B.tech ii unit-3 material multiple integration

Unit: 3 MULTIPLE INTEGRAL
RAI UNIVERSITY, AHMEDABAD 1
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 in Cartesian 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 ReferenceBook 21

MULTIPLE INTEGRALS
1.1 DOUBLE INTEGRALS:
We Know that
∫ 𝑓( 𝑥) 𝑑𝑥 = lim𝑛→∞
𝛿𝑥→0
[ 𝑓( 𝑥1) 𝛿𝑥1 + 𝑓( 𝑥2) 𝛿𝑥2 + 𝑓( 𝑥3) 𝛿𝑥3 + ⋯+
𝑏
𝑎
𝑓( 𝑥 𝑛) 𝛿𝑥 𝑛]
Let us consider a function 𝑓(𝑥, 𝑦) of two variables 𝑥 and 𝑦 defines in
the finite region A of 𝑥𝑦- plane. Divide the region 𝐴 into elementary areas.
𝛿𝐴1, 𝛿𝐴2, 𝛿𝐴3,… 𝛿𝐴 𝑛
Then ∬ 𝑓( 𝑥, 𝑦) 𝑑𝐴 =𝐴
lim𝑛→∞
𝛿𝐴→0
[ 𝑓( 𝑥1, 𝑦1) 𝛿𝐴1 + 𝑓( 𝑥2, 𝑦2) 𝛿𝐴2 + ⋯+
𝑓( 𝑥 𝑛, 𝑦𝑛) 𝛿𝐴 𝑛]
2.1 Evaluation of Double Integral:
Double integral over region A may be evaluated by
two successiveintegrations.
If A is described as 𝑓1(𝑥) ≤ 𝑦 ≤ 𝑓2(𝑥) [ 𝑦1 ≤ y ≤ 𝑦2]
And 𝑎 ≤ 𝑥 ≤ 𝑏,
Then ∬ 𝑓( 𝑥, 𝑦) 𝑑𝐴 = ∫ ∫ 𝑓(𝑥, 𝑦)
𝑦2
𝑦1
𝑑𝑥 𝑑𝑦
𝑏
𝑎𝐴

2.1.1 FIRST METHOD:
∬ 𝑓( 𝑥, 𝑦) 𝑑𝑥𝑑𝑦 = ∫ [∫ 𝑓( 𝑥, 𝑦) 𝑑𝑦
𝑦2
𝑦1
] 𝑑𝑥
𝑥2
𝑥1𝐴
𝑓(𝑥, 𝑦) is first integrated with respectto y treating 𝑥 as constant between the
limits 𝑦1 and 𝑦2 and then the result is integrated with respectto 𝑥 between
the limits 𝑥1 and 𝑥2.
In the region we take an elementary area 𝛿𝑥𝛿𝑦. Then integration w.r.t to 𝑦
(𝑥 keeping constant) converts small rectangle 𝛿𝑥𝛿𝑦 into a strip 𝑃𝑄(𝑦 𝛿𝑥).
While the integration of the result w.r.t 𝑥 correspondsto the sliding to the
strip, from 𝐴𝐷 to 𝐵𝐶 covering the whole region 𝐴𝐵𝐶𝐷.
2.1.2 SECOND METHOD:
∬ 𝑓( 𝑥, 𝑦) 𝑑𝑥𝑑𝑦 = ∫ [ ∫ 𝑓( 𝑥, 𝑦) 𝑑𝑥
𝑥2
𝑥1
] 𝑑𝑦
𝑦2
𝑦1𝐴
Here 𝑓(𝑥, 𝑦) is first integrated w.r.t 𝑥 keeping 𝑦 constant between the limits
𝑥1 and 𝑥2 and then the resulting expression is integrated with respectto 𝑦
between the limits 𝑦1 and 𝑦2 and vice versa.
NOTE:Forconstant limits, it does not matter whether we first integrate
w.r.t 𝑥 and then w.r.t 𝑦 or vice versa.

2.2 Examples:
Example 1: Find ∫ ∫ 𝑒 𝑦 𝑥⁄
𝑑𝑦 𝑑𝑥
𝑥2
0
1
0
Solution: Here, we have
∫ [∫ 𝑒 𝑦 𝑥⁄
𝑑𝑦
𝑥2
0
] 𝑑𝑥 = ∫ [
𝑒 𝑦 𝑥⁄
1 𝑥⁄
]
0
𝑥2
𝑑𝑥
1
0
1
0
= ∫
( 𝑒 𝑥−1)
1 𝑥⁄
𝑑𝑥
1
0
= ∫ 𝑥 𝑒 𝑥
𝑑𝑥 − ∫ 𝑥 𝑑𝑥
1
0
1
0
= [ 𝑥 𝑒 𝑥
− 𝑒 𝑥]0
1
− [
𝑥2
2
]
0
1
= 𝑒1
− 𝑒1
+ 1 −
1
2
=
1
2
∴ ∫ ∫ 𝒆 𝒚 𝒙⁄
𝒅𝒚 𝒅𝒙
𝒙 𝟐
𝟎
𝟏
𝟎
=
𝟏
𝟐
________ Answer
Example 2: Evaluate ∫ ∫ 𝑒−𝑥2(1+𝑦2)
𝑥 𝑑𝑥 𝑑𝑦
∞
0
∞
0
∫ ∫ 𝑒−𝑥2(1+𝑦2)
∞
0
∞
0
= ∫ 𝑑𝑦 ∫ 𝑒−𝑥2(1+𝑦2)
∞
0
𝑥 𝑑𝑥
∞
0
= ∫ 𝑑𝑦 [
𝑒−𝑥2(1+𝑦2)
−2(1+𝑦2)
]
0
∞
∞
0
= ∫ [0 +
1
2(1+𝑦2)
] 𝑑𝑦
∞
0
=
1
2
[tan−1
𝑦]0
∞
=
1
2
[tan−1
∞ − tan−1
0]
=
1
2
(
𝜋
2
)
=
𝜋
4

