Applications of Multiple Integrals -...

Applications of

Multiple Integrals

1. Introduction :

In this chapter we study some applications of muliple intgrals. We shall,

in particular, learn how to use multiple integrals to find (i) areas (ii) volumes

(iii) mass (iv) volumes of solids of revolutions (v) surfaces of solids of

revolution.

2. Area by single integrals

y

In some cases the following methods based

on single integrals enable us to find areas.

(a) Cartesian Coordinates

a b X

Consider the area enclosed by the curve

y = f (x), the x-axis and the lines x =a and x = b.

. Consider a small strip parallel to they-axis Fig. (10.1) at P (x, y) and of width ox. Then the area of the

strip = y dx. The strip sweeps the whole area when it moves parallel to itself

from x =a to x =b. Hence, the required area is given by

A= f: y� If the area is enclosed by the y = f (x),

they-axis and the line y = c and y = d ; we consider

a small strip parallel to the x-axis at P (x, y) and of

oy. Then the area of the strip= X dy. The strip

sweeps the whole area when it moves parallel to

itself from y = c toy = d. Hence the required area is

given by

A= f: xdy

y

Fig. (10.2)

.... (1)

X

(See Ex. 5) .... (2)

Engineering Mathematics · II (10-2) Appli. of Mult. lnte.

(b) Polar Coordinates y Consider a curve given in polar coordinates by Q

r = f (9). Consider two adjacent points P (r, 9) and Q (r + 8r, 9 + 89) and the area OPQ. This area is

approximately equal to=± ?- 8e

The strip sweeps the whole area when it moves from radius vector OA to the radius vector OB. If A is (a, a) and B is (b, 13) then the required area is given by

(c) Parametric Coordinates

0

Fig. (10.3)

When the curve is given in parametric coordinates x = f 1 (!), y = f 2 (t), it is convenient to use either ( 1) or (2) depending upon the boundries. (See Ex. 6)

x2 y2 y Ex. 1 : Find the area of ellipse -; + p = 1 Sol. : Let P (x, y) be any point on the curve in

the first quadrant. Then y = } a 2 - x 2

:. Area= 4 · area OAB

fa fa b =4 ydx=4

0 0

= 4 . E_ [ a 2 x 2 + a 2 sin- I .£]a a 2 a 0 b a 2 1t = 4 a . 2 . 2 = 1tab.

Fig. (10.4)

Cor. The area of the circle x2 + l = a2 (when b =a) is na2.

Ex. 2 : Find the area of the loop of the curve l = (x- a)(b-x)2 ; (b >a). (S.U. 2007)

Sol. : The curve intersects the x-axis at A, x = a and atB,x= b

:. Area= 2 area ACB b

=2 J ydx a b

= 2 J (b- x) x- a · dx a

Put x - a=r :. dx=2tdt,x=a+t2

0

Fig. (10.5)

Engineering Mathematics - II (1 0-3) Appli. of Mult. lnte.

When x ==a, t = 0, x == b, t a �

:. A = 2 J (b -a t 2) · t · 2t · dt 0

=2f [(b--a)·2t2-2t4]·dt 0 [ t3

=4 (b-a) 3-5 o

= (b a)512- (b- a)S/2 = 185 (b a)S/2

Ex. 3 : Find the area of the loops of the curve a2l = x2 (a2- x2) (S.U. 2006)

Sol. : The curve is shown in the figure and y = a 2 x 2

The required area

A = 4 ( y dx = 4 ( � 2 - x 2 dx o 0 a

:. A=% [- 1 (a 2 x 2)3/2 J: (Put a 2- x 2 = t)

.4 a 2 =3

Fig. (10.6)

Ex. 4: Find the area enclosed by the curve a4l =x5 (2a -x). (S.U. 2003) Sol. : The curve is shown in the figure.

x512� Now,y= a 2

J2a

J2a A =2 ydx=2 ·dx o o a

Put X= 2a sin2 e, dx = 4a sin e cos 8 d8. 1t

When X = 0, e = 0 ; when X = 2a, 8 = 2

Fig. (10.7)

2 rt/2 :. A= a 2 f0 (2a)512 sin5 8 · ..f2a ·cos 8 · 4a sin 8 cos 8 d8

rt/2 = 64a2 f0 sin6 8 cos2 8 d8

2 l. i7 1372 = 64a .

2 15

2 t· �-� ·fi-�fi = 32a 4 · 3 · 2 · 1

X

Engineering Mathematics - II (10-4)

32a 2 15 = "" . 24. 1t

[·: 11/2 = v'it ]

5 = 4 a 2n

Ex. 5 : Show that the area enclosed by the curve

a2 x2 = i (2a- y) is na2. (S.U. 1997, 2005) Sol. : The curve is shown in the figure. We take a

strip parallel to the x-axis. Also

y 312 :. x= a

2a 2a y 312 y :. A= 2 f x dy = 2 f a · dy 0

0

Appii. of Mutt. lnte.

y

0 X

Fig. (10.8)

As before put y = 2a sin2 8.

2 rt/2 :. A=-; f0 (2a)312 · sin3 8 · ffa ·cos 8 · 4a sin 8 cos 8 d8

rt/2 = 32a2 J0 sin

4 8 cos2 8 d8

2 1 i5i2 i372 = 32a 2 14

3 1 11 1 11 =16a· 2'2 2'2 2

3. 2. 1

Ex. 6: Find the area included between the cycloid

x =a (8 + sin8) andy = a( 1 -cos 8) and its base.

(S.U. 1997) Sol. : The curve is shown in the figure. Now

required area A is given by

A = 2 · area OAB

=2J:xdy

= 2 J: a(O +sinO) a sinO dO 1t

= 2a2 f (8 sin 8 + sin28) d8 0

6=-rr y e=rr


Now, (esin ede= [e(-cose)-- J (-cose)·l·deJ: = [-e cos e +sin et = 1t 0

and ( sin2 e de= Jlt ( 1 -cos 2e )

o 0 2 de

_l [e _ sin 2e )lt

= 2!.

-2 2 0 2

:. A = 2a2 ( 1t + ) = 3na2

Ex. 7: Find the area of the curve (xlai13 + (ylb)213 = 1.

a,

Sol. : The curve is shown below. Consider a

strip parallel to the y-axis.

A= 4f; y dx and y = b[1-(x I a)213r12

:.A=4bf; [1-(xla)213r'2 dx

Now, put X = a cos3 e

Fig. (10.10) :. dx =- 3a cos3 e sine de

When X= 0, e = nl2; X= a, e = 0.

:. A= 4bf 0 sin 3 0 · (-3a)cos 2 OsinO dO n/2

= 12abf: 12 sin 4 Ocos 2 0 dO

= 12 ab · _!_ B ( .?_ �) = 6ab · [5/2 [3/2 2 2'2 14 (3 I 2)(1 I 2) 1112 (I I 2) It I 2 3 = 6ab· =-nab. 3! 8

Ex. 8 : Find the area of the loop of the curve x (� + l) =a (�- i) Sol. : The curve is shown in the figure and

y = x Area of the loop

Engineering Mathematics - II (10-6) Appll. of Mult. lnte.

a A = 2 · area OAB = 2 f0 y dx

ra

= 2 .0 x dx A X

Ia M--x)

= 2 X 2 2 dx o a -x

Putting X = a sin e. dx = a cos e de. Fig. (10.11)

:. A I1t12 a sin e · a(l - sin e)

= 2 o acose ·acosede

= 2a2( 12 (sine-sin2e) de= 2a2 ( 12 [sin e- ( 1- 2e ) ]de

[ 1 sin 2e ]7t/2 =2a2 -cose 2 e+ 0

+o} (-1 0+o)] =2a2 Ex. 9 : Find the area of the loop x5 + i = 5ax2l .

Sol. : The curve is shown in the figure. Some times transformation to polar coordinates facilitates integration.

Putting X = r cos e. y = r sin e ; the equation becomes

5 sin2 e cos2 e r a

sin5 e + cos5 e

1t When r:: 0, e = 0 and e = 2 . Hence, the loop

lies between these limits.

