This document provides examples and explanations for evaluating double and triple integrals using Cartesian and polar coordinates. It begins by introducing double and triple integrals and their notation. It then discusses the evaluation of double and triple integrals, including the process of integrating inner integrals first and noting that integral limits should proceed from variable to constant. Several examples are worked through to demonstrate evaluating double and triple integrals over different regions of integration. The document also covers changing the order of integration and evaluating area integrals using double integrals in both Cartesian and polar coordinate systems.

1. SMT1105
ENGINEERING
MATHS- II
UNIT-I
MULTIPLE INTEGRALS

2. INTRODUTION
 When a function ( ) is integrated with respect to x between
the limits a and b, we get the double integral .
 If the integrand is a function , and if it is integrated
with respect to x and y repeatedly between the limits 0 and
1 (for x ) and between the limits 0 and 1 (for y ) we get a
double integral that is denoted by the symbol
,
1
0
1
0
.
 Extending the concept of double integral one step further,
we get the triple integral, denoted by
, ,
1
0
1
0
1
0
.

3. EVALUATION OF DOUBLE AND TRIPLE
INTEGRALS
 To evaluate ,
1
0
1
0
first integrate , with
respect to x partially, treating y as constant temporarily,
between the limits 0 and 1.
 Then integrate the resulting function of y with respect to y
between the limits 0 and 1 as usual.
 In notation ,
1
0
1
0
( for double integral)
, ,
1
0
1
0
1
0
( for triple
integral).
Note:
 Integral with variable limits should be the innermost integral
and it should be integrated first and then the constant limits.

4. REGION OF INTEGRATION
Consider the double integral ,
�2( )
�1( )
, varies from
�1 �2( ) and varies from . (i.e) �1 ≤ ≤
�2 and ≤ ≤ . These inequalities determine a region in the
− , which is shown in the following figure.This region
ABCD is known as the region of integration

5. EXAMPLE :1
Evaluate 2
2
0
1
0
Solution:
2
2
0
1
0
= /3
1
0 0
2
=
8
3
1
0
=
8
3
2
2 0
1
=
4
3

6. EXAMPLE :2
Evaluate
1
2
1
3
2
Solution:
1
2
1
3
2
= log 1
2 1
3
2
=( 2 − 1)
1
3
2
= 2 [ ]2
3
= 2( 3 − 2)
= 2. log⁡
(3/2)

7. EXAMPLE :3
Evaluate 2
2
1
3
1
2
0
Solution:
2
2
1
3
1
2
0
=
2
2 1
2
3
1
2
0
2
=
3
2
3
1
2
0
2
=
3
2
3
3 1
3
2
0
=
26
2
2
2 0
2
= 26

8. EXAMPLE :4
Evaluate 2
2
1
2
0
1
0
Solution:
2
2
1
2
0
1
0
=
2
0
1
0
2
2 1
2
2
=
3
2
2
2 0
2
1
0
2
=
3
2
2
1
0
2
=
3
3 0
2
=1

9. EXAMPLE :5
Evaluate 2
sin � � ∅
1
0
�
2
0
�
0
Solution:
2
sin � � ∅
1
0
�
2
0
�
0
= sin �
3
3 0
1
� ∅
�
2
0
�
0
=
1
3
sin � � ∅
�
2
0
�
0
=
1
3
− � 0
�
2
∅
�
0
=
1
3
∅
�
0
=
�
3

10. EXAMPLE :6
Evaluate 0
1
0
Solution:
0
1
0
=
1
0 0
=
1
0
=
2
2 0
1
=
1
2

11. EXAMPLE :7
Evaluate 0
0
0
Solution:
I = [ ]
0
0
0
=
2
2 0
0
0
=
2
2
0
0
=
3
2
0
0
=
4
8 0
0
=
6
48 0
=
6
48

12. EXAMPLE :8
Evaluate
1− 2− 2− 2
1− 2− 2
0
1− 2
0
1
0
Solution:
I = sin−1
1− 2− 2
0
1− 2− 2
1− 2
0
1
0
=
�
2
=
�
2
[ ]0
1− 2
1
0
1− 2
0
1
0
=
�
2
1 − 2
1
0
=
�
2 2
1 − 2 +
1
2
sin−1
0
1
=
�2
8

13. EXAMPLE :9
Evaluate �
sin �
0
�
0
Solution:
I=
2
2 0
sin �
�
�
0
=
1
2
2
� 2
� �
�
0
=
2
2
1− 2�
2
�
�
0
=
2
2
�
1
2
� −
� 2�
2
0
�
=
� 2
4

14. PROBLEMS FOR PRACTICE
Evaluate the following
1. 4
1
0
2
0
Ans: 4
2.
1
1
1
Ans: loga.logb
3. 0
1
0
Ans: 1/2
4. �
sin �
0
�
0
Ans: π/4
5.
3
0
2
0
1
0
Ans: 9/2
6.
+
0
0
1
0
Ans: ½

15. EXAMPLE :10
Sketch the region of integration for ( , )
2− 2
0
0
.
Solution:
Given = 0 = ; = 0 2 = 2 − 2
= 0 and 2 + 2 = 2

