Applications of Multiple Integrals

KOLHAPUR INSTITUTE OF TECHNOLOGY’S
COLLEGE OF ENGINEERING (AUTONOMOUS), KOLHAPUR
(AFFILIATED TO SHIVAJI UNIVERSITY, KOLHAPUR)

DEPARTMENT OF BASIC SCIENCES AND HUMANITIES
First Year B. Tech (Sem-II)
Engineering Mathematics-II (UBSH0201)
Unit No. 6
APPLICATION OF MULTIPLE INTEGRAL

1) Area:
 Cartesian coordinates -
Consider the area enclosed between two plane curves y  f1 ( x) and y  f 2 ( x )
intersecting at points A and B then the area obtained by,

A   dx dy

 Polar coordinates -
Consider the area enclosed between two plane curves r  f1 ( ) and r  f 2 ( )
intersecting at points A and B then the area obtained by,

A   r dr d


Solved Examples:-

1) Find by double integration the area enclosed by y 2  x 3 and y  x .

Solution: The two curves y 2  x 3 and y  x intersect at the origin O(0, 0) and A(1, 1).

Consider the strip parallel to y axis in the region of integration then y varies from
x to x and x varies from 0 to 1.
Then, area A   dx dy

1 x
A    dy dx
0 x3/ 2
1
A   y x dx
x

0
1
A   ( x  x ) dx
0


Department of Mathematics
KIT’s College of Engineering (Autonomous), Kolhapur Page 1

, Application of multiple integral

1
 x 2 x 
A  
2 0

1 2
A 
2 5
1
A
10

2) Find the area between the circles r  2a sin and r  2b sin , (b  a ).

Solution:
Consider the radial strip in region of integration.
Then the limits of r varies from r  2a sin  to
r  2b sin  and limits of  varies from   0 to
 .
We know that, Area A   r dr d
 2b sin 
A   r dr d
0 2 a sin 

 2b sin 
r2 
A    d
0 2
  2a sin 
  4b 2 sin 2   4a 2 sin 2  
A    d
0  2 

A  2b a 2 2
  sin 2
 d
0


A  2b a 2 2
   1  cos
2
2 
 d

0



A b a 2 2
 
 
sin 2 
2  0


A  b2  a 2  

Department of Mathematics
KIT’s College of Engineering (Autonomous), Kolhapur Page 2

, Application of multiple integral

3) Find the area of Cardioid r  a (1  cos ) by double integration.

Solution: Consider the radial strip in region of
integration then limits of r varies from r  0 to
r  a(1  cos  ) and limits of  varies from
  0 to    above the initial line.
(Note: The curve is symmetrical about initial line
Thus here we consider only upper part of cardioid
for finding limits. Then for total area take a 2 times
of area of upper part of cardioide)

We know that, Area A   r dr d
 a (1 cos )
A  2  r dr d
0 0

 a (1 cos )
r2 
A  2   d
0  2 0


A   a 2 1  cos 2 d
0

 2
  
A  a   2 sin 2  d
2

0 2

  
Aa   4 sin d
2 4

0 2
  0 
Put  t    2t  d  2dt
2 t 0  /2
 /2
A  8a 2  sin t dt
4

0

1 5 1
A  8a 2  , 
2 2 2

3a 2
A
2
4) Find the area between the curve (a  x ) y  a x and its asymptote.
2 2


Solution: Consider the strip parallel to y axis in region. In the area between the curve and
x
asymptote then y varies from y  0 to y  a and x varies from x  0 to x  a
ax
Department of Mathematics
KIT’s College of Engineering (Autonomous), Kolhapur Page 3