b.tech ii unit-3 material multiple integration

Unit: 3 MULTIPLE INTEGRAL

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 Reference Book 21

MULTIPLE INTEGRALS

RAI UNIVERSITY, AHMEDABAD 1


1.1 DOUBLE INTEGRALS:

We Know that

∫a

b

f ( x )dx= limn→ ∞δx →0

[f ( x1 ) δ x1+ f ( x2 ) δ x2+ f ( x3 ) δ x3+…+f ( xn ) δ xn ]

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.

δ A1, δ A2, δ A3, …δ An

Then ∬A

f ( x , y )dA=¿ limn→ ∞δA → 0

[ f ( x1 , y1 ) δ A1+f ( x2 , y2 ) δ A2+…+ f ( xn , yn ) δ An ] ¿

2.1 Evaluation of Double Integral:

Double integral over region A may be evaluated by two successive integrations.If A is described as f 1(x)≤ y ≤ f 2(x ) [ y1 ≤ y ≤ y2 ]And a ≤ x≤ b ,

Then ∬A

f ( x , y )dA=∫a

b

∫y1

y2

f (x , y )dx dy

2.1.1 FIRST METHOD:



∬A

f ( x , y )dxdy=∫x1

x2 [∫y1

y2

f ( x , y )dy ]dx

f (x , y ) is first integrated with respect to y treating x as constant between the limits y1 and y2 and then the result is integrated with respect to 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 stripPQ ( y δx). While the integration of the result w.r.t x corresponds to the sliding to the strip, from AD to BC covering the whole regionABCD.

2.1.2 SECOND METHOD:

∬A

f ( x , y )dxdy=∫y1

y2 [∫x1

x2

f ( x , y )dx ]dy

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 respect to y between the limits y1 and y2 and vice versa.

NOTE: For constant 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 ∫0

1

∫0

x2

ey / x dy dx

Solution: Here, we have

∫0

1 [∫0

x2

e y/ x dy ]dx=∫0

1

[ e y / x

1/ x ]0

x2

dx

¿∫0

1 (ex−1 )1/ x

dx

¿∫0

1

x ex dx−∫0

1

x dx

¿ [ x ex−ex ]01¿−[ x2

2 ]0

1

¿

¿e1−e1+1−12

¿12

∴∫0

1

∫0

x2

e y / x dy dx=12

________ Answer

Example 2: Evaluate ∫0

∞

∫0

∞

e−x2(1+ y2) x dxdy


∫0

∞

∫0

∞

e−x2(1+ y2) x dxdy=∫0

∞

dy∫0

∞

e−x2(1+ y2) x dx

¿∫0

∞

dy [ e−x2(1+ y2 )

−2(1+ y2) ]0∞

¿∫0

∞

[0+ 1

2 (1+ y2 ) ]dy

¿12

[ tan−1 y ]0∞¿¿

¿12

[ tan−1 ∞−tan−10 ]

¿12 ( π

2 )



¿π4

∴∫0

∞

∫0

∞

e− x2(1+ y2) x dxdy=π4

________ Answer

Example 3: Sketch the area of integration and evaluate∫1

2

∫−√2− y

√2− y¿

2 x2 y2 dxdy

¿¿.

Solution: Here we have

∫1

2

∫−√2− y

√2− y¿

2x2 y2 dxdy=2

∫1

2

y2 dy ∫−√2− y

√2− y¿

x2 dx ¿

¿¿

¿4∫1

2

y2dy ∫0

√2− y¿

x2 dx¿

¿

[ ∵∫−a

a

f ( x )dx=2∫0

a

f (x ) dx

w here x2is an even function ] ¿4∫

1

2

y2dy [ x3

3 ]0

√2− y

¿ 43∫

1

2

y2 dy (2− y )32

¿ 43∫

1

0

(2− y )2 t3 /2(−dt) [ put 2− y=t∴dy=−dt ]

¿ 43 [(2−t)2( 2 t

52

5 )−(−2 ) (2−t ) 25

.27

t72+(2) 2

5.