1 1t/2 :. A = 2 f0 r2de

- - 25a · · de - 2 0 (sinS e + cos5 e)2

Dividing Nand D by cos10 e.

_ 2Sa2 ( 12 tan 4 e sec2 6 A - 2 o (1 +tan s e)2

de

Putt= 1 + tans e, dt = 5tan4 e sec2 e d6

1t When e = 0, t = 1, when 6 = 2' t = co

A = 5a2 Joo dt = 5a2 [ 1. ]00 = 5a2

2 1 t2 2 t I 2

y

Fig. (10.12)

X

Engineering Mathematics - II (10·7) Appli. of Mult. lnte.

Ex. 10: Find the area of the loop:?+ y3 = 3a xy. Sol. : The curve is shown in the figure.

X

Fig. (10.13)

I rr./2 :. A= 2 I r2d8=

0 Dividing by cos6e

9 2 rr./2 A=.. I 2 0

Putting X = r cos e , y = r sin e 3a sine cos e

the equation becomes r = 3e + . 3e cos sm Putting r = 0 , we have, sine cose = 0

1t :. e = 0 and e = 2 .

9a 2 rc/2 sin2 e cos2 e - I de 2 0 (cos3 e + sin3 e)2

tan 2 e sec 2 e (I +tan 3 e)2

de

Putting I+ tan3 e = t. : . 3tan2esec2e de= dt

A = 3a 2 j i!J. = 3a 2 [-l]oo

= 3a 2 2 I 12 2 t I 2

Ex. 11 : Find the area common to the cardiodes r = a (1 - cos e) and r =a (I +cos e).

Sol. : The two curves are shown in the figure. It is easy to see that the point of

intersection P is (a, )

is Because of symmetry, the required area

rc/2 I A= 4 I 2 ? de where r =a (1 -cos e) 0

X

Fig. (10.14)

rr./2 rc/2 = 2 I a 2 ( 1 - cos e) 2 de = 2a 2 I [I - 2 cos e + cos 2 e) de

0 0

= 2 a 2 I orc/2 [(I - 2cos e + I + 2e

J dx

= 2 a 2 I rc/2 [ l. - 2cos e + cos 2e ) de 0 2 2

2 2 [ 3 e 2 . e + sin 2e ] rc/2 =a - - sm

--

2 4 0 2 [ 3tt ] a 2 = 2 a 4 - 2 = T [3tt - 8]

Engineering Mathematics - II (10-8) Appli. of Mult. lnte.

Ex. 12 :Find the area inside the cardioide r =a (1 +cos 8) and outside the circle r = 2a cos e .

Sol. : The curves are shown in the figure.

X

Fig. (10.12)

Now, the total area inside the cardioide

= 2 ( l. r 2 d8 = ( a 2 ( 1 + cos 8)2 d8 0 2 0

=a2 ( [ 1 + 2cos e + ( 1

= a 2 ( (1. + 2cos 8 + cos 28 ) d8

0 2 2

= a 2 1 8 + 2sin 8 + sm 2 = 1 [ . 8 ] lt 2 4 0 2

The area of the circle = rr.a2 . 3 ., 2 7ta2

:. The reqmred area= 2 1ta -- 1ta = .

Ex. 13 : Find the area of one loop of the Bernoulli's Lemniscate

r 2 =a 2 cos 28.

Sol. : The curve is shown in the figure. Here 8 varies from 0 to 1t/4.

Jtl4 I n/4 A = 2 I 2 r 2 d8 = I a2 cos 28 ·d8

0 0

=a 2 sin 28 =

� [ ] lt /4

2 0 2 .

Exercise -I

X

Fig. (10.16)

1. Find the area of the circle x2 + l = a2. [Ans.: rr.a2] ab

2. Find the area between the parabolas l =ax andx2 =by [Ans.: 31 (See fig. 10.43 page (10.27))

3. Find the area of the loop of the curve

(i) l = (x- 2)(4-x)2 (See fig. 10.5 page (10.2), a= 2, b = 4) (ii) l = (x- 1)(5 -x)2 (See fig. 10.5 page (10.2), a= I, b = 5)

32 28 (S.U. 2007) [Ans. : (i) lS .fi (ii) IS]

4. Find the area bounded by the curve y = � - 3x, the line y = 2x and 49

the x-axis. (See fig. 7.145 page (7.37)) [Ans. :3]

-••::t•u� .. a au� IWIQI.II'IIIIGI.I\.;, - II pu·l:IJ Appli. of Mult. lnte.

5. Find the area enclosed between the parabolas l = - 4(x - 1) and 8 l = - 2(x -2). (See fig. 7.146 page (7.37)) [Ans.: 3 ]

6. Find the area included between the cycloid x = a (8 - sin8), y = a (1 -cos 8) and its base. (See fig. 7.116 page (7 .29)) [ Ans. : 3 na2]

7. Find the area of one of the loops of the curve x = a sin 2t, y = a sin t.

4a 2 (See fig. 7.147 page (7.37)) [Ans.: 1

8. Find the area of the loop of the curve 3al = x (x- a)2. (S.U. 1999) 8a 2 (See fig. 7.62 page (7.15)) [Ans.: 15V3 1

9. Find the area of the loop of the curve r (cos 8 + sin 8) = 2a sin8cos 8

(See fig. 7.148 page (7.37)) [ Ans.: a 2 (1-%)] 10. Find the area enclosed by the curve x213 + l'3 = a213

(Hint: Use parametric equations x =a cos38, y =a sin3 8)

3 (See fig. 7.47 page (7.10), a= 1) (S.U. 2004) [ Ans.: g na2 1

11. Find the area of the loop of the curve al = x2 (a - x) 8

(See fig. 7.63 page (7.15)) [Ans: 15 a21 12. Find the area of the loop of the curve l = x2 ( 4 -x2)

(See fig. 10.6 page (10.3))

13. Find the area of the loop of the curve l = x2

32 [Ans.: 31

(See fig. 10.11 page (10.6)) [ Ans. : 2a2 - 1) ] 14. Find the whole area of the curve r = a + b cos 8. (a> b).

(See fig. 7.90 page (7.21)) (S.U. 2000) [ Ans. : (2a2 + b2)]

[a 2-x 2] 15. Find the area of one of the loops of the curve l = x2 · a 2 + x 2 (See fig. 7.77 page (7.18)) [ Ans.: a2 (n- 2) ]

16. Find the area of the cardioide r = a (1 +cos 8)

(See fig. 7.88 page (7 .19))

17. Find the area of the curve r = a sin 38

(See fig. 7.96 page (7.23))

3 2 [ Ans. : 2 na ]

1ta 2 [ Ans. :

Engineering Mathematics - II (1 0-1 0) Appli. of Mult. lnte.

18. Find the area of the curve r = a sin 28

(See fig. 7.98 page (7.24)) 1tQ 2 [ Ans.:

19. Find the area of one of the loops of the curve x4 + i = 2a2xy (See fig. 7.149 page (7.37))

(Hint: Transform to polar coordinates.) na 2 [ Ans.:

20. Find the area of one of the loops of the curve x6 + l = a2x2l (See fig. 7.150 page (7.37))

(Hint: Transform to polar coordinates.) na 2

[ Ans. : --r2 ]

21. Find the area of the loop of the curve (x + y )(x2 + l) = 2axy. (See fig. 7.148 page (7.37))

(Hint : Transform to polar coordinates.)

22. Find the area of the loop of the curve x4 - 2xy a2 + a2i = 0. (See fig. 7.151 page (7.37))

2 (Hint: Transform to polar coordinates.) [ Ans.: 3a ]

3. Area by double integrals In some problems it is more convient to use double integrals as explained

below.

(a) Cartesian Coordinates

Consider the area enclosed by two plane curves

y = f 1 (x) andy = f 2 (x) intersecting in

A (a, c) and B (b, d).

y

Consider a strip parallel to the y-axis and of width 8x and another strip parallel to the x-axis and 0 of width8 y .