∴ ∫ ∫ 𝑒−𝑥2(1+𝑦2)
∞
0
∞
0
=
𝜋
4
________ Answer
Example 3: Sketch the area of integration and evaluate ∫ ∫ 2𝑥2
𝑦2
𝑑𝑥 𝑑𝑦
√2−𝑦
−√2−𝑦
2
1
.
Solution: Here we have
∫ ∫ 2𝑥2
𝑦2
𝑑𝑥 𝑑𝑦
√2−𝑦
−√2−𝑦
2
1
= 2∫ 𝑦2
𝑑𝑦
2
1
∫ 𝑥2
𝑑𝑥
√2−𝑦
−√2−𝑦
= 4 ∫ 𝑦2
𝑑𝑦
2
1
∫ 𝑥2
𝑑𝑥
√2−𝑦
0
[
∵ ∫ 𝑓( 𝑥) 𝑑𝑥 = 2∫ 𝑓( 𝑥) 𝑑𝑥
𝑎
0
𝑎
−𝑎
𝑤ℎ𝑒𝑟𝑒 𝑥2
𝑖𝑠 𝑎𝑛 𝑒𝑣𝑒𝑛 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
]
= 4∫ 𝑦2
𝑑𝑦 [
𝑥3
3
]
0
√2−𝑦2
1
=
4
3
∫ 𝑦2
𝑑𝑦 (2− 𝑦)
3
2
2
1
=
4
3
∫ (2 − 𝑦)2
𝑡3 2⁄
(−𝑑𝑡)
0
1
[
𝑝𝑢𝑡 2 − 𝑦 = 𝑡
∴ 𝑑𝑦 = −𝑑𝑡
]
=
4
3
[(2 − 𝑡)2
(
2𝑡
5
2
5
) − (−2)(2− 𝑡)
2
5
.
2
7
𝑡
7
2 + (2)
2
5
.
2
7
.
2
9
𝑡
9
2]
0
1
=
4
3
[
2
5
+ 2 (
2
5
.
2
7
) + (2)(
2
5
.
2
7
.
2
9
)]
=
4
3
[
2
5
+
8
35
+
16
315
]
=
4
15
[2 +
8
7
+
16
63
]
=
4
15
[
126+72+16
63
]
=
4
15
(
214
63
)
=
856
945

∴ ∫ ∫ 𝟐𝒙 𝟐
𝒚 𝟐
𝒅𝒙 𝒅𝒚
√ 𝟐−𝒚
−√ 𝟐−𝒚
𝟐
𝟏
=
𝟖𝟓𝟔
𝟗𝟒𝟓
________ Answer
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 𝑑𝑥 𝑑𝑦. The integration w.r.t 𝑥
between the limits 𝑥1, 𝑥2 keeping 𝑦 fixed
indicates that integration is done, along 𝑃𝑄. Then
the integration of result w.r.t to 𝑦 correspondsto sliding the strips 𝑃𝑄 from
𝐵𝐶 to 𝐴𝐷 covering the whole region 𝐴𝐵𝐶𝐷.
We can also integrate first w.r.t ‘𝑦’ then w.r.t 𝑥, which ever is convenient.
Example 4: Evaluate ∬ 𝑥𝑦 𝑑𝑥 𝑑𝑦 over the region in the positive quadrant
for which 𝑥 + 𝑦 ≤ 1.
Solution: 𝑥 + 𝑦 = 1 represents a line AB in the
figure.
𝑥 + 𝑦 < 1 represents a plane 𝑂𝐴𝐵.
The region for integration is 𝑂𝐴𝐵 as shaded in
the figure.
By drawing 𝑃𝑄 parallel to y-axis, 𝑝 lies on the line 𝐴𝐵
𝑖. 𝑒. , (𝑥 + 𝑦 = 1) & Q lies on x-axis. The limit for 𝑦 is 1 − 𝑥 and 0.

