my
42
2sin
4
1
)2cos1(
4
1
2
sin
2
0
0
0
2
0
sin
0
2
0
sin
0























 




x
x
dxx
dx
x
dx
y
dydxydAy
xx
x
xy
yR

13. rahimahj@ump.edu.my
asdillustrateisRregionThe(iii)
Solution :

14. rahimahj@ump.edu.my
2
1
2
2
5
)49(
2
2
1
0
5
1
0
41
0
22
2
1
0
3
2
221
0
3
2
22
















 






y
dy
y
dyyy
y
dy
yx
dxdyxydAxy
yx
yx
y
y
yx
yxR

15. asdescribedissolidThe
.0and4,9
byboundedsolidtheofvolumetheFind
22
 zzyyx
922
 yx
yz  4
0z
R
rahimahj@ump.edu.my
Example 3

16. rahimahj@ump.edu.my
:RregiontheissolidtheofbaseThe
Solution :

17. 

 

















3
3
2
3
3
9
9
2
3
3
9
9
98
2
4
)4()4(
bygivenisvolumetheThus,
2
2
2
2
dxx
dx
y
y
dydxydAyV
V
xy
xy
x
x
xy
xyR
rahimahj@ump.edu.my

18. rahimahj@ump.edu.my










36
22
)sin(
22
sin
36
2
2sin
36
)12(cos36cos72
)cos3()sin1(9898
2/
2/
2/
2/
2/
2/
2
2/
2/
2
3
3
2

































t
t
dtttdt
tdttdxxV
Hence.cos3,sin3letNow tdtdxtx 

19. rahimahj@ump.edu.my
Example 4
 
2
0
1
20
2
(ii)
sin
(i)
Evaluate
y
x
x
dxdyedydx
y
y
 
asdillustrateis
RregionThe(i)
Solution :
Reversing The Order of
Integration

20. rahimahj@ump.edu.my
.integratedbecannot
sin
But  y
y
:nintegratiooforderthereverse,So dydx dxdy
:becomeregiontheThen,

21.  
 
21cos
cossin
)0(
sin
sin
sinsin
(i)
0
0
0
0
0
0 00








  



















 
yydy
dyy
y
y
dyx
y
y
dxdy
y
y
dydx
y
y
y
y
y
y
yx
x
y
y
yx
x
x
x
y
xy
rahimahj@ump.edu.my

22. rahimahj@ump.edu.my
asdillustrateisRregionThe(ii)
Solution :

23. rahimahj@ump.edu.my
.integratedbecannotBut
2
 dxex
:nintegratiooforderthereverseSo, dydx dxdy
:becomeregiontheThen,

24. rahimahj@ump.edu.my
 
  1
2
1
0
1
0
1
0
1
0
2
0
1
0
2
0
2
0
1
2/
2
2
22







  
















eedue
dxxe
dxye
dydxedxdye
u
u
u
u
x
x
x
x
x
xy
y
x
x
x
xy
y
x
y
y
x
yx
x

25. scoordinatepolarin the
)(ofintegraltheevaluatemaythen we)(
to)(functionaconvertcanthat weSuppose
the
inevaluateeasier toisitshape,circularinvolvingWhen
 r,fr,fz
x,yz

s.coordinatepolar
rahimahj@ump.edu.my

26. scoordinatePolar
r
θ
x
y
),( rP
cosrx 
sinry 
22
yxr 






 
x
y1
tan
 r0
 20 
rahimahj@ump.edu.my

27. R
dA
dV
A
V


R
dArfV
dArfdV
),(
),(


)( r,fz 
)( r,fz 

29. 
 



1
0 0
22
2
2
0
4
0
22
)((ii)
)cos((i)
scoordinatepolarto
changingbyintegralsfollowingtheEvaluate
2
y
x
dxdy
yx
y
dydxyx
rahimahj@ump.edu.my
Example 5

30. asdescribedisRnintegratioofregionThe(i)
rahimahj@ump.edu.my
Solution :

31. 0r
2r
0
2/ 
 
4
4sin
4sinsin
)(cos
)(cos
)cos(
Therefore
2/
0
2
1
2/
0
4
02
1
2/
0
4
0
2
1
2/
0
2
0
2
2
0
4
0
22
2













 
 