X

Fig. (10.17)

Then the area of the elementary rectangle (shaded area) is dx dy. The required area then is given by

b h(x) A=J J

1 dxdy

a 1 (x) (b) Polar Coordinates

Consider again the area enclosed by two plane curves r = !1 (8) and r = h (8) intersecting in A ( rl> a) and B (r2, [3).

Engineering Mathematics - II (10-11) Appll. of Muit. Inte.

We divide the area into small areas by taking

lines e =constant and circles r =constant.

Then the area of the elementary rectangle

(shaded area) is r d9 dr. The required area then is

given by

� /z(6) .

A=I I rd9dr ex I 1 (6)

X

Fig. (10.18)

Ex • 1 : Find by double integration the area bounded by the lines y = 2 + x, y = 2-x andx= 5. Sol.: Refer to the Example 2 of§ 5 of chapter 9 page (9.29). Consider a strip parallel to they-axis. On this stripx varies fromy = 2 -x toy= 2 +x. This strip

sweeps the area when it moves parallel to itself between x = 0 to x = 5. 5 2+x 5 2+x

:.A =I I dxdy=I [ y] 2 - dx 0 2 x 0 x 5 5

=I [ (2 +x)-(2-x) ] dx =I 2xdx 0 0 5

= [ � ]0 = 25.

Ex. 2 :Find by double integration the area common to the circle � + l = 10 and the parabola l = 9x.

Sol.: Refer to the Example 6 of§ 5 of chapter 9 page (9.32). Considerl strip

parallel to the x-axis. On this stripx varies from x = y 2 tox = T· This

strip sweeps the area when the strip moves parallel to itself from y = - 3 to y=3.

3 3 :. A =I I 2f9 dy dx = 2 I I 2f9 dy dx 3 y 0 y

3 3 [ 2] =2 J [x] 2 dy=2J dy 0 y 19 0 [ y 10 . y y3 ]3 -2 + sm-1 --- 2 2 {fO 27 0

= 2 [} + 5 sin 1 -1] = 2 (J + 5

Ex. 3 : Find by double integration the area included between the curves

l = 4a (x +a) and l = 4b (b-x). (S.U. 1998, 2003)


Sol. : The two parabolas are shown in the figure.

y The curves intersect at (b- a,± 2 {(i) Consider a strip parallel to the x-axis.

y2 On this strip x varies from x = - a to 4a

2 x = b- . This strip sweeps the area when

f= 4a (x +a) it moves parallel to itself from y = - 2 {(i to y = 2 {(i .

2 +2-.rab b � :. A= I I..t.:._ dydx 4a a

Fig. (10.19)

I+ 2 .. b y 2f4b I 2..;[ y 2 �]

=2 [y] 2f dx=2 b+a-4b -4a dx 0 Y 4a a 0

2..[ y2

] [ y3

]2..; = 2 (b +a) Io 1- 4ab dx = 2 ( b +a) y- 12ab o = 2 (b +a) [ 2-./ab- = 2 (b+a) 2-./ab [1 t )=t (a + b) .Jab

x2 Ex. 4 : Show that the area between the curves y = d andy = 1 - a ,

. 0 1s (a> ) (S.U.1999)

Sol. : The two curves are parabolas as shown in the figure. They intersect where

x2 . 2 2 1- a z.e. a =a -X

i.e. (a2+ 1)�=a i.e. 1 :. Required area A

I1 x2/a = 2 2 dy dx Fig. (10.20)

0 ax

/(a 2 +1) 1 x 2fa /(a 2 +1) x 2 2 =2 [y] 2 dx=2 [1 - --a ax ] dx 0 ax 0

[ x3 [ {1+a2) = 2 X- 3a - J 0

= 2 X- 3a . X 0


2 [ Va [ 1_ (I+ a 2) • a ) ] = 2 Va ( 1 _ l ) = Va2+} 3a (a 2 + 1) Va2+} 3

Ex. 5 : Find by double integration the area common to the circles x2 + y2 -4y = 0 and x2 + l-4x-4y + 4 = 0. (S.U. 1998) Sol. : The equations can be written as x2 + (y- 2 )2 = 22. Its centre is (0, 2) and

radius= 2. And (x- 2)2 + (y- 2)2 = 22 . Its centre is (2, 2) and radius= 2.

By subtraction, we see that the circles intersect

at points where x = l. Consider a strip parallel to they-axis. Then on

the circle on the left i.e. on x2 + l-4y = 0 i.e. on

. A X y = , y vanes from 2- -x to

2 Fig. (10.21) Required area = 2 area ABC

2 2+�2 = 2 f dxf dy

I 2 2+ �2 2

= 2 f [ y] • dx = 2 f 2 - x dx 1 1

2 = 4 [ � 4 - x 2 + j sin 1

Ex 6 : Find the area between the curves l = 4x and

2x-3y + 4 =0 Sol. : We first solve the first two equations to find the

points of intersection.

We get l = 2 (3y- 4) i.e. l-6y + 8 = 0 :.(y-4)(y-2)=0 :.y=2or4

Fig. (10.22) When y = 2, x = l; when y = 4, x = 4. Let the points

of intersection be A(l, 2) and B(4, 4). 4 2-Vx 4 2-Vx

:. A =f f dydx=f [y](2 +4)/3 dx I (2 + 4) /3 I

= fl4[2 .JX' 4) ]dx = [2 ·1 X 3/2- (X 4x)]:


Ex. 7: Find the area bounded by l = 4ax and x2 = 4by.

Sol. : The two parabolas are shown in the

figure. They intersect in

A (4a113 b213, 4a213 bl/3).

Now, consider a strip parallel to the

y-axis. On this stripy varies from y = x2!4b to

y

y = 2 .,J; · ,J; . And then x varies from x = 0 to x

x = 4al/3 b213.

Fig. (10.23)

Ex. 8 : Find by double integration the area encosed by the curve 9xy = 4 and

the line 2x + y = 2. (S.U. 1999, 2003) Sol.: The curve 9xy = 4 i.e. xy = 4/9 is a rectangular Y

hyperbola.

Now, 9x (2 - 2x) = 4

:. 18x- 18x2 = 4

:. 9x2-9x+2=0

:. (3x- 2) (3x-1) = 0

:. x = 2/3 or 1/3.

Hence, the points of intersection of the

hyperbola and the line are (2/3, 2/3) and (113, 4/3).

X y = 2 -2x

Fig. (10.24)

!7

Engineering Mathematics - II

f213J2 2x :

. A =

1/3 419x dy

dx

(10-15)

=J2/3[ ]2-2x dx=J2/3(2_2x _ _i_)dx 1/3 y 419x 1/3 • 9x

= 2x- x2 - Jogx

[ 4 ] 2/3

9 1/3

= ( i -i- i Jog 3. -3. +.! + i Jog .!.) 3 9 9 3399 3

I 4 =---log2. 3 9

Ex. 9 : Find by double integration the

area of the smaller region bounded by

� + i = 9 and x + y = 3. Sol. : The circle � + l = 9 and the line

x + y = 3 are shown in the figure.

:. A= d

ydx 0 y=3 x

= J3[

0 y 3-x dx

Appll. of Mult. lnte.

X

= s:[ -3+x Jdx Fig. (10.25)

[ 2 ]

3 X 2 9 . -1 X X = -� +-sm --3x+-

3 2 3 2 0

= [( 0+�·� 9+�) -(0)]= �(n-2).

Ex. 10 : Find the area of the curviliner triangle lying in

the first quadrant, bounded by the curves y2 = 4ax,

y= ax

�=4ay,�+l=5a2• x

Sol. : The required triangle is the curviliner triangle

ABC. The vertices are A(4a, 4a), B(2a, a), C(a, 2a). We

have to divide the area into two parts ABD and BDC. Fig. (10.26)


AreaABD I4a J2VaVx

= d dx 2a x 214a y

Appll. of Mutt. lnte.

I4a 2..[i .;x J4a [ x 2 ] = [ y ]

dx = 2Wi .fX 4a dx

2a x 2a - [2Wi .l

X 312 _.!2.]4a - 3 12a 2a s.J2

3·a2 + ja:?