27

.29

t92 ]

0

1

¿43 [ 2

5+2( 2

5.27 )+ (2 )( 2

5.27

.29 )]

¿43 [ 2

5+ 8

35+ 16

315 ] ¿

415 [2+ 8

7+16

63 ]RAI UNIVERSITY, AHMEDABAD 5


¿4

15 [ 126+72+1663 ]

¿4

15 ( 21463 )

¿856945

∴∫1

2

∫−√2− y

√2− y¿

2 x2 y2 dxdy=856945

¿¿ ________ 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 areadx dy. The integration w.r.t x between the limits x1 , x2 keeping y fixed indicates that integration is done, alongPQ. Then the integration of result w.r.t to y corresponds to sliding the strips PQ from BC to AD covering the whole regionABCD.

We can also integrate first w.r.t ‘ y’ then w.r.t x , which ever is convenient.

Example 4: Evaluate ∬ xy dxdy over the region in the positive quadrant for

whichx+ 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 =

∫0

1

dx ∫0

1−x

y dy=∫0

1

x dx [ y2

2 ]0

1− x

¿ 12∫

0

1

( xdx )(1−x)2

¿ 12∫

0

1

( x−2 x2+x3 ) dx

¿12 [ x2

2−

2 x3

3+

x4

4 ]0

1

¿ 12 [ 1

2−2

3+ 1

4 ]¿ 1

24 ________ Answer

Example 5: Evaluate ∬R

xy dxdy , where Rthe quadrant of the circle is

x2+ y2=a2 where x≥ 0∧ y ≥0.

Solution: Let the region of integration be the first quadrant of the circleOAB.

Let I=∬R

xy dx dy (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=∫0

a

x dx ∫0

√a2−x2

y dy

¿∫0

a

x dx [ y2

2 ]0

√a2−x2

¿ 12∫

0

a

x ( a2−x2 ) dx

¿12 [ a2 x2

2−

x4

4 ]0

a

¿ a4

8________ Answer

Example 6: Evaluate ∬A

x2dx dy , where A is the region in the first quadrant

bounded by the hyperbola xy=16 and the linesy=x , y=0∧x=8.

Solution: The line OP , y=x and the curve PS , xy=16 intersect atp(4,4).

The line SN ,x=8 intersects the hyperbola at S(8,2). Andy=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 16x and then x varies from 4 to 8.

∴∬A

x2dx dy=∫0

4

∫0

x

x2 dxdy+∫4

8

∫0

16x

x2 dxdy

¿∫0

4

x2dx∫0

x

dy+∫4

8

x2 dx∫0

16x

dy

¿∫0

4

x2 [ y ]¿0x

dx+∫4

8

x2 [ y ]¿0

16x dx



¿∫0

4

x3 dx+∫4

8

16 x dx

¿ [ x4

4 ]0

4

+16 [ x2

2 ]4

8

¿64+8(82−42)

¿64+384

¿448 ________ Answer

3.2 EXERCISE:

1) Find ∫0

1

∫0

y

x y e−x2

dx dy .

2) Evaluate the integral∫1

log 8

∫0

log y

ex+ y dx dy.

3) Evaluate∫0

a

∫xa

x¿

x dxdyx2+ y2

¿¿.

4) Evaluate∫0

1

∫0

1dx dy

√ (1−x2 ) (1− y2 ) .

5) Evaluate∬S

√xy− y2 dy dx, where S is a triangle with vertices (0, 0), (10,

1), and (1, 1).

6) Evaluate ∬ ( x2+ y2) dxdy over the area of the triangle whose vertices

are (0, 1), (1, 0), (1, 2).

7) Evaluate ∬ y dxdy 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+ y2−2 x=0 , y2=2 x , 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:


∞

∫x

∞e− y

ydx dy .

Solution: We have, ∫0

∞

∫x

∞e− y

ydx dy

Here the elementary strip PQ extends from y=x to y=∞ and this vertical strip slides fromx=0¿ 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 tox= y. To cover the given region, we then integrate w.r.t ' y ' fromy=0¿ y=∞.