16. EXAMPLE :11
Sketch the region of integration for ( , )
0
1
0
.
Solution:
Given = 0 ; = 1 = 0 ; = .
Y
x = y
x=1
X
y = 0

17. EXAMPLE :12
Evaluate �
where D is the region bounded
by the positive octant of the sphere 2 + 2 + 2 = 2
Solution:
� =
2− 2− 2
0
2− 2
0
0
=
2
2 0
2− 2− 2
2− 2
0
0

18. =
1
2
2− 2
0
0
( 2 − 2 − 2)
=
1
2
2− 2
0
0
( 2
− 2
− 3
)
=
1
2
2
2
2
− 2
2
2
−
4
4 0
2− 2
0
=
1
8 0
( 2 − 2)2
=
1
8
( 4 − 2 2 3 + 5)
0
=
1
8
4
2
2
− 2 2
4
4
−
6
6 0
=
6
48
.

19. PROBLEMS FOR PRACTICE
1.Sketch the region of integration for the following
(i) 2+ 2
2
4
4
0
(ii)
2− 2
−
0
(iii) 2+ 2
1
1
0
2.Evaluate + +
�
, where V is the region of
space bounded by x=0,x=1,y=0,y=2,z=0 and z=3.
Ans: 33/2
3. Evaluate
(1+ + + )3
�
, where V is the region of space
bounded by x=0,y=0,z=0 and x+y+z=1
Ans:
1
16
(8 2 − 5)
4. Evaluate �
, where V is the region of space bounded
by x=0,,y=0,,z=0 and 2x+3y+4z=12.
Ans: 12

20. CHANGE of ORDER OF INTEGRATION
If the limits of integration in a double integral are
constants, then the order of integration can be
changed, provided the relevant limits are taken
for the concerned variables.
When the limits for inner integration are
functions of a variable, the change in the order of
integration will result in changes in the limits of
integration.
i.e. ,
2
1
will take the form
,
ℎ2( )
ℎ1( )
This process of converting a given double
integral into its equivalent double integral by
changing the order of integration is called the
change of order of integration.

21. EXAMPLE :13
Evaluate
2−
1
0
by changing the order of
integration.
Solution: X
(1,1)
x=y x=2-y
Y
Given y : 0 to 1 and x : y to 2-y
By changing the order of integration,
In Region D1 x : 0 to 1 and y : 0 to x.
In Region D2 x : 1 to 2 and y : 0 to 2-x.
D1 D2

22. 2−
1
0
= 0
1
0
+
2−
0
2
1
=
2
2 0
1
0
+
2
2 0
2−
2
1
=
1
2
3
1
0
+
1
2
4 − 4 2
+ 3
2
1
=
1
2
4
4 0
1
+
1
2
2 2
−
4 3
3
+
4
4 1
2
=
1
8
+
5
24
=
1
3

23. EXAMPLE :14
Evaluate −
2
0
∞
0
by changing the order of integration.
Solution: Y
x = 0 x=y
X
Given x=0, x = y, y = 0, y = ∞ .
By changing the order of integration y: x to ∞, x : 0 to ∞

24. −
2
=
0
∞
0
−
2
∞
∞
0
= −
2 2
2
∞
∞
0
=
1
2
−
2
−1/
∞
∞
0
=
1
2
−
∞
0
� = , = − � � = , = − − ,
by integration by parts,
=
1
2
−
−1
− −
0
∞
=
1
2

25. EXAMPLE :15
Evaluate +
4−
1
3
0
by changing the order of integration.
Solution:
Y
y=3
x=1 y=4-x2
X
Given y=0,y=3 and x=1, x= 4 −
By changing the order of integration,
In region D, x : 1 to 2 and y : 0 to 4-x2
D

26. +
4−
1
3
0
= +
4−x2
0
2
1
= +
2
2 0
4−x2
2
1
= 4 − x2
+
(4−x2)2
2
2
1
=
4
4
− 3
− 4 2
+ 4 + 8
2
1
=
5
10
−
4
4
− 4
3
3
+ 2 2 + 8
1
2
=
241
8

27. EXAMPLE :16
Evaluate
2 −
2/
0
by changing the order of integration.
Solution:
Given y : 2
/ to 2 − and x : 0 to a
By changing the order of integration,
In Region D1 x : 0 to and y : 0 to a.
In Region D2 x : 0 to 2 − and y : a to 2a.

28. 2 −
2/
0
= 0
0
+
2 −
0
2
=
2
2 0
0
+
2
2 0
2 −
1
0
=
2
2
0
+
1
2
4 2
− 4 2
+ 3
2
=
2
3
3 0
+
1
2
2 2 2 −
4 3
3
+
4
4
2
=
4
6
+
5 4
24
=
3 4
8
.