Required integral =
∫ 𝑑𝑥 ∫ 𝑦 𝑑𝑦
1−𝑥
0
= ∫ 𝑥 𝑑𝑥 [
𝑦2
2
]
0
1−𝑥1
0
1
0
=
1
2
∫ ( 𝑥 𝑑𝑥) (1− 𝑥)21
0
=
1
2
∫ ( 𝑥 − 2𝑥2
+ 𝑥3) 𝑑𝑥
1
0
=
1
2
[
𝑥2
2
−
2𝑥3
3
+
𝑥4
4
]
0
1
=
1
2
[
1
2
−
2
3
+
1
4
]
=
1
24
________ Answer
Example 5: Evaluate ∬ 𝑥𝑦 𝑑𝑥 𝑑𝑦,𝑅
where 𝑅 the quadrant of the circle is
𝑥2
+ 𝑦2
= 𝑎2
where 𝑥 ≥ 0 𝑎𝑛𝑑 𝑦 ≥ 0.
Solution: Let the region of integration be the first quadrant of the
circle 𝑂𝐴𝐵.
Let 𝐼 = ∬ 𝑥𝑦 𝑑𝑥 𝑑𝑦𝑅
(𝑥2
+ 𝑦2
= 𝑎2
, 𝑦 = √ 𝑎2 − 𝑥2)
First we integrate w.r.t 𝑦 and then w.r.t 𝑥.
The limits for 𝑦 are 0 and √ 𝑎2 − 𝑥2 and for x, 0 to a.
𝐼 = ∫ 𝑥 𝑑𝑥 ∫ 𝑦 𝑑𝑦
√𝑎2−𝑥2
0
𝑎
0
= ∫ 𝑥 𝑑𝑥 [
𝑦2
2
]
0
√𝑎2−𝑥2
𝑎
0
=
1
2
∫ 𝑥 ( 𝑎2
− 𝑥2) 𝑑𝑥
𝑎
0
=
1
2
[
𝑎2 𝑥2
2
−
𝑥4
4
]
0
𝑎
=
𝑎4
8
________ Answer

Example 6: Evaluate ∬ 𝑥2
𝑑𝑥 𝑑𝑦𝐴
, where A is the region in the first quadrant
bounded by the hyperbola 𝑥𝑦 = 16 and the lines 𝑦 = 𝑥, 𝑦 = 0 𝑎𝑛𝑑 𝑥 = 8.
Solution: The line 𝑂𝑃, 𝑦 = 𝑥 and the curve 𝑃𝑆, 𝑥𝑦 = 16 intersect at 𝑝(4,4).
The line 𝑆𝑁, 𝑥 = 8 intersects the hyperbola at 𝑆(8,2). And 𝑦 = 0 is x-axis.
The area A is shown shaded.
Divide the area into two parts by PM perpendicular to OX.
For the area 𝑂𝑀𝑃, 𝑦 varies from 0 to 𝑥, and then 𝑥 varies from 0 to 4.
For the area 𝑃𝑀𝑁𝑆, 𝑦 varies from 0 to
16
𝑥
and then 𝑥 varies from 4 to 8.
∴ ∬ 𝑥2
𝐴
𝑑𝑥 𝑑𝑦 = ∫ ∫ 𝑥2
𝑑𝑥 𝑑𝑦 + ∫ ∫ 𝑥2
𝑑𝑥 𝑑𝑦
16
𝑥
0
8
4
𝑥
0
4
0
= ∫ 𝑥2
𝑑𝑥
4
0
∫ 𝑑𝑦
𝑥
0
+ ∫ 𝑥2
𝑑𝑥 ∫ 𝑑𝑦
16
𝑥
0
8
4
= ∫ 𝑥2 [ 𝑦]0
𝑥
𝑑𝑥 + ∫ 𝑥2[ 𝑦]
0
16
𝑥
𝑑𝑥
8
4
4
0
= ∫ 𝑥3
𝑑𝑥 + ∫ 16 𝑥 𝑑𝑥
8
4
4
0
= [
𝑥4
4
]
0
4
+ 16[
𝑥2
2
]
4
8
= 64 + 8 (82
− 42
)
= 64 + 384
= 448 ________ Answer
3.2 EXERCISE:
1) Find ∫ ∫ 𝑥 𝑦 𝑒−𝑥2
𝑑𝑥 𝑑𝑦.
𝑦
0
1
0
2) Evaluate the integral ∫ ∫ 𝑒 𝑥+𝑦
𝑑𝑥 𝑑𝑦
log 𝑦
0
log8
1
.
3) Evaluate ∫ ∫
𝑥2+𝑦2
𝑥
𝑥
𝑎
𝑎
0
.