18 2 s.J2 2 = 3a

2a Area BDC = J dy dx

= J2a

[ y ]2..fi¥x dx =J2a [ 2-Va..fX 2_x2] dx

a a

= [ 1 va · x 312 2_ x 2- 2 sin 1 ]:a

8-.fi 2a 2 Sa 2 • [ 2] 4 .f5 ja2

2a 2 Sa 2 . ( 1 ) + + --r sm-1 .f5

s..fi 4 Sa 2 [ . ( 1) . ( 2 )] sm-1 .f5 sm 1 .f5

:. Required area of the traingle ABC= area ABD + area BDC

= 13

4 a 2 + 2 [ sin-{ }s) -sin-1 (}s)] .

Ex. 11 : Find the area of the cardiode r =a (1 +cos 9). Sol. : For the cardioide r varies from 0 to

a (1 +cos 9) and e varies from 0 to 1t above

thex-axis.

. J7tJa(1+cos8) :. Area=2 0 0 r9d9dr

7t[ r2 I(l+cos8) =2J - de . 0 2

Fig. (10.27)

Engineering Mathematics - II (10·17) Appll. of Mult. lnte.

= a2 J: 4 cos4 e /2 ·de

= a2 J :12 4 cos 4 <I> • 2d<l>

2311t 3 2 =Sa ·-·-·- = -na . 4 2 2 2

Put 2e +<I>

Ex. 12 : Find the total area of the curve r=a sin 2e. Sol. : The curve is a four leaved rose as shown in the figure. Consider a radial strip. On this strip r varies from r = 0 to r = a sin 2e. Then e varies from 8 = 0 to e = 1T/2.

J'lt/2 Jasin26 :.Area=4 0 0 rdrd8

y

Fig. (10.28) [ 2 ]asin26 = 4 J;1

2 r2 0

de = 2 J:12 a2 sin2 2e de

= a2 Jn/2 (1- cos4e) de 0 2

=a2 [e- sin4e ]n'2 = na2• 4 0 2

(a, x

Ex. 13 : Find the area inside the circle r = a sin e and outside the cardioide r =a (l- cos e). (S.U. 2003)

Sol. : The circle and the cardioide intersect where a sin e = a (l - cos e) i.e. 2 sin (e/2) cos (e/2) = 2 sin 2 (e/2) i.e. sin e/2 [sin (e/2)-cos (e/2)] = 0.

When sine/2=0 :. e=O y . e e e n When sm--cos-=0 -=-

2 2 '2 4 :.e=�

2" Now, consider a radial strip in the region

of integration. On this strip r varies from r =a (l- cos e) tor= a sin e . Then e varies from 8 = 0 to 8 = 1T/2.

I7t/2

Jasin6

:.A= rdrd8 0 a(l-cos6) n/2 de

]asin6 � fo o(l-=91

Fig. (10.29)


= _!_ Jn12 [a2 sin2 8- a2 (l- cos8)2] d8 2 0

a2 Jn/2 2 2 = - (sin 8 -1 + 2 cos 8-cos 8) d8 2 0

a2 Jn/2 =- (-1+2 cos8-cos28)d8 2 0

2 [ . 28]7t/2 a . sm =- -8+2 sm8- --

2 2 0

= a2 [

-�+2]

= a2(4-1t)

. 2 2 4

Appli. of Mult. lnte.

Ex. 14: Find the area between the circles r = 2a sin 8, r = 2 b sin 8 (b >a). Sol. : We have r = 2a sin 8

. 2 2 y 1.e. + y = a·

+ l

i.e. x 2 + y 2 = 2ay i.e. x 2 + (y- a) 2 = a 2. Similarly, r= 2b sin 8 givesx2 + (y --b) 2 = b 2.

These are the circles with centres (0, a), (0, b) and radii a, b. Now, consider a radial strip. On this strip r varies from r = a sin 8 to r = b sin 8. Then 8 varies 8 = 0 to 8 = n/2 in the first quadrant. [ 2 ]bsinO J7t/2Jbsin0 J1t/2 r

:. A = r dr d8 = 2 - de 0 asinO 0 2 a sinO

Fig. (10.30)

= J�12 (b2 sin2 8-a2 sin2 8) d8

= (b2 -a2) J�12 sin28 d8

=(b2 -a2)._!_·�=(b2 -a2)�. 2 2 4

Ex. 15 : Find the area common to the circle r =a and the cardioide r =a (1 +cos 8).

Sol. : The two curves are shown in the figure.

Area = 2 area OAPC + 2 area OCDO

J7t/2 Ja J7t Ja(l+cosO) = 2 r dr d8 + 2 r dr d8

0 0 rr/2 0

Fig. (10.31)

Engineering Mathematics - II (10-19) Appll. of Mutt. lnte.

a [ 2l a(l+cos9)

=2J:12[�r2l d0+2J:12 r

2 0 dO

=a2Jnl2d0+Jn a2(1+cos0)2d0 0 n/2 =a 2 [ 0 ]�12 +a 2 J:12 a 2(1 + 2cos0 +cos 2 0) dO

na 2 zstr ( 1 2 1 + cos20) =--+a + 2 tr/2 2 ·na 2 n ( 3 1 ) =--+a2f -+2cos0+-cos20 dO 2 tr/2 2 2 na2 2. sin20 ]n 2 2 4 tr/2

=na2 +a2 [�n-3n_2] 2 2 4

= +a2( 2 )a2. Ex. 16 : Find the area outside the circle r = a and inside the cardiodide

r = a (1 + cos 8). y Sol. : The circle r = a and the cardioide

r =a (1 +cos 8) are as shown in the figure.

When they intersect

a =a (1 +cos 8) :. 1 = 1 +cos 8

:. cos 8 = o :. 8 = ± 1tl2. :. Area= 2 area ABQCPA

= 2 s:/2 s;<l+cos9) r d r dO

Fig. (10.32)

12[ l ]a(l+cos9) 12 =2J: 2r2 a

dO= J: [a2(1+cos0)2-a2Jdo

=a 2 J:12 (1 + 2cos0+cos2 0-l) dO

Enginee,ring Mathematics - II (10-20)

=a 2 s:'2 [ 2 cos 8 +( 1 + dB

a 2 J n/ 2 =- (l + 4cos 8+ cos28) dB

2 0

a . stn 2 [ • 28]17:/2 =- 8+4sm8+--

2 2 0

Ex. 17 : Find the area inside the cardioide r = 2a (1 + cos 0) and outside the parabolar

2a = (1 +cos 9). Sol. : The curves are shown in the figure. When

. 2a they mtersect, 2a (1 + cosO) = ( 1 + cos e)

:. (1 +cos 9)2 = 1 e e 1

:. 4 cos4 2 = 1 : . cos 2 = ± .fi

:. Area= 2 areaABQ CPA


Fig. (10.33)

= 2 ('2 f2a (I+ cosO) r dO dr = 2 ('2 [.1 r 2] 2a (I+ cosO) de

0 2a /(I +cosO) 0 2 2a I( I + cosG)

2 fn/2 [ 2 1 ] = 4a 0 (1 +cosO) - (1 + cose)2 de

7t/2 n/2 Now f0 (1 +cos 9)2 de = f0 (1 + 2 cos e + cos29) d e

= f:2 [ 1 + 2 cose + 1 + ) de

= f:2[! + 2 cosO+ cos 20) dO

= [! e + 2sin a + (2 = [ + 2] 0

I1tf2 de 1 n/2 de And o (1 + cose)2

= 4 Jo (cos2 9/2)2


1 rr/2 e 1 rr/2 [ e ] e = - f sec4 - de = -f 1 + tan2 - sec2 - de 4o 2 4o 2 2

1 ' [ e ) =4f0(I+t2)2dt Put tan2=t

[I+ ]: = � [ 1 + i] = j

Hence, the required area is

A

Ex. 18 : Find the are a between the curve y =a (sec e +cos 8) and its asymtote r =a sec e.

(S.U. 2003)

Sol. : The curve and its asymptote are as shown in the figure.

