SMT1105

ENGINEERING

MATHS- II

UNIT-I

MULTIPLE INTEGRALS

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 .

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.

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

EXAMPLE :1

Evaluate 22

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

EXAMPLE :2

Evaluate 12

1

3

2

Solution:

12

1

3

2 = log 1

2 13

2

=( 2 − 1) 13

2

= 2 [ ]23

= 2( 3 − 2)

= 2. log(3/2)

EXAMPLE :3

Evaluate 22

1

3

1

2

0

Solution:

22

1

3

1

2

0 = 2

2

1

23

1

2

02

= 3

2

3

1

2

02

= 3

2 3

3

1

32

0

= 26

2 2

2

0

2

= 26

EXAMPLE :4

Evaluate 22

1

2

0

1

0

Solution:

22

1

2

0

1

0 = 2

0

1

0 2

2

1

2

2

= 3

2 2

2

0

21

02

= 3

2 2

1

02

= 3

3

0

2

=1

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 � � ∅�

20

�0

=1

3 − �

0

�2 ∅�

0

= 1

3 ∅�

0

= �3

EXAMPLE :6

Evaluate 0

1

0

Solution:

0

1

0 = 1

0 0

= 1

0

= 2

2

0

1

= 1

2

EXAMPLE :7

Evaluate 000

Solution:

I = [ ]000

= 2

2

000

= 2

2

00

= 3

2

00 = 4

8

00

= 6

48

0 =

6

48

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− 21

0

1− 2

0

1

0

=�2 1 − 21

0

=�2

2 1 − 2 +

1

2sin−1

0

1

=�2

8

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

PROBLEMS FOR PRACTICE

Evaluate the following

1. 4 1

0

2

0 Ans: 4

2. 1

11 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. +

00

1

0 Ans: ½

EXAMPLE :10

Sketch the region of integration for ( , ) 2− 2

00.

Solution:

Given = 0 = ; = 0 2 = 2 − 2

= 0 and 2 + 2 = 2

EXAMPLE :11

Sketch the region of integration for ( , )0

1

0.

Solution:

Given = 0 ; = 1 = 0 ; = .

Y

x = y

x=1

X

y = 0

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

00

= 2

2

0

2− 2− 2 2− 2

00

= 1

2 2− 2

00( 2 − 2 − 2)

= 1

2 2− 2

00( 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.

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

11

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

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.

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

2−1

0=

0

1

0+ 2−

0

2

1

= 2

2

0

1

0+ 2

2

0

2−2

1

=1

2 31

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

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 ∞

− 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

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

+ 4−1

3

0= + 4−x2

0

2

1

= +2

2

0

4−x22

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

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.

2 −2/0

= 00

+ 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 .

EXAMPLE :17

Evaluate 2+ 2

2− 21

0 by changing the order of integration.

Solution:

Given x = 0, x = 1 and y = x, y

2 = 2-x

2

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

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

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

PLANE AREA USING DOUBLE

INTEGRAL

CARTESIAN FORM

EXAMPLE :18

Find by double integration, the area enclosed by the ellipse 2

2+

2

2= 1

Solution:

A = 4 = 4 1− 2

2

00

= 4 0

1− 2

2

0

= 4 2 − 2

0

= 4

2 2 − 2 +

2

2sin−1

0

= 4

x2

2x�2 = � sq.units.

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 − 23

0

= (3 − 2)3

0

= 3 2

2− 3

3

0

3

= 9

2

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 +33−1

= (2 + 3 − 2)3−1

= 2 2

2+ 3 − 3

3 −1

3

= 32

3

PLANE AREA USING DOUBLE

INTEGRAL

POLAR FORM

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 − cos 2�) � �

0

= 3 � − sin 2�2

0

� = 3� .

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.

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.

A = 2 � = 2 2

2 (1+cos �) ��

20

(1+cos �)�2

0

= [ (1 + cos �) 2 − 2] ��2

0

= 2 [2 cos� + cos� 2] ��2

0

=2

2 [4 cos � + 1 + cos 2�] ��

20

= 2

2 � +

sin 2�2

+ 4 sin � 0

�2

=2

2 � + 8 .

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

A = 2 �

= 2 � + 2 �2 cos �0

�2�

30

�3

0

= 2 2

2

0��

30

+ 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

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 . � .