4) Evaluate ∫ ∫
𝑑𝑥 𝑑𝑦
√(1−𝑥2)(1−𝑦2)
1
0
1
0
.
5) Evaluate ∬ √ 𝑥𝑦 − 𝑦2 𝑑𝑦 𝑑𝑥𝑆
, where S is a triangle with vertices (0,
0), (10, 1), and (1, 1).
6) Evaluate ∬( 𝑥2
+ 𝑦2) 𝑑𝑥 𝑑𝑦 over the area of the triangle whose
vertices are (0, 1), (1, 0), (1, 2).
7) Evaluate ∬ 𝑦 𝑑𝑥 𝑑𝑦 over the area bounded by 𝑥 = 0, 𝑦 = 𝑥2
,
𝑥 + 𝑦 = 2 in the first quadrant.
8) Evaluate ∬ 𝑥𝑦 𝑑𝑥𝑑𝑦 over the region R given by 𝑥2
+ 𝑦2
− 2𝑥 =
0, 𝑦2
= 2𝑥, 𝑦 = 𝑥.
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 ∫ ∫
𝑒 −𝑦
𝑦
𝑑𝑥 𝑑𝑦
∞
𝑥
∞
0
.
Solution: We have, ∫ ∫
𝑒−𝑦
𝑦
𝑑𝑥 𝑑𝑦
∞
𝑥
∞
0
Here the elementary strip 𝑃𝑄 extends from 𝑦 = 𝑥 to 𝑦 = ∞ and this vertical
strip slides from
𝑥 = 0 𝑡𝑜 𝑥 = ∞. The shaded portion of the figure is, therefore, the region of
integration.
On changing the order of integration, we first integrate w.r.t 𝑥 along a
horizontal strip 𝑅𝑆 which extends from 𝑥 = 0 to 𝑥 = 𝑦. To cover the given
region, we then integrate w.r.t ′𝑦′ from
𝑦 = 0 𝑡𝑜 𝑦 = ∞.

Thus
∫ 𝑑𝑥
∞
0
∫
𝑒−𝑦
𝑦
∞
𝑥
𝑑𝑦 = ∫
𝑒−𝑦
𝑦
𝑑𝑦 ∫ 𝑑𝑥
𝑦
0
∞
0
= ∫
𝑒−𝑦
𝑦
𝑑𝑦 [ 𝑥]0
𝑦∞
0
= ∫ 𝑦
∞
0
𝑒−𝑦
𝑦
𝑑𝑦
= ∫ 𝑒−𝑦
𝑑𝑦
∞
0
= [
𝑒−𝑦
−1
]
0
∞
= −[
1
𝑒 𝑦
]
0
∞
= −[
1
∞
− 1]
= 1 ________ Answer
Example 2: Change the order of integration in 𝐼 = ∫ ∫ 𝑥𝑦 𝑑𝑥 𝑑𝑦
2−𝑥
𝑥2
1
0
and hence
evaluate the same.
Solution: We have 𝐼 = ∫ ∫ 𝑥𝑦 𝑑𝑥 𝑑𝑦
2−𝑥
𝑥2
1
0
The region of integration is shown by shaded portion in the figure bounded by
parabola 𝑦 = 𝑥2
, 𝑦 = 2 − 𝑥, 𝑥 = 0 (𝑦 − 𝑎𝑥𝑖𝑠).
The point of intersection of the parabola 𝑦 = 𝑥2
and the line 𝑦 = 2 − 𝑥 is 𝐵(1,1).
In the figure below (left) we draw a strip parallel to y-axis and the strip y, varies
from 𝑥2
to 2 − 𝑥 and 𝑥 varies from 0 to 1.

On changing the order of integration we have taken a strip parallel to x-axis in the
area 𝑂𝐵𝐶 and second strip in the area 𝐶𝐵𝐴. The limits of 𝑥 in the area 𝑂𝐵𝐶 are 0
and √ 𝑦 and the limits of 𝑥 in the area 𝐶𝐵𝐴 are 0 and 2 − 𝑦.
So, the given integral is
= ∫ 𝑦 𝑑𝑦 ∫ 𝑥 𝑑𝑥 + ∫ 𝑦 𝑑𝑦 ∫ 𝑥 𝑑𝑥
2−𝑦
0
2
1
√ 𝑦
0
1
0
= ∫ 𝑦 𝑑𝑦[
𝑥2
2
]
0
√ 𝑦
+ ∫ 𝑦 𝑑𝑦[
𝑥2
2
]
0
2−𝑦2
1
1
0
=
1
2
∫ 𝑦2
𝑑𝑦 +
1
2
∫ 𝑦(2 − 𝑦)2
𝑑𝑦
2
1
1
0
=
1
2
[
𝑦3
3
]
0
1
+
1
2
∫ (4𝑦 − 4𝑦2
+ 𝑦3
)
2
1
=
1
6
+
1
2
[
96−128+48−24+16−3
12
]
=
1
6
+
5
24
=
9
24
=
3
8
________ Answer
4.3 EXERCISE:
1) Change the order of the integration ∫ ∫ 𝑒−𝑥𝑦
𝑦 𝑑𝑦 𝑑𝑥
𝑥
0
∞
0
.
2) Evaluate ∫ ∫ 𝑑𝑥 𝑑𝑦
𝑒 𝑥
1
2
0
by changing the order of integration.
3) Change the order of integration and evaluate ∫ ∫
𝑥2 𝑑𝑥 𝑑𝑦
√𝑥4−4𝑦2
2
√2𝑦
2
0
.
5.1 CHANGE OF VARIABLE:
Sometimes the problems of double integration can be solved easily by
change of independent variables. Let the double integral be
as ∬ 𝑓( 𝑥, 𝑦) 𝑑𝑥𝑑𝑦𝑅
. It is to be changed by the new variables 𝑢, 𝑣.
The relation of 𝑥, 𝑦 with 𝑢, 𝑣 are given as 𝑥 = ∅( 𝑢, 𝑣), 𝑦 = 𝜓(𝑢, 𝑣). Then
the double integration is converted into.
1. ∬ 𝑓( 𝑥, 𝑦) 𝑑𝑥 𝑑𝑦𝑅
= ∬ 𝑓{ 𝜙( 𝑢, 𝑣), 𝜓(𝑢, 𝑣)}𝑅′ | 𝐽| 𝑑𝑢 𝑑𝑣,
[𝑑𝑥 𝑑𝑦 =
𝜕 (𝑥,𝑦)
𝜕 (𝑢,𝑣)
𝑑𝑢 𝑑𝑣]