Thus

∫0

∞

dx∫x

∞e− y

ydy=∫

0

∞e− y

ydy∫

0

y

dx

¿∫0

∞e− y

ydy [ x ]0

y¿ ¿

¿∫0

∞

ye− y

ydy

¿∫0

∞

e− y dy

¿ [ e− y

−1 ]0

∞

¿−[ 1e y ]

0

∞



¿−[ 1∞

−1]¿1 ________ Answer

Example 2: Change the order of integration in I=∫0

1

∫x2

2−x

xy dx dy and hence evaluate

the same.

Solution: We have I=∫0

1

∫x2

2−x

xy dx dy

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 areaCBA. The limits of x in the area OBC are 0 and √ y and the limits of x in the area CBA are 0 and2− y.

So, the given integral is

¿∫0

1

y dy∫0

√y

x dx+∫1

2

y dy ∫0

2− y

x dx

¿∫0

1

y dy [ x2

2 ]0

√ y

+∫1

2

ydy [ x2

2 ]0

2− y

¿ 12∫

0

1

y2 dy+ 12∫1

2

y (2− y )2dy

¿12 [ y3

3 ]0

1

+12∫1

2

(4 y−4 y2+ y3)

¿ 16+ 1

2 [ 96−128+48−24+16−312 ]

¿ 16+ 5

24



¿ 924

¿ 38 ________ Answer

4.3 EXERCISE:

1) Change the order of the integration∫0

∞

∫0

x

e−xy y dydx.

2) Evaluate ∫0

2

∫1

ex

dxdy by changing the order of integration.

3) Change the order of integration and evaluate∫0

2

∫√2 y

2x2 dx dy

√x 4−4 y2 .

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∬R

f ( x , y )dxdy.

It is to be changed by the new variablesu , v.The relation of x , y with u , v are given asx=∅ (u , v ) , y=ψ (u , v ). Then the double integration is converted into.

1. ∬R

f ( x , y )dx dy=∬R'

f {ϕ (u , v ) , ψ (u , v )}|J|du dv ,

[dx dy=∂(x , y )∂(u , v )

du dv] ∬

R

f ( x , y )dx dy=∬D

f {x (u , v ) , y (u , v)}|∂(x , y)∂(u , v)|du dv

5.2 Example 1: Using x+ y=u , x− y=v , evaluate the double integral over the square R

∬R

( x2+ y2) dxdy

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=12

(u+v ) , y=12

(u−v )

J=∂( x , y )∂ (u , v)

=|∂ x∂ u

∂ x∂ v

∂ y∂ u

∂ y∂ v

|=|12

12

12

−12

| ¿−14−1

4 ¿−12

∬R

( x2+ y2) dxdy=∫0

2

∫0

2

[ 14(u+v)2+ 1

4(u−v)2]|∂(x , y )

∂(u , v )|du dv

¿∫0

2

∫0

212

(u2+v2 )|−12 |du dv

¿−14∫

0

2

dv∫0

2

(u2+v2 ) du

¿−14∫0

2

dv [ u3

3+u v2]

0

2

¿−14∫

0

2

dv ( 83+2v2)

¿−14∫

0

2

( 83+2 v2)dv

¿−14 [ 8

3v+ 2

3v3]

0

2

¿−14 [ 8

3(2 )+ 2

3(2 )3]

¿−14 ( 16

3+16

3 )¿−1

4 ( 323 )



¿−83 ________ Answer

5.3 EXERCISE:

1) Using the transformation x+ y=u , y=uv show that ∫0

1

∫0

1− x

e y/(x + y)dy dx=12(e−1)

2) Evaluate ∬R

(x+ y )2dx dy , 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−2 y.

