1. Unit: 3 MULTIPLE INTEGRAL
RAI UNIVERSITY, AHMEDABAD 1
Course: B.Tech- II
Subject: Engineering Mathematics II
Unit-3
RAI UNIVERSITY, AHMEDABAD
2. Unit: 3 MULTIPLE INTEGRAL
RAI UNIVERSITY, AHMEDABAD 2
Unit-III: MULTIPLE INTEGRAL
Sr. No. Name of the Topic Page No.
1 Double integrals 2
2 Evaluation of Double Integral 2
3 To Calculate the integral over a given region 6
4 Change of order of integration 9
5 Change of variable 11
6 Area inCartesian co-ordinates 13
7 Volume of solids by double integral 15
8 Volume of solids by rotation of an area
(Double Integral)
16
9 Triple Integration (Volume) 18
10 Reference Book 21
3. Unit: 3
RAI UNIVERSITY, AHMEDABAD
MULTIPLE INTEGRALS
1.1 DOUBLE INTEGRALS
We Know that
( ) = lim→∞
→
[
Let us consider a function
the finite region A of
Then ∬ ( , )
( , ) ]
2.1 Evaluation of Double Integral
Double integral over region A may be evaluated by
two successive integrations.
If A is described as
And ≤
Then ∬ ( , )
MULTIPLE INTEGRAL
MULTIPLE INTEGRALS
OUBLE INTEGRALS:
[ ( ) + ( ) + ( ) + ⋯ +
Let us consider a function ( , ) of two variables and
the finite region A of - plane. Divide the region into elementary areas.
, , , …
) = lim →∞
→
[ ( , ) + ( , )
Evaluation of Double Integral:
Double integral over region A may be evaluated by
two successive integrations.
( ) ≤ ≤ ( )[ ≤ y ≤ ]
≤ ≤ ,
) = ∫ ∫ ( , )
MULTIPLE INTEGRAL
3
+ ( ) ]
and defines in
into elementary areas.
) + ⋯ +
4. Unit: 3
RAI UNIVERSITY, AHMEDABAD
2.1.1FIRST METHOD
∬ ( , )
( , ) is first integrated with res
limits and and then the result is integrated with respect to
the limits and .
In the region we take an elementary area
keeping constant) converts small rectangle
the integration of the result w.r.t
from to covering the whole
2.1.2 SECOND METHOD
Here ( , ) is first integrated w.r.t
and and then the resulting expression is integrated with respect to
between the limits
NOTE: For constant limits, it does not matter whether we first integrate
w.r.t and then w.r.t
MULTIPLE INTEGRAL
FIRST METHOD:
) = ∫ ∫ ( , )
is first integrated with respect to y treating as constant between the
and then the result is integrated with respect to
In the region we take an elementary area . Then integration w.r.t to
keeping constant) converts small rectangle into a strip
the integration of the result w.r.t corresponds to the sliding to the
covering the whole region .
SECOND METHOD:
( , ) = ( , )
is first integrated w.r.t keeping constant between the limits
and then the resulting expression is integrated with respect to
and and vice versa.
For constant limits, it does not matter whether we first integrate
and then w.r.t or vice versa.
MULTIPLE INTEGRAL
4
as constant between the
and then the result is integrated with respect to between
Then integration w.r.t to (
( ). While
corresponds to the sliding to the strip,
constant between the limits
and then the resulting expression is integrated with respect to
For constant limits, it does not matter whether we first integrate
7. Unit: 3
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=
=
∴ ∫ ∫ =
3.1 To Calculate the integral over a given region
Sometimes the limits of integration are not given but the area of the
integration is given.
If the area of integration is given then we proceed
as follows:
Take a small area
between the limits
indicates that integration is done, along
the integration of result
to covering the whole
We can also integrate first w.r.t ‘
Example 4: Evaluate
which + ≤ 1.
Solution: + = 1 represents a line
figure.
+ < 1 represents a plane
The region for integration is
the figure.
By drawing parallel to y
. . , ( + = 1)&Q
MULTIPLE INTEGRAL
= ________Answer
To Calculate the integral over a given region:
Sometimes the limits of integration are not given but the area of the
If the area of integration is given then we proceed
. The integration w.r.t
between the limits , keeping fixed
indicates that integration is done, along . Then
the integration of result w.r.t to corresponds to sliding the strips
covering the whole region .
We can also integrate first w.r.t ‘ ’ then w.r.t , which ever is convenient.
Evaluate ∬ over the region in the positive quadrant for
represents a line AB in the
represents a plane .
