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MODULE-4
Multivariable Calculus- I
(Multiple Integration)
Syllabus: Multiple integral: Double integral, Triple integral, Change of order of integration, Change of
variables, Applications: Area, Volume, Centre of Mass & Centre of Gravity (constant and variable
densities)
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TABLE OF CONTENTS
S
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Topic

Page No
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1

Course Objectives of Module 4

1

4
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3

Multiple Integration

2

4
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1

Double Integral

3

4
...
2

Triple Integral

12

4
...
3

Change of order of integration

15

4
...
4

Change of variables

19

4
...
4
...
4
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4
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5

Related Links

46

1

4
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4
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1 APPLICATIONS

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Double and triple integrals are useful in finding Area
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Double and triple integrals are useful in finding Mass
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In finding Average value of a function
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Useful in calculating Kinetic energy and Improper Integrals
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The most important application of Multiple Integrals involves finding areas bounded by a curve and
coordinate axes and area between two curves
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In mechanics, the moment of inertia is calculated as the volume integral (triple integral) of
the density weighed with the square of the distance from the axis
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The gravitational potential associated with a mass
distribution given by a mass measure on three-dimensional Euclidean space R3 is calculated by triple
integration
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We can determine the probability of an event if we know the probability density function using double
integration
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3 MULTIPLE INTEGRAL
4
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1 Double integral
A double integral is its counterpart in two dimensions
...

Then double integral of f(x, y) over the region R is denoted by,

Also we can express as

or

2

4
...
1
...
Let the region
be bounded by the curves

...
)
(ii) When

are functions of x and

variable x [

are constants: If we have functional limits of y in terms of dependent

and constant limits of variable x then we will first integrate with respect to variable y

in case of double integral, as follows:

=

3

(Here we have drawn the strip parallel to y axis, because variable limits [

(iii) When

are provided
...
)

From case no
...
r
...
the variable
limits first and then w
...
t
...


4
...
1
...


=

(Here we have constant limits for both x and y variables,
so we may integrate w
...
t
...
)
For the ellipse we may write

or

The region of integration R can be expressed as
-a

,

where we have chosen variable limits of y and constant limits of x
...
r
...
y,

5

=

=

=

+0

[using the property of even and odd functions
...


intersect each other at two distinct points O (0,0) and A (3,3)
...

=

6

=

dx

=
=
=
Example 4: Evaluate
Solution: Solving

-

] dx

-

] dx
=

(Answer)

over the region R bounded by the parabolas
, we have

x = 0, 4
When x = 4, y = 4
Co-ordinates of A (intersection point of parabolas) are (4, 4)
The region R can be expressed as
0

x

4,

7


...


Solution: Let OAB be the triangle formed by given vertices (0, 0), (10, 1) and (1, 1) as shown in the figure through shaded
area
...

Hence the region of integration can be expressed as

8

dxdy =

=

= 18

dxdy

dy =

= 18

(Answer)

Example 6: Let D be the region in the first quadrant bounded by the curves xy = 16, x = y, y = 0 and x = 8
...

Solution: In this question we have to integrate the given function within the region bounded by the straight line x = y,
hyperbola xy = 16, y = 0 and x = 8, so first we will draw the figure for clarity finding all intersection points of curves
provided in question
...

Now we are to decide with respect to which variable we should first integrate, we construct strips in such a manner that
the complete area may be covered
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Because area is changing from dotted lines, if we plot strip parallel to y-axis and also area is changing from lines drawn, if
we plot the strip parallel to x-axis
...

Here we are splitting the area OABNO in two parts by AM as shown in figure and plotted strips parallel to y- axis from x
= 0 to x = 4 and from x= 4 to x = 8
Then,

=
=

10

=
=

=

= 64 +8(64 -16) = 64 +384 = 448 (Answer)

4
...
1
...
Evaluate

2 x dydx
1 0
2
2

Ans :

x +y

2
...
8( log8-2)+e


...
Evaluate:
(i)

Ans
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41/210

(iii)

Ans
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3/ 35

4
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Evaluate
6
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...

Ans
...


Ans
...
3
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5 Evaluation of Double Integrals in Polar Coordinates
In polar coordinates we know that, x = r cos

and y = r sin

Sometimes integration can be easier by converting Cartesian form to polar form
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Here we draw radial strip to decide the
limit in order to cover the whole area
...
3
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6 Solved Examples
Example 1: Evaluate

, over the area bounded between the circles r = 2 cos

11

and r = 4 cos
...
Here r varies from r = 2 cos
from

to r = 4 cos


...


Solution: I =

=

12

(Answer)

while

varies

=

=

=

(Answer)

4
...
1
...
Evaluate





2

0



a cos 

0

r sin  drd

Ans:

2
...
Evaluate

Ans:

, over the area bounded between the circles r = 2 sin and r = 4 sin
...
3
...

The triple integral of f(x, y, z) over the region and is denoted by
For purpose of evaluation, it can be expressed as the repeated integral

………………………
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Working rule:
Let
Let


...
t
...
x (keeping y and x constant)between the limits
13

and


...
r
...
y (keeping
z constant) between the limits

and


...
r
...
z between the limits

and


...


Limits involving two variables are to be kept innermost, then the limits involving
one variable and finally the constant limits
...
Thus
=

=

4
...
2
...


Solution: Let I =
=
=
=
=
=
=

Example 2: Evaluate

=

=

(Answer)


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Example 3: Evaluate

Solution: Here limits are constant but different for x,y,z
...
r
...
x,y or z, but limits are
fixed for x,y,z [0 ≤ x ≤ 1, 1 ≤ y ≤ 2, 2 ≤ z ≤ 3]
=
=
=
=

dx

=
=

= ½+4 – 0 = 9/2(Answer)

=

4
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2
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Evaluate the integral:
2
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1/48
, where S is the region bounded by the surfaces
and z = x+3
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Evaluate

Ans
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,
Ans
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Evaluate

Ans
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3
...
Thus,

But if the limits of integration are variable, then in order to change the order of limits of integration we have to construct
the rough figure of given region of integration and re construct the strip parallel to that axis with respect to which we want
to first integrate
...


4
...
3
...


Solution: The given limits show that the region of integration is bounded by the curves y =

, y = e, x = 0, x = 1
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In given problem we had variable limits of y in terms of x, so we had to integrate w
...
t
...
But we are instructed to
solve this problem changing the order of integration
...
r
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x we have to find variable limits of x in terms of y
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From the strip we can see lower limit lies on x = 0 and y =

in between the constant limits of y from y = 1

to y = e

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