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Problem :-02
...


x0
xa

yx
ya

i
...
[ Corresponds
to vertical strip method
...

i
...


x2
4

Solution:
From the given data, first integral with respect to x and second with respect to y
x0
x4

x2
4
y2 x

y

i
...
x 2  4 y
i
...
y 2  4 x

i
...
[corresponds to
vertical strip method]
 The region of integration is

y2 = 4x
x2 = 4y
(4,4)

(0,0)

x=4

x=0

By changing the order,
Let us consider horizontal strip method, we have the following limits
x=y2/4, x=2(y)1/2, y=0, y=4

4

I 

2 y

dx dy

 
0

y2
4

4



=

y2
dy
4

2 y 

0

4

1

=  2 y2 
0

=

y2
dy
4
4

 4 3 y3 
=  y2  
12  0
3
 4 32 64 
=  4 

12 
3

I=

16
3

Problem:-4
a

Change the order of integration and hence evaluate

2 a x

 
0

xy dy dx

2

x
a

Solution:
From the given data, first integral with respect to x and second integral with respect
to y
...
e
...
e
...
e the given order of integration is with respect to y first and x second [corresponds
to vertical strip method]
 The region of integration is

x2 = ay

(0,2a)

(a,a)

(0,a)

y=a

(0,0)

x=a
x + y = 2a

x=0

By changing the order,
Let us consider the horizontal strip method,
i
...
therefore we must divide the whole region R into
two regions R1(region below the line y=a) and R2 (region above the line y=a)
...
therefore we have R=R1+R2
I   xydxdy   xydxdy
R1

R2

R1 (region below the line y=a)

R2(region above the line y=a)
...


a

I 

 
0

a



ay


0

2a

xy dx dy 

0

 x2 
y 
 2 0

 
1

ay

xy dx dy

0

2a

dy 

2a y


a

2 a y

 x2 
y 
 2 0

aa 2
1 2a
2

y dy   y  2a  y  dy

20
2a

dy

aa 2
1
I
y dy 

20
2

2a



4a 2 y  y 3  4ay 2 dy

a

a

2a

a  y3  1 
y 4 4ay 3 
2
     2ay 


2  3 0 2 
4
3 a


I

a 4 1  4
32a 4   4 a 4 4a 4  
   8a  4a 4 
   2a  

6 2 
3  
4
3 
9 4
a
24

Problem:-5
...
e the given order of integration is with respect to y first and x second [corresponds to
vertical strip method]
The region of integration is

By changing the order of integration,
Let us consider horizontal strip method
i
...

x

y, x  a

y  0,

y  a2

a x2

  x( x
0 0

a2 a
2

2

 y )dydx  
0

a



 x( x

2

 y 2 )dxdy

y
2

a

  (x
0

3

 xy 2 )dxdy

y

a2

a

 x4 x2 y2 
  
 dy
4
2  y
0 
a2

 a 4 a 2 y 2   y 2 y 3  
   
     dy
2   4
2  
 4
0 
a2

 a4 y a 2 y3 y 3 y 4 


  
6
12 8 0
 4
a 6 a8 a 6 84
  
4 6 12 8
6a 6  4a8  2a 6  3a8

24
6
8
4a  a

24


a x2
2
2
  x( x  y )dydx 
0 0

a 6 (a 2  4)
24

Problem:-06
a

Evaluate

a 2  x2

 
0

xydydx by changing the order of integration

0

Solution:
From the given, first integral with respect to x and second integral with respect to y
...
line parallal y  axis)

i
...

i
...


a

x  a2  y2

y 0

ya

a 2  x2

 
0

x0

0

a 2  y2

a

xydydx  



0

xydxdy

0

a

 x2 
  y 
 2 0
0

a 2  y2

dy

a

1
  y a 2  y 2 dy
20





a



1
a 2 y  y 3 dy

20





a

1  a2 y2 y 4 
 
 
2 2
4 0
1  a4 a4 
   
2 2 4 
a

a 2  x2

 
0

0

a4
xydydx 
8

Problem:-7
a

Evaluate by changing the order of integration

a2  y 2

 
0

y dy dx

a y

Solution:
From the given, first integral with respect to y and second integral with respect to x
...
e
...
e
...
e the given order of integration is with respect to x first and y second
...