Area = 2 r d r d() fn/2 Ja(sec8+cos8) 0 asec8 0� �0

� � [

fn/2 r =2 - rdrd() 0 2 asecll

Fig. (l0.34)

= J;12 [a 2 (sec 2 () + 2 +cos 2 8)-a 2 sec 2 () J d()

=a (2+cos 8) d8= a 2·-+-·- =-a . 2Jn/2 2 2 [ 1l 1 11:] 51l 2 0 2 2 2 4

o"" )(c;

Ex. 19 : Find the area enclosed by the curve x(x2 + i) = a (x2 - i) and its asymtotes. (S.U. 2005) Sol. : See the figure (10.11) 10·6. The asymptote is x =-a

fo (a x)l(a + x) :. Area= 2 dy dx

a 0 o _ 0 =2Cx To find the integral we put X=-a sine, dx =-a cos e de when X=-a,

1t 8 = l ; When X = 0, 8 = 0

. _ fo -a sin +a . . . Area - - 2 a cos (-a cos 8) d8 rr/2

Engineering Mathematics - II (i0-22) Appll. of Mult. lnte.

'It/2 =- zf a 2 {sin 8+ sin 2 8) de

0

2 2Jrt/2 . e + -cos - 2 2 [ e +.! sin 28]1t'2 =- a sm -- a - cos 2 - 2 0 0

= _ 2a 2 [ i + 1] = 2a 2 [ i + 1] (Numerically)

Ex. 20 : Find the area bounded by the curve l (2a- x) = � and its asymptote. Sol. : The curve is shown in the figure. Its asymptote is y x=2a.

I 2a X x) :. Area= 2

0 0 dy dx x

J2a

= 2 0

Put X = 2a sin28, dx = 4a sin 8 COS 8 d8 when X = 0, 8 = 0; when X = 2a, 8 = rrJ2.

f1t12 sin2 Area= 2 . 2 . 4a sine cos e de

o - 2a sm 8) 'It/2 'It/2

= 16a 2 f sin4 8 de = 16 a 2 f sin4 8 cosO 8 de 0 0

_ 16a 2 . 1 i572 il72 _ Sa 2 . . 3/2 · 112 · ii72 · ii72 - 2 13 - 2 · I

=Sa 2 ·in= 3 1t a 2

Ex. 21 : Find the area common of the circles

Fig. (10.35)

xZ + l = a2 and xZ + l = 2ax i.e. r = a and r = 2a cos e Sol. : First we note that r = a is a circle with centre at the origin and radius = a and r = 2a cos 8 is the circle with centre at (a, 0) and radius = a.

To find the points of intersection we solve the two equations.

:. a=2a cos8 1 1t :. cos e = 2 :. e = 3

:. Required area= 2 [area OCB +area OBA ]

[ I'It/3 Ja f'It/2 J2a cos 6 ] = 2 rd rd8+ rdrd8

0 0 'It/30

= 2 [('3[ r22 ]: de+ c::[ r22 ]:a cos 6 de]

Fig. (10.36)


[ 7t/3 a 2 rt/2 ) = 2 f. 2 d e + J 2a 2 cos 2 e de

0 7t/3 7t/3 rt/ 2

= a 2 J d 8 + 2a 2 J (I + cos 28 )d 8 0 rt/3

= a 2 [ 8 ] 1t/\ 2a 2 [8 + sin 28 )rt/2

0 2 rt/3

= a 2 � + 2a 2 [ � + 0 - � - ± ·1 ) = 2 1 a 2

Ex. 22 : Find by double integration the area between the curve y 2x = 4a 2 (2a- x) and its asymptote. (S.U. 2005)

Sol. : The curve is shown in the figure. They-axis is its asymptote. The curve is known as "Which of Agnesi".

2 a--

Area= 2 s:a f0 .[; dx dy

=2f2a·2a-� dx 0 ..[; Now, put x = 2a t :. d.t = 2a dt

f I ( -1/2

( ) 1/? Area =2 02a· 2at) 2a

=8a2 J�t-112(1-t)112dt = 8a2 B ( 112, 3/2)

= 8a 2 lif2 [3/2 = 8a 2 �� . _!_I_!_ 12 2 2 2

= 4a2 (7t/2)2 = 4a2 n.

(2a, 0)

Fig. (10.37)

Alternatively : The above imtegral can also be evaluated by putting x = 2a sin 2 8, dx = 4a sin 8 cos 8 d8

f;r;/2 cos 2 8 . :.1=4a . 2 ·4asmfJcosfJdfJ 0 2asm 8

= 16a2 f:12 cos28 dfJ

1 1C 2 = ·-·-= 4a 1C. 2 2

Engineering Mathematics - II (10·24) _Appll. of Mult. lnte.

Exercise - II

1. Find by double integration the area between the circle

x 2 + y 2 -- 2ax = 0 and the parabola y 2 =ax. [Ans. : 2a 2 ( * - �)] (Fig. similar to fig. 9.54 page (9.55))

2. Find by double integration the area bounded by the parabolas

y 2 = 4ax and x 2 = 4ay. (See fig. 10.43 page (10.27)) [ Ans.: a 2] 3. Find the area betweenn the curves y = x 2 - 6x + 3 and y = 2x- 9.

(See fig. 7.152 page (7.38)) (S.U. 2000, 02) [ Ans.: 32/3]

4. Find the area between the curves y = 3x 2 - x- 3 andy = 7 + 4x- 2.l-.

(See fig. 7.153 page (7.38)) (S.U.1997, 2003) [ Ans.: 5. Find by double integration the area bounded by the parabola y 2 = 4x

and the line y = 2x- 4. (See fig. 7.154 page (7 .38)) [ Ans. : 9 ] 6. Find by double integration the area enclosed by the rectangular

hyperbola .xy = 4 and the line 2x + y = 6. (S.U. 1999, 2003) (Fig. similar to fig. 10.24 page (10.14)) [ Ans.: 3- log 16]

7. Sketch the region R where

J!Jx

J J 0 0 dx dy + I 0 dx .xy = R dx dy.

Hence, find the area of the region. (S.U. 2001) (See fig. 9.28 page (9.33)) [ Ans. : rr14 ]

8. Find the area between the parabola y = 4x - x 2 and the line y = x.

(See fig. 7.155 page (7.38)) (S.U. 2006) [ Ans.: 912]

4. Mass of a Lamina If the surface density p of a plane varies from point to point of the lamina

and if it can be an expressed as a function of the coordinates of a point then the

mass of an elementary area dA is p dA . In cartesian coordinates if p = f (x, y), since dA = dx dy, mass of the

lamina = JJ f(x,y) dx dy.

In polar coordinates if p = f (r, e), since dA = r de dr mass of the lamina

= JJ f(r, e) r de dr where the double integral is to evaluated over the area of

the lamina.

Ex. 1 : Find the mass of the lamina bounded by the curve ay 2 = x 3 and the line


y by= x if the density at a point varies as the distance of the

point from the x-axis.

x Sol. : The two curves intersect at A [ ba 2 ' ba 3) .

The lamina is the area OAB. On the curve OBA, xY2 X

y = 40 and on the line OA, y = b. The surface

Fig. (10.38) density is given by p = ky. Taking the elementary strip parallel to they-axis, mass of the lamina.

alb2 M=kf ydxdy

0 x3t2fya k alb2[x2 x3] -k dx=- --- dx - 2 x312Jva 2 � b 2 a

k [ x 3 x 4 aJb2 k [ a 3 a 3 ] = 2 3b 2 4a ]0 = 2 3b 8 - 4b 8 ka3[1 1] k a3 = 2 7)8 3 - 4 = 24 . 7)8

Ex. 2 : The lamina in the form of parabolic segment of mass M, height h and

base 2k has density at a poit given by Apq3 per unit area where p, q are distances

from the base and axis respectively. Find the value of A. Sol. : Let the parabolic segment be as shown in the y figure. Let the equation of the parabola be i = 4ax. Since the point B(h, k) lies on the parabola k2 = 4ah;

k2 k2 4a = h ; the equation is l = h x.

:. If P (x, y) is any point on the lamina, then the

distances p, q are as shown in the figure.

