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UNIT- I
MULTIVARIABLE CALCULUS (INTEGRATION)

DOUBLE INTEGRATION
We know that the double integral over the region R of a function
f(x,y) is

 f ( x, y)dxdy
R

Case(i)
let R be the region bounded by the lines x  c1 , x  c2 , y  c3 , y  c4 where
c1 , c2 , c3 , c4 are constant
...


We know the region integration R, the double integral

 f ( x, y)dxdy
R

c4 c2

can be written as

  f ( x, y)dxdy

c3 c1

Case (ii)
c 2 x2

Consider the double integral

  f ( x, y)dxdy

c1 x1

Suppose x1 and x2 are the function of y say x1  f ( y ), x2   ( y ) and
c1 and c2 are constants then the region of integration R is bounded by curve
x1  f ( y ) and x2  ( y ) and the lines y  c1 and y  c2
...
Here we integrate f ( x, y ) first w
...
to x
...
r
...

i
...


The region shown in figure
...
r
...

Keeping x as a constant and integrate the resulting expression w
...
to x
...
e first integration is along the vertical strip RS and then slide this
strip RS horizontally
Problem:-01
...
e region corresponding to the above integral is rectangular region
bounded by the line x=0,x=1,y=1 and y=2]
=∫

+

=2
=1

=∫


...
1+

Note:If all the limits of double integrals are numbers, then the integrals are
indentified using rectangle box, and any order of integration (x first y
second or y first x second) can be followed, both will give same answer
...
e region corresponding to the above integral is rectangular region
bounded by x=1, x=2, y=2 and y=3]
=2
=1

= ∫

(log )

= ∫

(log 2 − log 1)

=∫

(log 2)

= log 2 [log ]

= log 2 [ log 3 – log 2]
= log2
...

Problem:- 3
Evaluate: ∫

∫

(

+

)

=3
=2

Solution
Let I = ∫ ∫
=∫

(
(

∫

=∫
=
=

+

)
+

)

= ∫

+

+
+

∗

+

5
0

∗

=5

+

=5


...

Suppose if the integral limit is a function of x say f(x), then it is
corresponding to y integral
...
e y=f(x)
...

Suppose if the integral limit is a function of y say f(y), then it is
corresponding to x integral
...
e x=f(y)
...


Problem:-04
1 y


...

1 y

y

1

 x3 
x
dydx

0 0
0  3  dy
0
2

1

1
  y 3dy
30
1

1  y4 
  
3  4 0
1 1
  
3 4
1

12
Note
The region corresponding to the above integral limits x=0, x=y , y=0 and y=1 is

Problem:-05
Evaluate ∫ ∫

√

Solution:
Since y=0 and y=(1+x2)1/2, therefore the order of integration with
respective y first and x second
Let I = ∫ ∫

√

= ∫ ∫

=∫

√
√

√

= ∫

(1) −

√

=∫

(0)

√

= ∫

√

=

[

ℎ

( )]1
0

=

[

ℎ

(1) −

= [

ℎ

(1) − 0]

=

√1 +
0

√

log( 1 + √2
...


Solution:Given that x = 2a
...
To find the
limit for y, we take a strip PQ parallel to the y – axis, it’s lower end P lies on
y = 0 and upper end Q lies on
x2 = 4ay = y =

∫∫

=∫

∫
4
0

=∫
= ∫
=
=
=
=

∫
2
0

CHANGE OF ORDER OF INTEGRATION
The evaluation of some double integrals may be very difficult
...
When we change the order of integration the limits
are also changed but the there will be no change in final answer
...

(i)

If the limits of the inner integral is a function of x (or function of y)
the first integration should be w
...
to y (or w
...
to x)

(ii)

Draw the region of integration by using the given limits
...
r
...
r
...
0)
x=4

Given integral limits are corresponds to horizontal strip method, So
By changing the order, we have consider vertical strip method
4 x

I 


0 0

4




0

x
dx dy
x  y2
2

x

 1 y 
 tan x  dx

0

4



  tan

1

1  tan 1 0  dx

0



4

4



dx

0


4
 x0
4

  4  0
4

  4
4





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