B.Tech-II_Unit-III

Unit: 3 MULTIPLE INTEGRAL

RAI UNIVERSITY, AHMEDABAD 1

Course: B.Tech- IISubject: Engineering Mathematics II

Unit-3RAI UNIVERSITY, AHMEDABAD

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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


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.

) + ⋯ +


Unit: 3


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 stripthe 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




2.2Examples:

Example 1: Find ∫ ∫ ⁄Solution: Here, we have

∫ ∫ ⁄ = ∫ ⁄⁄

= ∫ ( )⁄

= ∫ − ∫= [ − ] −= − + 1 −=

∴ ∫ ∫ ⁄ = ________Answer

Example 2:Evaluate ∫ ∫ ( )∞∞

Solution: Here, we have

( )∞∞

= ( )∞∞

= ( )−2(1 + )

∞∞

= ∫ 0 + ( )∞

= [tan ]∞

= [tan ∞− tan 0]=




=∴ ∫ ∫ ( )∞∞ = ________Answer

Example 3: Sketch the area of integration and evaluate∫ ∫ 2 .

Solution: Here we have

2 = 2

= 4 ∫ ∫∵ ∫ ( ) = 2 ∫ ( )ℎ

= 4 ∫= ∫ (2 − )= ∫ (2 − ) ⁄ (− ) 2 − =∴ = −= (2 − ) − (−2)(2− ) . + (2) . .= + 2 . + (2) . .= + += 2 + +=


Unit: 3


==

∴ ∫ ∫ =

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 areabetween the limits indicates that integration is done, alongthe 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.




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.

= ∫ ∫√

= ∫ √


Unit: 3


= ∫ (==

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

+ 1664 + 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.




= 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.


Unit: 3


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 yfrom 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

∞

= ∞

∫ [ ]∞

∫∞∫∞

∞

− ∞

−∞

− 11 ________

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


Unit: 3


On changing the order of integration we have taken a strip parallel to xarea and second strip in the areaand 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 xand 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

.


Unit: 3


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 ( + ) + 14 ( − ) ( , )( , )= ∫ ∫ ( + ) −= − ∫ ∫ ( + )

= − ∫ += − ∫ + 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

= += − +=




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


Unit: 3


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 rectanglesthe of height .Volume of each vertical rod =

MULTIPLE INTEGRAL

: Find the area between the parabolas = and

= 4 ________ (1)= 4 ________ (2)


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

16

= .


varies from

).

Answer

. Erect vertical rod on


Unit: 3


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)




= 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 × 42 sin 2 + 13 (4 − ) ⁄




= 4 6 × 2 + 6 × 2= 24 ________ Ans.

8.2 EXERCISE:

1) Find the area of the ellipse + = 12) 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

( , , )




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 ≤ ≤ 10 ≤ ≤0 ≤ ≤ +Solution:∭ ( − 2 + )

= ∫ ∫ ∫ ( − 2 + )= ∫ ∫ − 2 += ∫ ∫ ( + ) − 2 ( + ) + ( )

= ∫ ∫ + − 2 − 2 + ( )

= ∫ ∫ − − 2 + + +




= ∫ ∫ −= ∫ ∫ ( − )= ∫ −= ∫ −= −= −= ________Answer

Example 2: Evaluate ∫ ∫ ∫Solution: = ∫ ∫ [ ]

= ∫ ∫ ( − 1)= ∫ ∫ ( ) −= ∫ . − .= ∫ − − += − − += − − + − − − + 1= − − + − − − + 1




= − − + 2 − − − + 1= ________Answer

9.2 EXERCISE:1) Evaluate

∭ ( + + ) , ℎ : 0 ≤ ≤ 1, 1 ≤ ≤ 2, 2 ≤ ≤ 3.2) Evaluate∫ ∫ ∫ .3) Evaluate ∭ ( + + ) where

: = 0, = 0, = 0+ + = , ( > 0)

10.1 REFERENCE BOOK:1) Introduction to Engineering Mathematics

By 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.pdf




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


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