∬ 𝒇( 𝒙,𝒚) 𝒅𝒙 𝒅𝒚 = ∬ 𝒇{ 𝒙( 𝒖, 𝒗), 𝒚(𝒖, 𝒗)} |
𝝏 (𝒙,𝒚)
𝝏 (𝒖,𝒗)
|𝑫𝑹
𝒅𝒖 𝒅𝒗
5.2 Example 1: Using 𝑥 + 𝑦 = 𝑢, 𝑥 − 𝑦 = 𝑣, evaluate the double integral over
the square R
∬ ( 𝑥2
+ 𝑦2) 𝑑𝑥 𝑑𝑦𝑅
Integration being taken over the area bounded by the lines 𝑥 + 𝑦 = 2, 𝑥 +
𝑦 = 0, 𝑥 − 𝑦 = 2, 𝑥 − 𝑦 = 0.
Solution: 𝑥 + 𝑦 = 𝑢 ________(1)
𝑥 − 𝑦 = 𝑣 ________(2)
On solving (1) and (2), we get
𝑥 =
1
2
( 𝑢 + 𝑣), 𝑦 =
1
2
( 𝑢 − 𝑣)
𝐽 =
𝜕 (𝑥,𝑦)
𝜕 (𝑢,𝑣)
= |
𝜕𝑥
𝜕𝑢
𝜕𝑥
𝜕𝑣
𝜕𝑦
𝜕𝑢
𝜕𝑦
𝜕𝑣
| = |
1
2
1
2
1
2
−
1
2
| = −
1
4
−
1
4
= −
1
2
∬ ( 𝑥2
+ 𝑦2) 𝑑𝑥 𝑑𝑦 = ∫ ∫ [
1
4
(𝑢 + 𝑣)2
+
1
4
(𝑢 −
2
0
2
0𝑅
𝑣)2
] |
𝜕 (𝑥,𝑦)
𝜕 (𝑢,𝑣)
| 𝑑𝑢 𝑑𝑣
= ∫ ∫
1
2
( 𝑢2
+ 𝑣2)|−
1
2
| 𝑑𝑢 𝑑𝑣
2
0
2
0
= −
1
4
∫ 𝑑𝑣∫ ( 𝑢2
+ 𝑣2) 𝑑𝑢
2
0
2
0
= −
1
4
∫ 𝑑𝑣 [
𝑢3
3
+ 𝑢𝑣2
]
0
22
0
= −
1
4
∫ 𝑑𝑣(
8
3
+ 2𝑣2
)
2
0
= −
1
4
∫ (
8
3
+ 2𝑣2
)
2
0
𝑑𝑣

= −
1
4
[
8
3
𝑣 +
2
3
𝑣3
]
0
2
= −
1
4
[
8
3
(2) +
2
3
(2)3
]
= −
1
4
(
16
3
+
16
3
)
= −
1
4
(
32
3
)
= −
8
3
________ Answer
5.3 EXERCISE:
1) Using the transformation 𝑥 + 𝑦 = 𝑢, 𝑦 = 𝑢𝑣 show that
∫ ∫ 𝑒 𝑦 (𝑥+𝑦)⁄
𝑑𝑦 𝑑𝑥 =
1
2
(𝑒 − 1)
1−𝑥
0
1
0
2) Evaluate ∬ (𝑥 + 𝑦)2
𝑑𝑥 𝑑𝑦𝑅
, where R is the parallelogram in the xy-plane
with vertices (1,0), (3,1), (2,2), (0,1), using the transformation 𝑢 = 𝑥 + 𝑦
and 𝑣 = 𝑥 − 2𝑦.
6.1 AREA IN CARTESIAN CO-ORDINATES:
Area = ∫ ∫ 𝒅𝒙 𝒅𝒚
𝒚 𝟐
𝒚 𝟏
𝒃
𝒂
6.2 Example 1: Find the area bounded by the lines
𝒚 = 𝒙 + 𝟐
𝒚 = −𝒙 + 𝟐
𝒙 = 𝟓
Solution: The region of integration is bounded by the lines
𝑦 = 𝑥 + 2 _________(1)
𝑦 = −𝑥 + 2 _________(2)
𝑥 = 5 _________(3)
On solving (1) and (2), we get the point 𝐴(0,2)
On solving (2) and (3), we get the point 𝐶(5,−3)
On solving (1) and (3), we get the point 𝐸(5,7)