The region for integration is as shaded in
parallel to y-axis, lies on the line
lies on x-axis. The limit for is1 − and 0.
MULTIPLE INTEGRAL
7
Sometimes the limits of integration are not given but the area of the
corresponds to sliding the strips from
which ever is convenient.
over the region in the positive quadrant for
and 0.
8. Unit: 3 MULTIPLE INTEGRAL
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Required integral =
=
2
= ∫ ( )(1 − )
= ∫ ( − 2 + )
= − +
= − +
= ________Answer
Example 5: Evaluate ∬ , where the quadrant of the circle is
+ = where ≥ 0 ≥ 0.
Solution: Let the region of integration be the first quadrant of the
circle .
Let = ∬ ( + = , = √ − )
First we integrate w.r.t and then w.r.t .
The limits for are 0 and √ − and for x, 0 to a.
= ∫ ∫
√
= ∫
√
9. Unit: 3
RAI UNIVERSITY, AHMEDABAD
= ∫ (
=
=
Example 6: Evaluate ∬
bounded by the hyperbola
Solution: The line , =
The line , = 8 intersects the hyperbola at
The area A is shown shaded.
Divide the area into two parts by
For the area , varies from 0 to
For the area ,
∴ ∬ = ∫ ∫
= ∫
=
=
=
=
=
MULTIPLE INTEGRAL
( − )
−
________Answer
, where A is the region in the first quadrant
= 16 and the lines = , = 0 =
and the curve , = 16 intersect at
intersects the hyperbola at (8,2). And = 0 is x
The area A is shown shaded.
Divide the area into two parts by PM perpendicular to OX.
varies from 0 to , and then varies from 0 to 4.
varies from 0 to and then varies from 4 to 8.
∫ + ∫ ∫
∫ + ∫ ∫
= [ ] + [ ]
∫ + ∫ 16
+ 16
64 + 8 (8 − 4 )
64 + 384
MULTIPLE INTEGRAL
9
Answer
, where A is the region in the first quadrant
= 8.
(4,4).
is x-axis.
varies from 0 to 4.
varies from 4 to 8.
10. Unit: 3 MULTIPLE INTEGRAL
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= 448 ________Answer
3.2 EXERCISE:
1) Find ∫ ∫ .
2) Evaluate the integral∫ ∫ .
3) Evaluate∫ ∫ .
4) Evaluate∫ ∫ ( )( )
.
5) Evaluate∬ − , where S is a triangle with vertices (0,
0), (10, 1), and (1, 1).
6) Evaluate ∬( + ) over the area of the triangle whose
vertices are (0, 1), (1, 0), (1, 2).
7) Evaluate ∬ over the area bounded by = 0, = ,
+ = 2 in the first quadrant.
8) Evaluate ∬ over the region R given by + − 2 =
0, = 2 , = .
4.1 CHANGE OF ORDER OF INTEGRATION:
On changing the order of integration, the limits of the integration change. To
find the new limits, draw the rough sketch of the region of integration.
Some of the problems connected with double integrals, which seem to be
complicated can be made easy to handle by a change in the order of
integration.
4.2 Examples:
Example 1: Evaluate ∫ ∫
∞∞
.
Solution: We have, ∫ ∫
∞∞
Here the elementary strip extends from = to = ∞ and this vertical
strip slides from
= 0 = ∞. The shaded portion of the figure is, therefore, the region of
integration.
11. Unit: 3
RAI UNIVERSITY, AHMEDABAD
On changing the order of integration, we first integrate w.r.t
horizontal strip which extends from
region, we then integrate w.r.t
= 0 = ∞.
Thus
∞
=
=
=
=
=
=
=
Example 2:Change the order of integration in
evaluate the same.
Solution: We have =
The region of integration is shown by shaded portion in the figure bounded by
parabola = , = 2 − ,
The point of intersection of the parabola
In the figure below (left) we draw a strip parallel to y
from to 2 − and varies from 0 to 1.
MULTIPLE INTEGRAL
On changing the order of integration, we first integrate w.r.t
which extends from = 0 to = . To cover
then integrate w.r.t ′ ′ from
∞
=
∞
∫ [ ]
∞
∫
∞
∫
∞
∞
−
∞
−
∞
− 1
1 ________
Change the order of integration in = ∫ ∫
∫ ∫
The region of integration is shown by shaded portion in the figure bounded by
, = 0 ( − ).
The point of intersection of the parabola = and the line = 2 −
In the figure below (left) we draw a strip parallel to y-axis and the strip y, varies
varies from 0 to 1.