29. EXAMPLE :17
Evaluate 2+ 2
2− 2
1
0
by changing the order of integration.
Solution:
Given x = 0, x = 1 and y = x, y2
= 2-x2
By changing the order of integration
In Region D1, y : 0 to 1,x : 0 to y
In Region D2, y : 1 to 2 , x : 0 to 2 − 2

30. I = 2+ 2
0
+ 2+ 2
2− 2
0
2
1
1
0
= 2 + 2
0
2
+ 2 + 2
0
2− 2
2
1
1
0
= 2 − + 2 −
2
1
1
0
= ( 2 − 1)
2
2 0
1
+ 2 −
2
2 1
2
= 1-
1
2

31. PROBLEMS FOR PRACTICE
Evaluate the following by changing the order of integration
1. ( 2
+ 2
)
0
Ans:
4
3
2.
2 −
2
0
Ans:
3 4
8
3.
2− 2
−
0
Ans:
3
6
4.
2−
1
0
Ans:
1
3

32. PLANE AREA USING DOUBLE
INTEGRAL
CARTESIAN FORM

33. EXAMPLE :18
Find by double integration, the area enclosed by the ellipse
2
2
+
2
2
= 1
Solution:
A = 4 = 4
1−
2
2
0
0
= 4 0
1−
2
2
0
=
4 2 − 2
0
=
4
2
2 − 2 +
2
2
sin−1
0
=
4
x
2
2
x
�
2
= � sq.units.

34. EXAMPLE :19
Find the area between the parabola = 4 − 2
and the line = .
Solution:
Given = 4 − 2
= , solving for x,
= 4 − 2
=> 0 = 3 − 2
=> 0 = 3 − => = 0,3
A = =
4 − 2
3
0
4 − 2
3
0
= (3 − 2
)
3
0
=
3 2
2
−
3
3 0
3
=
9
2

35. EXAMPLE :20
Find the area between the parabola = 2
and the line
= 2 + 3.
Solution:
Given = 2
and = 2 + 3.
solving for , 2
= 2 + 3 => = −1,3
A = =
2 +3
2
3
−1
2
2 +3
3
−1
= (2 + 3 − 2
)
3
−1
=
2 2
2
+ 3 −
3
3 −1
3
=
32
3

36. PLANE AREA USING DOUBLE
INTEGRAL
POLAR FORM

37. EXAMPLE :21
Find the area bounded by the circle
= 2 sin� = 4 sin�.
Solution: � = �/2
� = � � = 0
A = � =
2
2 2 sin �
4 sin �
�
�
0
4 sin �
2 sin �
�
0
= 6 sin � 2
�
�
0
= 3 (1 − cos2�) �
�
0
= 3 � −
sin 2�
2 0
�
= 3� .

38. EXAMPLE :22
Find the area enclosed by the leminiscate 2
= 2
cos 2� by double
integration.
Solution:
If r = 0 then cos 2� = 0 implies � =
�
4
.
A = 4 �
2 cos 2�
0
�
4
0
= 4
2
2 0
2 cos 2�
�
�
4
0
= 4 2 cos 2�
2
�
�
4
0
= 4
2 sin 2�
4 0
�
4
= 2
.

39. EXAMPLE :23
Find the area that lies inside the cardioids = (1 + cos �) and outside
the circle = , by double integration.
Solution:
Solving = (1 + cos �) =
=> (1 + cos �) =
=> cos � = 0
=> � =
�
2
.

40. A = 2 � = 2
2
2
(1+cos �)
�
�
2
0
(1+cos �)
�
2
0
= [ (1 + cos�) 2
− 2
] �
�
2
0
= 2
[2cos� + cos� 2
] �
�
2
0
=
2
2
[4cos� + 1 + cos2�] �
�
2
0
=
2
2
� +
sin 2�
2
+ 4sin�
0
�
2
=
2
2
� + 8 .

41. EXAMPLE :24
Find the common area to the circles = , = 2 cos �.
Solution:
Given = , = 2 cos � , solving
 = 2 cos �
 cos � =
1
2
 θ=π/3
when = 0 => cos � = 0 => � = �/2

42. A = 2 �
= 2 � + 2 �
2 cos �
0
�
2
�
3
0
�
3
0
= 2
2
2 0
�
�
3
0
+ 2
2
2 0
2 cos �
�
�
2
�
3
= 2
�
�
3
0
+ 2 2
cos� 2
�
�
2
�
3
= 2
� 0
�
3
+ 2 2
� +
sin 2 �
2 �
3
�
2
= 2 �
3
+ 2 2 �
2
−
�
3
− 2 3
2
= 2 2�
3
−
3
2

43. PROBLEMS FOR PRACTICE
1.Find by double integration, the area bounded by the parabolas
2
= 4 and 2
= 4 .
Ans:
16 2
3
. � .
2.Find by double integration, the smallest area bounded by the
circle 2
+ 2
= 9 and the line + = 3.
Ans:
9
4
� − 2 . � .
3.Find by double integration, the area common to the parabola
2
= and the circle 2
+ 2
= 2.
Ans:
1
3
+
�
2
� .
4.Find by double integration, the area lying inside the circle
= sin � and outside the coordinate = (1 − �).
Ans: 2
1 −
�
4
. � .