:. x + p = h i. e. p = h -x; q = y h k./Xlh

Mass of the lamina M = 2f0 f0 A pq3 dxdy h k./Xlh

:. M =2 f0 f0 A(h-x)y3 dxdy h [ 4 ]k./Xlh

= 2A I (h x) L dx 0 4 0

Fig. (10.39)

h A h 4 x2 A.k4 [ x3 x4] =2 f0 (h x)·k "hf dx= 2h2 h3 4 o A.k4 [h4 h4] A.k4 h2 24M

:. M = 2h2 T 4 = 24 :. A=I4JI


Ex. 3: The density of a uniform circular lamina of radius a varies as the square of its distance from a fixed point on the circumference of the circle. Find the mass of the lamina.

Sol. : Let the fixed point on the circumference of the circle be the origin and the diameter through it be the x-axis. Then the polar equation of the circle is r = 2a cos e. The density at any point P (r, e) is= ktJ. Hence,

rr/2 2a cos e Mass of the lamina = 2 f0 f0 (k?) r de dr

rr/2

[ ,.4 ]2a cos e

=2kl de 0 4 0 k rr/2

= 2 f0 16 a4 cos4 e de

4311t 3 4 = 8ka . 4

. 2 . 2 = 2 ka 1t.

y

Fig. (10.40)

Ex. 4 : The density at any point of a cardioide = a(l +cos e) varies as the square of its distance from its axis of symmetry. Find its mass.

(S.U. 1999,2006, 07)

Sol. : Let P (r, e) be any point on the given cardioide, The distance of P from the axis is r sin e. The density at any point P (r, e) is p = k? sin2 e

Mass of the lamina y rr a (I +cos 8)

M = 2 J0 J0 (k? sin2 e) r de dr

x [ 4 ]a(l +cos e) = 2kf0 sin 2e r4 0 de

k 4 2 4 = 2 a sin 8 (1 + cos e) de

= ((2 sin � cos � r (2 cos 2 � r d8

= 32ka4( sin2 � cos 10 � de.

e = o x

Fig. (10.41)

[ e ] 1 13/2 11 112 = 64 ka2 sin2 t cos10 t dt where 2 = t = 64ka4

• 2 17 lll9753lll

=64 ka4-l 2 2"2"2"2"2"2 2 2 6·5·4·3·2·1

21 4 = ka 1t

Engineering Mathematics - II (10-27) Appli. of Mutt. Inte.

Ex. 5 : A lamina is bounded by y = x2- 3x andy= 2x. If the density at any point is given by (24/25) xy. Find the mass of the lamina. (S.U. 1997,2001, 05)

. 2

Sol. :The curve y = x2- 3x i.e. y + � = ( x- %) is a parabola intersecting the

x-axis in x = 0 and x = 3. The line y = 2x intersects y this parabola at 2 - 3x = 2t i.e. 2 - 5x = 0 i.e. at x = 0, x = 5. Therefore, points of intersection are (0, 0) and (5, 10). The lamina is the area OAB. Taking the elementary strip parallel to the y-axis, mass of the lamina

JSJ2x (24) M = 2 - xydxdy 0 x 3x 25 dx

= ( s:[ 4x 3 -x(x4 -6x3 + 9x2) J dx

24 Js s 4 3 = - (-x + 6x - 5x ) dx 50 0

= 24[-�+ 6x5 _ 5x4 ] 5

50 6 5 4 0

= = 87.5. 50 24

X

Fig. (10.42)

Ex. 6 : Find the mass of the lamina bounded by the curves l = ax and 2 = ay if the density of the lamina at any point varies as the square of its distance from the origin.

(S.U. 1997)

Sol. : The two curves intersect A (a, a). The lamina is the area OBACO. On the curve OCA, y = ,J; and on the curve OBA, y = 2ta. The surface density is given by p = k (x2 + l>. Taking the elementary strip parallel to the y-axis, the mass of the lamina

M = kfaJJf (x2 + y2) dx dy 0 x Ia

x2 =ay y

Fig. (10.43)

Engineering Mathematics - II (10-28) Appll. of Muit. lnte.

= X 7/2 + a.ra. XS/2 ..!._�

7/2 3 5/2 a 5 3a3 7 0

= k[3.a4 +I.a4 -�-�]= 6ka4. 7 15 5 21 35

2 2 Ex. 7 : Find the mass of the lamina in the form of an ellipse � + � = 1 if

a b the density at any point varies as the product of the distance from the axes of

the ellipse. (S.U.1999, 2004)

Sol. : Mass of the lamina = 4 Jf p dx dy = 4k JJ xy dx dy

M = 4kJ: xy dx dy

Fig. (10.44)

Ja b2 2 2 2kb2 Ja 2 3 =4k x·-(a -x )dx=-- (a x-x ) dx o a2 a2 o

2kb2 a

Ex. 8 : Find by double integration the mass of a thin plate bounded by i = x


and y = x3 if the density at any point varies as the Y

square of its distance from the origin. Sol. : Clearly the curves intersect at A (1 , 1 )

Mass = Jf p dx dy = f�f.;'f k(x2 + y2)dx dy

[ =kf x2y+L d:t 0 3 3 X

1 [ 2 X 3/2 2 3 X 9] =kJ0 x ·x -3 dx

7/2 X5!2 X6 X 10 = k 7 I 2 + 3 · (5/2) 6 30 0

= k[% + �- = k.

Fig. (10.45)

Ex. 9: A lamina is bounded by the cycloid x = a (8 +sin 8), y = a (1- cos 8), the ordinates at the two cusps and the tangent at the vertex. If the density at a point varies as the square of its ordinate, find the mass of the lamina. Sol. : Mass of the lamina by symmetry Y

M=2ff:ky2dydx

dx

2kf 3 =3 y dx

X

Fig. (10.46)

Now, we use parametric equations of the cycloid. y = a (1 - cos 8), dx = a (1 +cos 8) de and in the first quadrant e varies from e = 0 toe= 1t.

:. Mass = 2k JTr a \1 - cosO) 3 · a (1 +cosO) dO 3 0

= 2k a 4J1f 8 sin 6(0 I 2) · 2cos 2(0 I 2) dO 3 0

Now, put 8/2 = t :. dO= 2 dt


32 4Jn/2 . 6 2 :. Mass = -ka sm tcos t · 2 dt 3 0


= 64 ka4 .!.B(� �) = 32 ka4 . [7/2 f3J2 3 2 2 '2 3 15

32 4 (5 I 2)(3 I 2)(1 I 2) I 2 (1/2) I 2 =- ka · 3 4·3·2·1 ==2ka47r. 12

Ex. 10 : The density of a circular lamina is k times

its distance from a given diameter. Find its mass.

(S.U. 2001) Sol. : Let the given diameter be the x-axis and a

line perpendicular to it from one end of the diameter

be they-axis. Let a be the radius of the circle. Then

the polar equation of the circle is r == 2a cos e. The

density at any point P (r, e) is k y = k r sin e.

Fig. (10.47) .

Jn/2J

2acos8 . :. Mass of the lamma = 2 ·

0 0 (k r sm8) r dr d8

Exercise - Ill

X 2/3 2/3 1. Find the mass of the lamina in the form of astroid -- + E._ = 1 if

a 2!3 b 2!3 the density varies as xy. (See fig. 10.10 page (10.5))

[Ans. : A.a2b2!20. Hint Put X= a cos3 e, y = b sin3 e] 2. Find the mass of the lamina of the region included between the curves

y =Jog x, y = 0, x = 2, having uniform density. (S.U. 2000) (Draw y =log x) [ Ans.: 2log 2- l]

Engineering Mathematics - II (10-31) Appli. of Mutt. lnte.

3. Find the mass of the lamina bounded by the curve i = x3 and the line

y = x, if the density at a point varies as the distance of the point from the x-axis.