 

ddu
dudu
rdrdr
dydxyx
x
rahimahj@ump.edu.my

32. Solution :
asdescribedisRnintegratioofregionThe(ii)

33. 8422
1
2
1
sin
2
csc
sin
2
sin
)sin(
Therefore
2/
4/
2/
4/
2
2
2/
4/
2
csc
0
22/
4/
csc
0
2
2/
4/
csc
0
2
2
1
0 0
22
2















































 
 



dd
d
r
drdr
rdrd
r
r
dxdy
yx
y
r
r
y
rahimahj@ump.edu.my

34. asdescribedisbaseitsandsolidThe
3.ExamplesolvetoscoordinatepolartheUse
R
3r
R
y
x
3 3
rahimahj@ump.edu.my
Example 6

35.   








36cos918)sin918(
sin
3
2
)sin4()sin4(
)sin4()4(
bygivenisVvolumetheThus,
2
0
2
0
2
0
3
0
3
2
2
0
3
0
2
2
0
3
0












 







d
d
r
r
drdrrrdrdr
dArdAyV
r
r
r
r
RR
rahimahj@ump.edu.my

36. rahimahj@ump.edu.my

37. A lamina is a flat sheet (or plate) that is so thin as to
be considered two-dimensional.
Suppose the lamina occupies a region D of the xy-
plane and its density (in units of mass per area) at a
point (x, y) in D is given by ρ(x, y), where ρ is a
continuous function on D. This means that
A
m
yx


 lim),(
where Δm and ΔA are the mass and area of a small
rectangle that contains (x, y) and the limit is taken as
the dimensions of the rectangle approach 0.
Laminas & Density

38. Definition mass of a planar lamina
of variable density

39. rahimahj@ump.edu.my
   
   
 
1 1
0 0
11
2
00
A triangular lamina with vertices 0,0 , 0,1
and 1,0 has density function , .
Find its total mass.
Solution :
,
1 1
...
2 24
x
R
x
x y xy
m x y dA xy dydx
m xy dx unit of mass


 
 

 
 
    
  

Example 7

40. rahimahj@ump.edu.my
The moment of a point about an axis is the product of
its mass and its distance from the axis.
To find the moments of a lamina about the x- and y-
axes, we partition D into small rectangles and assume
the entire mass of each subrectangle is concentrated
at an interior point. Then the moment of Rij about the
x-axis is given by
and the moment of Rk about the y-axis is given by
  ****
),())(mass( ijijijij yAyxy  
  ****
),())(mass( ijijijij xAyxx  
Moment

41. rahimahj@ump.edu.my
  


m
i
n
j D
ijijij
nm
x dAyxyAyxyM
1 1
***
,
),(),(lim 
The moment about the x-axis of the entire
lamina is
The moment about the y-axis of the entire
lamina is
  


m
i
n
j D
ijijij
nm
y dAyxxAyxxM
1 1
***
,
),(),(lim 

42. rahimahj@ump.edu.my
Center of Mass
The center of mass of a lamina is the “balance point.”
That is, the place where you could balance the lamina
on a “pencil point.” The coordinates (x, y) of the center
of mass of a lamina occupying the region D and having
density function ρ(x, y) is
where the mass m is given by
 
D
x
D
y
dAyxy
mm
M
ydAyxx
mm
M
x ),(
1
),(
1


D
dAyxm ),(

43. Moments and Center of Mass of A
Variable Density Planar Lamina

45.      2
Find the mass and center of mass of the lamina
that occupies the region and has the given
density of function .
a) , 0 2, 1 1 ; ,
4 4
: , ,0
3 3
) is bounded by , 0, 0, and 1;x
D
D x y x y x y xy
Ans
b D y e y x x
x



      
 
 
 
   
 
 
 
 
 
32
2
2 2
,
4 11 1
: 1 , ,
4 2 1 9 1
y y
ee
Ans e
e e

 
 
   
Example 8

46. rahimahj@ump.edu.my
Example 9
Find the surface area of the portion of the surface
that lies above the rectangle R in the xy-
plane whose coordinates satisfy
2
4 xz 
40and10  yx

47. rahimahj@ump.edu.my
Example 10
Find the surface area of the portion of the paraboloid
below the plane
22
yxz  1z