i
...

a 2  x2

a

I 

 

y dy dx

ax

0

a 2  x2
1a
2


=
y
dx
2 0   a  x

1
=
2

a

 a

2

 x 2   (a  x )2 dx

0

a

1
x 3 ( a  x )3 
 a 2 x  

2
3
3  0

I

1
2

I

a3
6

 3 a3 a3 
a  3  3 



Problem:-8
1

Change the order of integration and hence evaluate

2 y

 
0

y

xy dx dy

2

Solution:
From the given, first integral with respect to y and second integral with respect to x
...
e
...
e integration with respect to x first and y second [ corresponds to horizontal strip
method]
 The region of integration is bounded by

(0,2)
y2 = x
y=1

(1,1)

(2,0)
y=0
(0,0)

x=1

x+y=2

By changing the order,
Let us consider the vertical strip method,
i
...
therefore we must divide the whole region R into two
regions R1(region in left side of the line x=1) and R2 (region in right side of the line x=1)
...
therefore we have R=R1+R2
I   xydxdy   xydxdy
R1

R2

R1 (region in left side of the line x=1)

R2( region in right side of the line x=1)

y=0, y=(x)1/2

y=0, y=2-x

x=0,x=1

x=1,x=2

1

I 

x

 
0

1




0

2 x

2

xy dy dx 

0

 
1

x

 y2 
x   dx 
 2 0

xy dy dx

0

2


1

2 x

 y2 
x   dx
 2 0

11 2
12
2
  x dx   x  2  x  dx
20
21
11 2
1
  x dx 
20
2
1

2



4 x  x3  4 x 2 dx

1

2

1  x3  1  2 x 4 4 x 3 
     2x  

2  3 0 2 
4
3 1



1 1 
32  
1 4 
  8  4     2    
6 2 
3 
4 3 

I 

23
12

TRIPLE INTEGRAL
Consider a function f(x,y,z) defined at every point of the three dimensional finite
region V
...
 Vn
...
Consider the sum

 f ( x , y , z )V
r

r

r

r

r 1

The limits of this sum, if it exists, as n   and  Vr   is called the triple integral
of f(x,y,z)over the region V and is denoted by

 f ( x, y, z )dV
For purposes of evaluation it can also be expressed as the repeated integral
x2 y2 z2

   f ( x, y, z) dxdydz

x1 y1 x z

If x1 , x2 are constants: y1 , y2 are either constants or functions of x and z1 , z2 are
either constants or functions of x and y then this integral is evaluated as follows:
First f(x,y,z)is integrated w
...
to z between the limits z1 and z2 keeping x and y fixed
...
r
...
The result just obtained is finally integrated w
...
to x between the limits x1 and x2
x2

Thus

I

y2 ( x )

 
x1

y1 ( x )

z2 ( x , y )



f ( x, y, z ) dz dy dx

z1 ( x , y )

Where the integration is carried out from the innermost rectangle to the outermost rectangle
...
Also any order of integration can be followed for evaluation
...


a  x2  y 2  z2

Solution:The equation of sphere x2+y2+z2=a2 and the limits are

x0

xa

y0

y  a2  x2

z0

z  a2  x 2  y 2



2

2

2

a x y z

a 2  x2

a 2  x2  y2







0

0

0

a

dzdydx
2

a 2  x2

a





0

0

a

a 2  x2



(a  x 2  y 2 )  z 2




z
sin 1 

 (a 2  x 2  y 2 )  


 0
sin 1 1  sin 1  0   dydx



0

dzdydx
2

0

a


 
20

a 2  x2



dydx

0

a




a2  x 2
y
dx


0
2 0




a 2  x 2 dx

20

a

a

  x 2 2 a 2 1  x  
 
a  x  sin  
2 2
2
 a  0

  a 2 1
  0  sin 1  0 
2
2


  a2  
2  2 2 
 2a2

8


a 2  x2  y 2

dydx



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