(See fig. 10.38 page (10.25)) [Ans.:

4. Find the ma�s of the lamina bounded by the curve 16l = x3 and the

line 2y = x, if the density at a point varies as the square of the distance of the

point from the x-axis. (Same as above) [Ans.:

5. The density at any point of a uniform circular lamina of radius a varies

as its distance from a ixed point on the circumference of the circle. Find the

mass of the lamina. (See fig. 10.40 page (10.26)) [Ans.: ka3]

6. The boundaries of a plate can be given by x = 0, y = 0, x = l andy= ex.

If the density at any point varies as the square of its distance from the origin,

find the mass of the plate. (Draw y = ex) [ Ans. : k ( e + e92 - ) ] 7. Find the mass of the lamina bounded by the curves i = ax and x2 =

ay, if the density of the lamina at any point varies as the square of its distance

from the x-axis. (See fig. 10.43 page (10.27)) (S.U.1997) [Ans.: 365 ka4] 8. Find the mass of the lamina in the form of a cardioide r = a ( 1 +cos 8);

if the deusity of mass at a point varies as its distance from the pole.

(See fig. 10.27 page (1 0.16))

5. Volumes of Solids Let z = f(x, y) be the equation of the surfaceS. Let

z R be the orthogonal projection of this surface on the xy-

plane. Let the equation of this projection be f (x, y) = 0. Consider an elementary parallelepiped with dxdy on the

[ Ans.: �kna3]

xy-plane as the base and bounded by surface S on the , , top. Its volume is z dxdy = f(x, y) dxdy. ' Y

The summation of all such terms over the region

R gives the volume of the cylinder bounded by the

surface S and the xy-plane.

Volume= J J f(x, y) dx dy R

Fig. (10.48)

The volume of a solid can also be expressed as a triple integral. If we

consider an elementary cuboid then its volume is dxdydz and hence the required

volume is


= fJf dx dy Ex. 1 : Find the volume bounded by l = x, x2 = y and the planes z = 0 and x + y + z = 1. Sol. : The solid is bounded by the parabolas l = x and � = y in the xy-plane which is its base and by

the plane x + y + z = 1 at the top.

:. V = J J Z dx dy = J J (1 -X- y) dx dy R R


Fig. (10.49)

Now, R is bounded by parabolas l = x and x2 ::; y in the plane. They

intersect at (0, 0) and (1,1). I JVx

:. V = J0 x2 (1-x-y)dxdy

I [ y2 ]Vx = f y-xy-- dx 0 2 x2

= J� [[;x-x3/2 -1)- [x2 -x3- x24 )] dx

[ 2x3/2 _

2xs12 _ £. _ � + x4 +.£. ]1

= 2 2 4 3 4 10 0 2 2 1 1 1 1 1 = 3_5_4_3+

4+10=30

y

X

Fig. (10.50)

Ex. 2 : Find the volume cut off from the paraboloid x2 + l + z = 1 by the

plane z = 0. Sol. : The paraboloid cuts the xy-plane in the ellipse

'1. l:. '

�+ 9 = 1.

Hence, the volume

V = JRJ zdxdy = JRJ (1-x2-f )dxdy

where R is the area of the ellipse x2 + y2 19 = 1.

Jl J+3� [ 2 .i.] :. V = 1-x - dx dy -1 -3� 9

I [ 3 ]+3� = f (1-x2)y-L ·dx -1 27 -3..'f:7"

I = I_ l 4 ( 1 x2)312 dx [ Put X = sin 8 ]

Jx/2 4 = 4 1t12 cos e d8

z

Fig. (10.51)


Irr./2 = 4 · 2 0 cos4 B dB

= 4 . 2 ·[ t. t. = 1t

Ex. 3 : Find by double integration the volume of the sphere;(- + 1 + z2 = a2 cut off by the plane z = 0 and the cylinder :l- + 1 = ax.

z Sol. : The base of the cylinder :l- + 1 = ax is the circle

( x-�)2 + i =( �)

2. The volume is bounded by this

circle in xy-plane by the cylinder on one side and by the sphere :l- + i + i = a2 on the top.

Taking polar coordinates elementary area at Pis F" (10 52) r dB dr. On the circle ;(- + 1 = ax i. e. on r = a cos e. Ig. •

y

r varies from 0 to a cos e and e varies from - Tri2 to Tri2. y

Also z varies from 0 to + from 0 to

� :. v I1t/2 I

a cos a =2 0 0

Fig. (10.53) To find the first integral put a2 - ? = t.

dt :. rdr=-z When r = 0, t = a2; when r = a cos B, t = a2 sin2 B

I 1t/2 I

a2 sin2 a In dt I rr./2 [ t312 ]a2 sin2 a

:. V = 2 0 a2 - t 2 dB= 2 0 - 3 a2 dB

2 I7t/2 2a3Irr./2 = 3 0 [ - a3 sin3 e + a3 ] dB = 3 0 (1 - sin3 B) dB

_ 2a3 [I 1[12 dB_ J rr./2

sin3 e dB) - 3 0 0

2a3 [

.n.- ..£ t ] = 3 2 3 2a3 = 2a

93 (37t- 4). = 3 6

Ex. 4 : Find the volume cut off from the sphere :l- + i + i = a2 by the cone :l- + 1 = i. (S.U. 2004)

Sol. : Using triple integral V = JJJ dz dx dy. Consider the intersection of the sphere and the

z

Fig. (10.54)

y


2 cone. On this intersection we have� + l =

a2 . In polar

coordinates it is a circle r = a I .J2 . On this circle r varies

from 0 to a I fi and e varies from 0 to 2n.

y

Consider an elementary parallelepiped (as shown in the figure) and change dx dy tor d9 dr. Fig. (10.55)

I 21t I at.J'I I� :.V=

0 0 r rd9drdz

X

I 21t I a/.J'I I 21t Ia/.J'I = [ z ]r rd9dr= rd9dr

0 0 0 0

I 21t [ (a2 _ r2)3/2 r3 ]ai.J'I

= o -3 o de li21t(L L 3)de =-3 o 2..+2VI a

=t a3I o21t(I-- d9= (I-

1ta3 = (2 -¥2).

Ex. 5 : Find the volume common to the right circular cylinders x2 + l = a2 and x2 + z2 = a2.

(S.U. 1997, 2002)

Sol. : By symmetry the required volume = 8 volume in the first octant.

:. V = Jff dx dy dz

On the first octant z varies from 0 to �:. V =8Jf -x2 dx dy

Now in the circle x2 + y2 = a2, y varies from 0 to

� and x varies from 0 to a. a

:. V = 8 I0 I0 ..;dx dy

= -x2 )·y I ·dx = 8 J� (a2 -x2) dx

= s[ a2 x- = 8. 3 = 3

. .

.

Fig. (10.56)

y

Fig. (10.57)

X


Ex. 6 : Find the volume bounded by the cylinder x2 + l = 4 and the planes

y + z = 3, z = 0.

Sol. : It is clear from the figure that z varies from z = 0 to z = 3 - y. It we use polar coordinates then z varies from z = 0

to z = 3- r sin e, r varies from 0 to 2 and e varies from 0 to 2n. Note that the top of the cylinder is not symmetrical.

J21t J2 J3-rsin9 .. V = dz rdrde

0 0 0

J

21!J2 =

0 0[3-rsin8]rdrd8

2n r2 r3 [

]

2

=

J

0 32-3sin8

0

de

J21![

8 . J =

0 6-3sm8 d8 [ 8

]21!

= 68+ 3 cos8 0

= ( 6 ·2n + �) - ( 0- �) = 12n

Ex. 7 : Find the volume of the solid bounded

by the cylinder x2 + l = 2ay, the paraboloid

x2 + j = az and plane z = 0.

Sol. : The base of the cylinder x2 + j = 2ay is the

circle x2 + (y -a )2 = a2. And z varies from z = 0 to 2 z = + y ) Ia.