We draw a strip parallel to 𝑦-axis.
On this strip the limits of 𝑦 are 𝑦 = −𝑥 + 2 and 𝑦 = 𝑥 + 2, and the limit of
𝑥 are 𝑥 = 0 and 𝑥 = 5.
Required area = Shaded portion of the figure
= ∬ 𝑑𝑥 𝑑𝑦
= ∫ 𝑑𝑥 ∫ 𝑑𝑦
𝑥+2
–𝑥+2
5
0
= ∫ 𝑑𝑥 [ 𝑦]−𝑥+2
𝑥+25
0
= ∫ 𝑑𝑥 [ 𝑥 + 2 − (−𝑥 + 2)]
5
0
= ∫ [2𝑥] 𝑑𝑥
5
0
= [
2𝑥2
2
]
0
5
= [ 𝑥2]0
5
= [25− 0]
= 25 Sq. units ________ Answer
Example 2: Find the area betweenthe parabolas 𝒚 𝟐
= 𝟒𝒂𝒙 and 𝒙 𝟐
= 𝟒𝒂𝒚.
Solution: We have, 𝑦2
= 4𝑎𝑥 ________ (1)
𝑥2
= 4𝑎𝑦 ________ (2)
On solving the equations (1) and (2) we get the point of intersection (4a, 4a).
Divide the area into horizontal strips of width 𝛿𝑦, 𝑥 varies from
𝑃,
𝑦2
4𝑎
𝑡𝑜 𝑄,√4𝑎𝑦 and then 𝑦 varies from 𝑂 ( 𝑦 = 0) 𝑡𝑜 𝐴 (𝑦 = 4𝑎).
∴ The required area = ∫ 𝑑𝑦 ∫ 𝑑𝑥
√4𝑎𝑦
𝑦2 4𝑎⁄
4𝑎
0

= ∫ 𝑑𝑦 [ 𝑥]
𝑦2
4𝑎
√4𝑎𝑦4𝑎
0
= ∫ 𝑑𝑦 [√4𝑎𝑦 −
𝑦2
4𝑎
]
4𝑎
0
= [√4𝑎
𝑦3 2⁄
3
2
−
𝑦3
12𝑎
]
0
4𝑎
=
4√𝑎
3
(4𝑎)3 2⁄
−
(4𝑎)3
12𝑎
=
16
3
𝑎2
________ Answer
7.1 VOLUME OF SOLIDS BY DOUBLE INTEGRAL:
Let a surface 𝑆′
be 𝑧 = 𝑓(𝑥, 𝑦)
The projection of 𝑠′ on 𝑥 − 𝑦 plane be 𝑆.
Take infinite number of elementary rectangles 𝛿𝑥 𝛿𝑦. Erect vertical rod on
the 𝛿𝑥 𝛿𝑦 of height 𝑧.
Volume of each vertical rod = Area of the base × height
= 𝛿𝑥 𝛿𝑦 . 𝑧
Volume of the solid cylinder on S = lim
𝛿𝑥→0
𝛿𝑦→0
∑ ∑ 𝑧 𝑑𝑥 𝑑𝑦
= ∬ 𝑧 𝑑𝑥 𝑑𝑦
= ∬ 𝑓( 𝑥, 𝑦) 𝑑𝑥 𝑑𝑦
Here the integration is carried out over the area S.

Example 1: Find the volume bounded by the 𝒙𝒚-plane, the paraboloid
𝟐𝒛 = 𝒙 𝟐
+ 𝒚 𝟐
and the cylinder 𝒙 𝟐
+ 𝒚 𝟐
= 𝟒.
2𝑧 = 𝑥2
+ 𝑦2
⇒ 2𝑧 = 𝑟2
⇒ 𝑧 =
𝑟2
2
(Paraboloid) ______(1)
𝑥2
+ 𝑦2
= 4 ⇒ 𝑟 = 2, 𝑧 = 0, (circle) ______(2)
Volume of one vertical rod = 𝑧. 𝑟 𝑑𝑟 𝑑𝜃
Volume of the solid = ∬ 𝑧 𝑟 𝑑𝑟 𝑑𝜃
= 2 ∫ 𝑑𝜃 ∫
𝑟2
2
𝑟 𝑑𝑟
2
0
𝜋
0
=
2
2
∫ 𝑑𝜃 ∫ 𝑟3
𝑑𝑟
2
0
𝜋
0
= ∫ 𝑑𝜃 (
𝑟4
4
)
0
2𝜋
0
= ∫ 𝑑𝜃 (
16
4
)
𝜋
0
= 4∫ 𝑑𝜃
𝜋
0
= 4[ 𝜃]0
𝜋
= 4𝜋 ________ Answer
8.1 VOLUME OF SOLID BY ROTATION OF AN AREA (DOUBLE
INTEGRAL):
When the area enclosed by a curve 𝑦 = 𝑓(𝑥) 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 = ∫ ∫ 2𝜋 𝑃𝑄 𝑑𝑥 𝑑𝑦
𝑦2(𝑥)
𝑦1(𝑥)
𝑏
𝑎