MULTIPLE INTEGRAL
11
On changing the order of integration, we first integrate w.r.t along a
cover the given
________Answer
and hence
The region of integration is shown by shaded portion in the figure bounded by
− is (1,1).
axis and the strip y, varies
12. Unit: 3
RAI UNIVERSITY, AHMEDABAD
On changing the order of integration we have taken a strip parallel to x
area and second strip in the area
and and the limits of in the area
So, the given integral is
=
=
=
=
=
=
=
=
4.3 EXERCISE:
1) Change the order of the
2) Evaluate ∫ ∫
3) Change the order of integration and
5.1 CHANGE OF VARIABLE
Sometimes the problems of double integration can be solved easily by
change of independent variables. Let the double
as∬ ( , ) . It is to be changed by the new variables
MULTIPLE INTEGRAL
On changing the order of integration we have taken a strip parallel to x
and second strip in the area . The limits of in the area
in the area are 0 and 2 − .
So, the given integral is
∫ ∫ + ∫ ∫
√
∫
√
+ ∫
∫ + ∫ (2 − )
+ ∫ (4 − 4 + )
+
+
________Answer
Change the order of the integration∫ ∫
∞
.
by changing the order of integration.
Change the order of integration and evaluate∫ ∫
CHANGE OF VARIABLE:
Sometimes the problems of double integration can be solved easily by
change of independent variables. Let the double integral be
. It is to be changed by the new variables ,
MULTIPLE INTEGRAL
12
On changing the order of integration we have taken a strip parallel to x-axis in the
in the area are 0
Answer
by changing the order of integration.
.
Sometimes the problems of double integration can be solved easily by
.
13. Unit: 3
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The relation of , with
the double integration is converted into.
1. ∬ ( , )
( ,
5.2 Example 1: Using
the square R
∬ ( +
Integration being taken over the area bounded by the lines
= 0, − = 2, −
Solution: +
−
On solving (1) and (2), we get
= (
=
( ,
( ,
MULTIPLE INTEGRAL
with , are given as = ∅( , ), = (
the double integration is converted into.
) = ∬ { ( , ), ( , )}′ | | ,
=
( , )
( , )
( ) = { ( , ), ( , )}
( ,
( ,
+ = , − = , evaluate the double integral
( + )
Integration being taken over the area bounded by the lines +
− = 0.
+ = ________(1)
− = ________(2)
On solving (1) and (2), we get
( + ), = ( − )
)
)
= =
−
= − − = −
MULTIPLE INTEGRAL
13
( , ). Then
)
)
evaluate the double integral over
+ = 2, +
−
14. Unit: 3 MULTIPLE INTEGRAL
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( + ) =
1
4
( + ) +
1
4
( − )
( , )
( , )
= ∫ ∫ ( + ) −
= − ∫ ∫ ( + )
= − ∫ +
= − ∫ + 2
= − ∫ + 2
= − +
= − (2) + (2)
= − +
= −
= − ________Answer
5.3 EXERCISE:
1) Using the transformation + = , = show that
∫ ∫ ( )⁄
= ( − 1)
2) Evaluate ∬ ( + ) , where R is the parallelogram in the xy-plane
with vertices (1,0), (3,1), (2,2), (0,1), using the transformation = +
and = − 2 .
6.1AREA IN CARTESIAN CO-ORDINATES:
Area = ∫ ∫
6.2Example 1: Find the area bounded by the lines
= +
= − +
=
15. Unit: 3 MULTIPLE INTEGRAL
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Solution: The region of integration is bounded by the lines
= + 2 _________(1)
= − + 2 _________(2)
= 5 _________(3)
On solving (1) and (2), we get the point (0,2)
On solving (2) and (3), we get the point (5, −3)
On solving (1) and (3), we get the point (5,7)
We draw a strip parallel to -axis.
On this strip the limits of are = − + 2 and = + 2, and the limit of
are = 0 and = 5.
Required area = Shaded portion of the figure
= ∬
= ∫ ∫–
= ∫ [ ]
= ∫ [ + 2 − (− + 2)]
= ∫ [2 ]
=
= [ ]
= [25 − 0]
= 25Sq. units ________Answer
16. Unit: 3
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Example 2: Find the area between the parabolas
Solution: We have,
On solving the equations (1) and (2) we get the point of intersection (4a, 4a).
Divide the area into horizontal strips of width
, , 4 and then
∴The required area = ∫
=
=
=
=
=
7.1 VOLUME OF SOLIDS BY DOUBLE INTEGRAL
Let a surface ′
be =
The projection of ′ on
Take infinite number of elementary rectangles
the of height .