6.1 AREA IN CARTESIAN CO-ORDINATES:

Area = ∫a

b

∫y1

y2

dxdy

6.2 Example 1: Find the area bounded by the lines

y=x+2

y=− x+2x=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

¿∬dxdy

¿∫0

5

dx ∫– x+2

x+2

dy

¿∫0

5

dx [ y ]−x+2

x+2¿ ¿

¿∫0

5

dx [ x+2−(−x+2)]

¿∫0

5

[ 2 x ] dx

¿ [ 2 x2

2 ]0

5

¿ [ x2 ]05¿¿

¿ [ 25−0 ]

¿25 Sq. units ________ Answer

Example 2: Find the area between the parabolas y2=4 ax andx2=4 ay.

Solution: We have, y2=4 ax ________ (1)

x2=4 ay ________ (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¿Q ,√4 ay and then y varies from O ( y=0 )¿ A( y=4 a).

∴The required area ¿∫0

4 a

dy ∫y2 /4 a

√4 ay

dx



¿∫0

4 a

dy [ x ]¿ y2

4 a

√4 ay

¿∫0

4 a

dy [√4 ay− y2

4 a ]

¿ [√4 ay3 /2

32

− y3

12 a ]0

4 a

¿ 4√a3

(4 a)3 /2−(4 a)3

12 a

¿ 163

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 rectanglesδx δy. Erect vertical rod on the δx δy of height z .

Volume of each vertical rod ¿ Areaof the base× height

¿δx δy . z

Volume of the solid cylinder on S ¿ lim

δx → 0δy → 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 2 z=x2+ y2 and the cylinderx2+ y2=4.


2 z=x2+ y2⇒2 z=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∫0

π

dθ∫0

2r 2

2r dr

¿ 22∫

0

π

dθ∫0

2

r3 dr

¿∫0

π

dθ (r 4

4 )0

2

¿∫0

π

dθ (164 )

¿4∫0

π

dθ

¿4 [θ ]0π¿¿

¿4 π ________ 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 ¿∫a

b

∫y1( x)

y2( x)

2 π PQ dxdy



Example 1: Find the volume of the torus generated by revolving the circle x2+ y2=4 about the linex=3.

Solution: x2+ y2=4

V=∬ (2 π PQ ) dxdy

¿2 π∬ (3−x ) dx dy

¿2 π∫−2

2

dx (3 y−xy )¿−√4− x2

+√4−x2

(3− x ) dy

¿2 π∫−2

+2

dx [3√4−x2−x √4−x2+3√4−x2−x √4−x2 ]

¿4 π∫−2

+2

[ 3√4−x2−x √4−x2 ] dx

¿4 π [3 x2

√4−x2+3×42

sin−1 x2+ 1

3(4−x2 )3 /2]

−2

+2

¿4 π [6×π2

+6 ×π2 ]

¿24 π2 ________ 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 byxy−plane, the cylinder x2+ y2=1 and the planex+ y+z=3.

4) Evaluate the volume of the solid generated by revolving the area of the parabola y2=4 ax 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 δ s1 , δ s2 , δ s3 , …. δ sn of the space S.If (xr , yr , zr) be a point in the rth subspace.

The limit of the sum ∑r=1

n¿

f (x r , y r , zr)δ sr ,as n →∞ ,δ sr →0¿

¿ is known as the triple integral of

f (x , y , z) over the space S.

Symbolically, it is denoted by

∭S

f (x , y , z ) dS

It can be calculated as∫x1

x2

∫y1

y2

∫z 1

z 2

f (x , y , z ) dx dy dz. First we integrate with respect to

z treating x , y as constant between the limitsz1∧z2. The resulting expression (function ofx , y) is integrated with respect to y keeping x as constant between the limits y1∧ y2. At the end we integrate the resulting expression (function of x only) within the limitsx1∧x2.

First we integrate from inner most integral w.r.t z, and then we integrate w.r.ty, and finally the outer most w.r.tx.



But the above order of integration is immaterial provided the limits change accordingly.