Example 1: Find the volume of the torus generatedby revolving the circle
𝒙 𝟐
+ 𝒚 𝟐
= 𝟒 about the line 𝒙 = 𝟑.
Solution: 𝑥2
+ 𝑦2
= 4
𝑉 = ∬(2𝜋 𝑃𝑄) 𝑑𝑥 𝑑𝑦
= 2𝜋 ∬(3 − 𝑥) 𝑑𝑥 𝑑𝑦
= 2𝜋 ∫ 𝑑𝑥 (3𝑦 − 𝑥𝑦)
−√4−𝑥2
+√4−𝑥22
−2
(3 − 𝑥) 𝑑𝑦
= 2𝜋 ∫ 𝑑𝑥 [3√4 − 𝑥2 − 𝑥√4 − 𝑥2 + 3√4 − 𝑥2 − 𝑥√4 − 𝑥2]
+2
−2
= 4𝜋∫ [3√4 − 𝑥2 − 𝑥√4 − 𝑥2]
+2
−2
𝑑𝑥
= 4𝜋[3
𝑥
2
√4 − 𝑥2 + 3 ×
4
2
sin−1 𝑥
2
+
1
3
(4 − 𝑥2)3 2⁄
]
−2
+2
= 4𝜋 [6 ×
𝜋
2
+ 6 ×
𝜋
2
]
= 24𝜋2
________ Ans.
8.2 EXERCISE:
1) Find the area of the ellipse
𝑥2
𝑎2
+
𝑦2
𝑏2
= 1
2) Find by double integration the area of the smaller region bounded by
𝑥2
+ 𝑦2
= 𝑎2
and𝑥 + 𝑦 = 𝑎.
3) Find the volume bounded by 𝑥𝑦 − 𝑝𝑙𝑎𝑛𝑒, the cylinder 𝑥2
+ 𝑦2
= 1
and the plane 𝑥 + 𝑦 + 𝑧 = 3.
4) Evaluate the volume of the solid generated by revolving the area of
the parabola 𝑦2
= 4𝑎𝑥 bounded by the latus rectum about the tangent
at the vertex.

9.1 TRIPLE INTEGRATION (VOLUME) :
Let a function 𝑓(𝑥, 𝑦, 𝑧) be a continuous at every point of a finite region 𝑆 of
three dimensional spaces. Consider 𝑛 sub-spaces 𝛿𝑠1, 𝛿𝑠2, 𝛿𝑠3,… . 𝛿𝑠 𝑛 of the
spaceS.
If (𝑥 𝑟, 𝑦𝑟, 𝑧 𝑟) be a point in the rth subspace.
The limit of the sum ∑ 𝑓(𝑥 𝑟, 𝑦𝑟, 𝑧 𝑟)
𝑛
𝑟=1 𝛿𝑠𝑟, 𝑎𝑠 𝑛 → ∞, 𝛿𝑠𝑟 → 0 is
known as the triple integral of 𝑓(𝑥, 𝑦, 𝑧) over the spaceS.
Symbolically, it is denoted by
∭𝑓( 𝑥, 𝑦, 𝑧) 𝑑𝑆
𝑆
It can be calculated as ∫ ∫ ∫ 𝑓( 𝑥, 𝑦, 𝑧) 𝑑𝑥 𝑑𝑦 𝑑𝑧
𝑧2
𝑧1
𝑦2
𝑦1
𝑥2
𝑥1
. First we integrate with
respect to 𝑧 treating 𝑥, 𝑦 as constant between the limits 𝑧1 𝑎𝑛𝑑 𝑧2 . The
resulting expression (function of 𝑥, 𝑦) is integrated with respect to 𝑦 keeping
𝑥 as constant between the limits 𝑦1 𝑎𝑛𝑑 𝑦2. At the end we integrate the
resulting expression (function of 𝑥 only) within the limits 𝑥1 𝑎𝑛𝑑 𝑥2.
First we integrate from inner most integral w.r.t z, and then we integrate
w.r.t 𝑦, and finally the outer most w.r.t 𝑥.
But the above order of integration is immaterial provided the limits change
accordingly.
Example 1: Evaluate ∭ ( 𝑥 − 2𝑦 +𝑅
𝑧) 𝑑𝑥 𝑑𝑦 𝑑𝑧, 𝑤ℎ𝑒𝑟𝑒 𝑅:
0 ≤ 𝑥 ≤ 1
0 ≤ 𝑦 ≤ 𝑥2
0 ≤ 𝑧 ≤ 𝑥 + 𝑦