48. GregionclosedD-3aon
)(variablesthreeoffunctionaofnintegratio
anisscoordinateCartesianinintegralsTriple
x,y,zf
dzdAdzdydxdV 

49. rahimahj@ump.edu.my

50. rahimahj@ump.edu.my
Example 11
Gregiontheofvolume(ii)The
6),,(wheredV),,((i)
evaluateplane,-andplane-,1
byboundedoctantfirstin theregiontheisGIf
G
2
zzyxfzyxf
yzxyyz,xy




51. belowshownas
isplane-xyon theRprojectionitsandGregionThe
rahimahj@ump.edu.my
Solution:

52. obtainweThus.1and
10byboundedRregionaisplane-
on theGofprojectionThe1)(zand
0)(6z,)(havewecaseIn this
2
2
1




yxy
,, xxxy
– y.x,y
x,yzx,y,zf
 


 















R
R
yz
z
R
yz
zGG
dAy
dAz
dAzdzzdVdVzyxf
2
1
0
2
1
0
)1(3
3
66),,(
rahimahj@ump.edu.my
(i)

53.  
35
16
75
3
)331()1(
)1(
)1(3
1
0
7
53
1
0
642
1
0
32
1
0
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2
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

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



x
xxx
dxxxxdxx
dxy
dydxy
x
x
y
xy
x
x
y
xy
rahimahj@ump.edu.my

54. ♣
  



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





 RR
yz
zG
dAydAdzdVV )1(
1
0
15
4
103222
1
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)1(
1
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xxx
dx
x
x
dx
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x
x
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rahimahj@ump.edu.my
(ii)

55. )( rdrddzdzdAdV 
x
z
y
dAr
dr
dz
dƟ
rdƟ
rahimahj@ump.edu.my

56. rahimahj@ump.edu.my
sCoordinatePolarlCylindrica
cosrx 
sinry 
22
yxr 






 
x
y1
tan
 r0
 20 
 z

57. .
areplane-
on theRprojectionitsandGregiontheofsolidThe
.0and,25,9bybounded
solidtheofvolumethefindtoscoordinatelcylindricaeUs
2222
xy
zyxzyx 
922
 yx
22
25 yxz 
0z
Example 12

58.  
3
122
3
61
3
)25(
)25(
2
1
25
25
Thus,0.zhavealsoWe.25z
obtainwe,relationUsing
2
0
2
0
9
0
2/3
2
0
9
0
2/1
2
0
3
0
2
2
25
0
22
222
2
1



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
dd
u
dudurdrdr
dArdAdzdVV
-r
ryx
r
r
RR
rz
zG
rahimahj@ump.edu.my
Solution:

59. rahimahj@ump.edu.my
Example 13
areplane-xyon theRprojectionitsandGregiontheofsolidThe
9.zplanetheandparaboloidby the
boundedsolidtheofvolumethefindtoscoordinatelcylindricaUse
22
 yxz

60. 2
81
4
81
42
9
)9(
)9(
)9(
isvolumerequiredtheThus
2
0
2
0
3
0
422
0
3
0
3
2
0
3
0
2
2
9
2








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 






d
d
rr
drdrr
rdrdr
dArdAdzdVV
r
r
RR
z
rzG
rahimahj@ump.edu.my
Solution:

61. ρ
dρ
Ɵ
Ɵ
dƟ
ϕ
dϕ
dρ
x
y
z
rahimahj@ump.edu.my
 dsin
d
 dsin

62. rahimahj@ump.edu.my
sCoordinatePolarSpherical

63. 

ddd
ddddV
sin
)sin)()((
2


rahimahj@ump.edu.my

64. belowdillustrateasGregiontheofsolidThe
.3spheretheinsideand
3
1
conetheaboveliesthatsolidtheofvolumetheFind




Example 14

65.  














9
2
9
)1cos(9cos9
sin9sin
3
sin
isvolumerequiredtheThus
2
0
2
0
3
2
0
0
2
0 0
2
0 0
3
0
3
2
0 0
3
0
2
3
33
3














d
dd
dddd
ddddVV
G
rahimahj@ump.edu.my
Solution:

66. rahimahj@ump.edu.my
“In order to succeed, your desire for success
should be greater than your fear of failure. ”
rahimahj@ump.edu.my