Example 1: Evaluate ∭R

( x−2 y+z ) dx dydz , w h ere R :0 ≤ x ≤10≤ y≤ x2

0 ≤ z≤ x+ y

Solution: ∭R

( x−2 y+z ) dx dydz

¿∫0

1

dx∫0

x2

dy ∫0

x+ y

(x−2 y+z)dz

¿∫0

1

dx∫0

x2

dy (xz−2 yz+ z2

2 )0

x+ y

¿∫0

1

dx∫0

x2

dy [ x ( x+ y )−2 y ( x+ y )+ (x+ y)2

2 ] ¿∫

0

1

dx∫0

x2

dy [ x2+xy−2xy−2 y2+(x+ y )2

2 ] ¿∫

0

1

dx∫0

x2

dy [ x2−xy−2 y2+ x2

2+xy+ y2

2 ] ¿∫

0

1

dx∫0

x2

dy [ 3 x2

2−3 y2

2 ] ¿ 3

2∫0

1

dx∫0

x2

( x2− y2 ) dy

¿32∫0

1

dx (x2 y−y3

3 )0

x2

¿ 32∫0

1

dx (x4− x6

3 ) ¿ 3

2 ( x5

5− x7

21 )0

1



¿ 32 ( 1

5− 1

21 ) ¿ 8

35 ________ Answer


log 2

∫0

x

∫0

x+ y

ex+ y+z dx dy dz

Solution: I=∫0

log 2

∫0

x

ex+ y [ez ]0x+ y¿

dx dy¿

¿ ∫0

log 2

∫0

x

ex+ y(ex+ y−1)dx dy

¿ ∫0

log 2

∫0

x

[e2(x+ y)−ex + y ] dxdy

¿ ∫0

log 2

[e2 x .e2 x

2−ex . e y ]

0

x

dx

¿ ∫0

log 2

[ e4 x

2−e2x−

e2 x

2+ex]

0

x

dx

¿ [ e4 x

8−

e2x

2−

e2x

4+ex ]

0

log 2

¿ [ e4 log 2

8− e2 log 2

2− e2 log2

4+e log2]−( 1

8−1

2−1

4+1)

¿ [ e log16

8− e log 4

2− elog 4

4+e log2]−( 1

8−1

2−1

4+1)

¿( 168

− 42−4

4+2)−( 1

8−1

2− 1

4+1)

¿ 58 ________ Answer

9.2 EXERCISE:



1) Evaluate ∭R

¿ ¿( x+ y+z ) dxdy dz ,where R : 0≤ x ≤1 , 1≤ y ≤2 , 2≤ z≤ 3.

¿¿

2) Evaluate∫0

log 2

∫0

x

∫0

x+ log y

ex+ y +z dz dy dx.3) Evaluate ∭

R

( x2+ y2+z2 ) dxdy dz where R :x=0 , y=0 , z=0

x+ y+z=a ,(a>0)

10.1 REFERENCE BOOK:1) Introduction to Engineering MathematicsBy H. K. DASS. & Dr. RAMA VERMA2) www.bookspar.com/wp-content/uploads/vtu/notes/1st- 2nd-sem/m2-21/Unit-5-Multiple-Integrals.pdf3) http://www.mathstat.concordia.ca/faculty/cdavid/ EMAT212/solintegrals.pdf4) http://studentsblog100.blogspot.in/2013/02/anna- university-engineering-mathematics.html


http://studentsblog100.blogspot.in/2013/02/anna-university-engineering-mathematics.html

http://studentsblog100.blogspot.in/2013/02/anna-university-engineering-mathematics.html

http://www.mathstat.concordia.ca/faculty/cdavid/EMAT212/solintegrals.pdf

http://www.mathstat.concordia.ca/faculty/cdavid/EMAT212/solintegrals.pdf

http://www.bookspar.com/wp-content/uploads/vtu/notes/1st-2nd-sem/m2-21/Unit-5-Multiple-Integrals.pdf

http://www.bookspar.com/wp-content/uploads/vtu/notes/1st-2nd-sem/m2-21/Unit-5-Multiple-Integrals.pdf

b.tech ii unit-3 material multiple integration

Education