Solution: ∭ ( 𝑥 − 2𝑦 + 𝑧) 𝑑𝑥 𝑑𝑦 𝑑𝑧𝑅
= ∫ 𝑑𝑥 ∫ 𝑑𝑦 ∫ (𝑥 − 2𝑦 + 𝑧)𝑑𝑧
𝑥+𝑦
0
𝑥2
0
1
0
= ∫ 𝑑𝑥 ∫ 𝑑𝑦 (𝑥𝑧 − 2𝑦𝑧 +
𝑧2
2
)
0
𝑥+𝑦𝑥2
0
1
0
= ∫ 𝑑𝑥 ∫ 𝑑𝑦 [𝑥 ( 𝑥 + 𝑦) − 2𝑦( 𝑥 + 𝑦) +
(𝑥+𝑦)2
2
]
𝑥2
0
1
0
= ∫ 𝑑𝑥 ∫ 𝑑𝑦 [𝑥2
+ 𝑥𝑦 − 2𝑥𝑦 − 2𝑦2
+
(𝑥+𝑦)2
2
]
𝑥2
0
1
0
= ∫ 𝑑𝑥 ∫ 𝑑𝑦 [𝑥2
− 𝑥𝑦 − 2𝑦2
+
𝑥2
2
+ 𝑥𝑦 +
𝑦2
2
]
𝑥2
0
1
0
= ∫ 𝑑𝑥 ∫ 𝑑𝑦 [
3𝑥2
2
−
3𝑦2
2
]
𝑥2
0
1
0
=
3
2
∫ 𝑑𝑥
1
0
∫ ( 𝑥2
− 𝑦2) 𝑑𝑦
𝑥2
0
=
3
2
∫ 𝑑𝑥
1
0
(𝑥2
𝑦 −
𝑦3
3
)
0
𝑥2
=
3
2
∫ 𝑑𝑥
1
0
(𝑥4
−
𝑥6
3
)
=
3
2
(
𝑥5
5
−
𝑥7
21
)
0
1
=
3
2
(
1
5
−
1
21
)
=
8
35
________Answer
Example 2: Evaluate ∫ ∫ ∫ 𝑒 𝑥+𝑦+𝑧
𝑑𝑥 𝑑𝑦 𝑑𝑧
𝑥+𝑦
0
𝑥
0
𝑙𝑜𝑔 2
0
Solution: 𝐼 = ∫ ∫ 𝑒 𝑥+𝑦 [ 𝑒 𝑧]0
𝑥+𝑦
𝑑𝑥 𝑑𝑦
𝑥
0
𝑙𝑜𝑔 2
0
= ∫ ∫ 𝑒 𝑥+𝑦
(𝑒 𝑥+𝑦
− 1)𝑑𝑥 𝑑𝑦
𝑥
0
𝑙𝑜𝑔 2
0
= ∫ ∫ [𝑒2(𝑥+𝑦)
− 𝑒 𝑥+𝑦
] 𝑑𝑥 𝑑𝑦
𝑥
0
𝑙𝑜𝑔 2
0

= ∫ [𝑒2𝑥
.
𝑒2𝑥
2
− 𝑒 𝑥
. 𝑒 𝑦
]
0
𝑥
𝑑𝑥
𝑙𝑜𝑔 2
0
= ∫ [
𝑒4𝑥
2
− 𝑒2𝑥
−
𝑒2𝑥
2
+ 𝑒 𝑥
]
0
𝑥
𝑑𝑥
𝑙𝑜𝑔 2
0
= [
𝑒4𝑥
8
−
𝑒2𝑥
2
−
𝑒2𝑥
4
+ 𝑒 𝑥
]
0
log2
= [
𝑒4 𝑙𝑜𝑔2
8
−
𝑒2 log2
2
−
𝑒2 log2
4
+ 𝑒log 2
] − (
1
8
−
1
2
−
1
4
+ 1)
= [
𝑒 𝑙𝑜𝑔16
8
−
𝑒log4
2
−
𝑒log4
4
+ 𝑒log 2
] − (
1
8
−
1
2
−
1
4
+ 1)
= (
16
8
−
4
2
−
4
4
+ 2) − (
1
8
−
1
2
−
1
4
+ 1)
=
5
8
________ Answer
9.2 EXERCISE:
1) Evaluate
∭ ( 𝑥 + 𝑦 + 𝑧) 𝑑𝑥 𝑑𝑦 𝑑𝑧, 𝑤ℎ𝑒𝑟𝑒 𝑅: 0 ≤ 𝑥 ≤ 1, 1 ≤ 𝑦 ≤ 2, 2 ≤ 𝑧 ≤ 3.𝑅
2) Evaluate ∫ ∫ ∫ 𝑒 𝑥+𝑦+𝑧𝑥+log 𝑦
0
𝑑𝑧 𝑑𝑦 𝑑𝑥
𝑥
0
log 2
0
.
3) Evaluate∭ ( 𝑥2
+ 𝑦2
+ 𝑧2) 𝑑𝑥 𝑑𝑦 𝑑𝑧𝑅
where
𝑅:
𝑥 = 0, 𝑦 = 0, 𝑧 = 0
𝑥 + 𝑦 + 𝑧 = 𝑎, (𝑎 > 0)

10.1REFERENCE BOOK:
1) Introduction to EngineeringMathematics
By H. K. DASS. & Dr. RAMA VERMA
2) www.bookspar.com/wp-content/uploads/vtu/notes/1st-2nd-
sem/m2-21/Unit-5-Multiple-Integrals.pdf
3) http://www.mathstat.concordia.ca/faculty/cdavid/EMAT212/sol
integrals.pdf
4) http://studentsblog100.blogspot.in/2013/02/anna-university-
engineering-mathematics.html

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