Volume of each vertical rod
=
MULTIPLE INTEGRAL
: Find the area between the parabolas = and
= 4 ________ (1)
= 4 ________ (2)
On solving the equations (1) and (2) we get the point of intersection (4a, 4a).
Divide the area into horizontal strips of width , varies from
and then varies from ( = 0) ( = 4
∫ ⁄
∫ [ ]
∫ 4 −
√4
⁄
−
√
(4 ) ⁄
−
( )
________Answer
VOLUME OF SOLIDS BY DOUBLE INTEGRAL:
= ( , )
on − plane be .
Take infinite number of elementary rectangles . Erect vertical rod on
Volume of each vertical rod = Area of the base × height
.
MULTIPLE INTEGRAL
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= .
On solving the equations (1) and (2) we get the point of intersection (4a, 4a).
varies from
).
Answer
. Erect vertical rod on
17. Unit: 3
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Volume of the solid cylinder on S
=
=
Here the integration is carried out over the area S.
Example 1: Find the volume bounded by the
= + and the cylinder
Solution: Here, we have
2 = + ⇒ 2
+ = 4 ⇒ =
Volume of one vertical rod
Volume of the solid =
= 2 ∫
= ∫
= ∫
= ∫
= 4 ∫
MULTIPLE INTEGRAL
Volume of the solid cylinder on S = lim →
→
∑ ∑
∬
∬ ( , )
Here the integration is carried out over the area S.
: Find the volume bounded by the -plane, the paraboloid
and the cylinder + = .
Here, we have
= ⇒ = (Paraboloid)
2, = 0, (circle)
Volume of one vertical rod = .
= ∬
∫
∫
= 4[ ]
MULTIPLE INTEGRAL
17
plane, the paraboloid
______ (1)
______ (2)
18. Unit: 3 MULTIPLE INTEGRAL
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= 4 ________Answer
8.1 VOLUME OF SOLID BY ROTATION OF AN AREA (DOUBLE
INTEGRAL):
When the area enclosed by a curve = ( ) is revolved about an axis, a
solid is generated; we have to find out the volume of solid generated.
Volume of the solid generated about x-axis = ∫ ∫ 2
( )
( )
Example 1: Find the volume of the torus generated by revolving the circle
+ = about the line = .
Solution: + = 4
= (2 )
= 2 (3 − )
= 2 (3 − )
√
√
(3 − )
= 2 3 4 − − 4 − + 3 4 − − 4 −
= 4 3 4 − − 4 −
= 4 3
2
4 − + 3 ×
4
2
sin
2
+
1
3
(4 − ) ⁄
19. Unit: 3 MULTIPLE INTEGRAL
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= 4 6 ×
2
+ 6 ×
2
= 24 ________ Ans.
8.2 EXERCISE:
1) Find the area of the ellipse + = 1
2) Find by double integration the area of the smaller region bounded by
+ = and + = .
3) Find the volume bounded by − , the cylinder + = 1
and the plane + + = 3.
4) Evaluate the volume of the solid generated by revolving the area of
the parabola = 4 bounded by the latus rectum about the tangent
at the vertex.
9.1TRIPLE INTEGRATION (VOLUME) :
Let a function ( , , ) be a continuous at every point of a finite region of
three dimensional spaces. Consider sub-spaces , , , … . of the
space S.
If ( , , ) be a point in the rth
subspace.
The limit of the sum ∑ ( , , ) , → ∞, → 0 is known
as the triple integral of ( , , ) over the space S.
Symbolically, it is denoted by
( , , )
20. Unit: 3 MULTIPLE INTEGRAL
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It can be calculated as∫ ∫ ∫ ( , , ) . First we integrate with
respect to treating , as constant between the limits . The
resulting expression (function of , ) is integrated with respect to keeping
as constant between the limits . At the end we integrate the
resulting expression (function of only) within the limits .
First we integrate from inner most integral w.r.t z, and then we integrate
w.r.t , and finally the outer most w.r.t .
But the above order of integration is immaterial provided the limits change
accordingly.
Example 1: Evaluate
∭ ( − 2 + ) , ℎ :
0 ≤ ≤ 1
0 ≤ ≤
0 ≤ ≤ +
Solution:∭ ( − 2 + )
= ∫ ∫ ∫ ( − 2 + )
= ∫ ∫ − 2 +
= ∫ ∫ ( + ) − 2 ( + ) +
( )
= ∫ ∫ + − 2 − 2 +
( )
= ∫ ∫ − − 2 + + +