X Fig. (10.58)

Fig. (10.59)

y

X

(x2 +y2)fa :. V =

J J J dx dy dz R 0

where R is the region of the base X Fig. (10.60)

JJ JJ(x2+y2)

:. V = R[ z ]0 dx dy = R · dx dy

Changing to polar coordinates.

x2+y2 r2 2 2

a' dx dy = rdr d 8, the base circle X + y = 2ay is

r2 = 2ar sin e i.e. r = 2a sin 8

Jn/2 J

2a sine r 3 2 Jrt/2 [r4 ]2a sine :. v = 2 -a dr de= -a -4 d e 0 0 0 0


7tl2 = 8a3 Jo sin4 de

_ SaJl.l.K - 4 2 2

= � 1ta3. z

Ex. 8 : Find the volume enclosed by the cone

+ = z and the paraboloid x2 + i = z. JJ J..;xr+yt

Sol. : V = x2 + y2 dx dy dz = J J [zr;:; dx dy = H + y2- (x2 + i> 1 dx dy

Fig. (10.61)

The intersection of the cone and the paraboloid is + y2 = x2 + i i.e. �+i=L

Changing to polar coordinates r varies from 0 to 1 and e from 0 to 2n.

J1t/2JI 2 y V = 4 0 0 l r- r ] r dr de

= 4 I ;12 J � [ ? - ? 1 dr de - J 1t/2 (.2. - r4 ]I de = 4J 1t/2 .J_ de -4 0 3 4 0 0 12

Fig. (10.62)

Ex. 9 : Find the volume of the region bounded by z = � + i and z = 2x.

Sol. : � + l = z is a paraboloid and z = 2x is a plane

:. z varies from x2 + i to 2x.

V =If J�+y2dxdydz = JJ [ 2x -x2 -l] dx dy

Now the intersection of the two surfaces is

the circle x2 + y2 = 2x i.e. (x- 1)2 + l = l. The

elementary area at p is r d e dr. Now r varies from 0 to ? = 2r cos e i.e.

r = 2 cos e and e varies from -1t/2 to 1t/2

J'lt/2 J2 cos e :. v = 2 0 0 (2 r cos e- r dr de

y

Fig. (10.63)

X


= 2J;I2 [ 2 cos e r44 J: cos �e

= 2J;12 [ 136 cos4 e-

146

cos4 e] de

- .§. J 7t14 cos 4 de = ! · l · l · E. = E. -3o 3 4 2 2 2 · Ex. 10 : Find the volume of the sphere x2 + i + z2 = a2 by using triple integration. (S.U. 2003) Sol. : Using cylindrical polar coordinates X= r COS e, y = r sin e, Z = Z, We See that in the first quadrant z varies from z = 0 to

rvaries from 0 to a, and e varies from 0 to rrl2


y

X

Fig. (10.64)

Y

fn/2J"JJa2-r2 Fig. (10.65) .. V=8 dzdrrde 0 0 0

J7t/2J a[ = 8 z 0 rdr de 0 0

= 8 J ; 12 J; a2 - r2 r dr de X

_ Jnl2[ (a2-r2)312 (-.!_)]"de Fig. (10.66) -8 0 3/2 2 0

8 Jn/2 3 e =- a d 3 0

[Put a2 +? = t]

3 = � a3 [ e r/2 = 8a • � = .i 1ta3

3 ° 3 2 3 Exercise - IV

1. Find the volume of the tetrahedron bounded by the planes x = 0, y = 0,

= 0 and .!. + � + .£. -1 a b c- .

(See fig. 10.49 page (10.32)) . J aJ b(l -xla)

J c(l xla xlb)

[ Hmt : V = 0 0 0 dx dy dz ]

(S.U. 2006)

(Ans.:

2. A cylindrical hole of radius b is bored symmetrically through a sphere

of radius a show that the volume of the remaining solid is 1t (a2- b2)312•


(See fig. 7.156 page (7.38)) " Jx/2 Ja [ Hint : V = 8 0 0 � · r dr de]

3. Find the volume bounded by the cylinder :l- + l = 4 and the planes

y + z = 4 and z = 0. (S.U. 1998, 2004)

(See fig. 10.58 page (10.35)) [ Ans. : 16 - 332 ]

4. Find the volume bounded by l = x, :l- = y and the planes z = 0 and

x + y + z = 2. (See fig. 10.49 page (10.32)) (S.U. 2005) [ Ans.: 11/30 J

5. Find the volume cut off from the paraboloid :l- + l + z = 1 by the

plane z = 0. (See fig. 10.51 page (10.32)) [ Ans.: 1t] 6. Find the volume enclosed by the paraboloid x2 + l = az and the plane

6 0 3 6 [Ans. : rt2a3 ] z = a. (See fig. 10. 3 page (1 . ))

" Jaix/2 Ivaz [ Hint : V = 0 0 0 4 r dz d e dr. ] 7. Find the volume of region bounded by paraboloid z = :l- + l and the

plane z = kx. (See fig. 10.6 3 page (10. 3 6)) [ Ans.: k41t/32] kx

[Hint: V = IJ I dx dy dz = IRI Ix2 + Tdx dy dz

Ix/2 Jkcose

= _ x/2 0 [ kr cos e - ? ] r dr where R is the region

x2 + l = 2x i. e. r = kcos e.] 8. Find the volume of the region above the xy plane enclosed by the

paraboloid az = :l- + y2 and the cylinder :l- + i = 2 ax. [ Ans. : 321t a 2]

(See fig. 10.60 page (10.35)) [Hint: v = 2 IR I Io

(x2 + r)la dx dy dz = I I R(;;. + y2) dx dy - 1.

Ix/2I2acose ed - a o o d r.]

9. Find the volume of the region enclosed by the cone z = k +

and the paraboloid z = k (x2 + l1 above the .xy-plane. [Ans. :

(See fig. 7.157 page (10.38)) • I f

[ Hmt: V = 4 IRI Ik(x2+r>dxdy dz = 4 RJ [ r- rde dr

= 4 I;'2 I� <l- ?) r de drl


10. Find the volume of the region enclosed by the paraboloid

az = x2 + / and the cylinder x2 + i = a2 above the xy-plane. [Ans. : a3] (See fig. 7.158 page (7.38))

• (x2+i)la [ Hmt : V = 4 f R f f 0 d-r dy dz

r2 rt/2 a r3 =4f f -·rd8dr=4f f -d8dr.]

R a o 0 a 2

11. Find the volume enclosed by the paraboloids z = 4 - x2 - and

z = 3x2 + t /. (See fig. 7.159 page (7.38)) (Ans.: 4¥2 · rc]

[ Hint : Intersection of the paraboloids is the ellipes 8x2 + l = 8 .

f f f 4 x2 (y214)

V = 4 R 3x2 + (y2/4) dx dy dz

J I J� 4 -x2- (y2/4) = 4 0 0 [ z) Jx2 + (y2/4) dx dy

64fl J1 ') 3/2 . = 3 0 (1- x"' ) dx. Put x = sm 8

64{2J"12 = 0 cos 4 8 d 8. ]

6. Volume of a Solid of Revolution

(a) Volume of a Solid of Revolution : Suppose the area Y

bounded by the curve y = f (x), the ordinates at x = a and

x = b and the x -axis is rotated the x -axis. To find

the volume of the solid so generated, consider a small

strip at P (x, y) on the curve. This strip generates a disc of

volume nldx. Hence, the total volume is given by

0 a dx b x

Fig. (10.67)

V=n f: y2dx •••• (1)

If the curve is rotated about they-axis then clearly

v = 1t s: x2 dy •.•• (2)

Polar Coordinates : If the curve is given in polar coordinates r = f (8) then the above formula can be suitably changed by putting X = r COS 8,

y = r sin 8. (b) When a closed curve is revolved: Suppose the area bounded by a closed

simple curve such as shown in the figure is rotated about x-axis. Now, consider


an elementary area dx dy at P (x, y). This area when Y revolved about x-axis will generate a ring of volume

2 1ty dx dy. By taking the intrgral over the given area we

get the total volume as

•••• (3) 0 X

If the area is rotated about they-axis then clearly

Polar Coordinates : If the closed curve is given in Y

polar coordinates as r = f (8), we consider an elementary

area r dr d8 at P (r, 8.) When the area rotates about

the initial line i.e. about the x-axis, the elementary

Fig. (10.69)

•••• (4)

area generates a ring of radius r sin 8. Its volume is o x

2n r sin 8 · r dr d 8. Hence, the total volume Fig. (10.70)

= 2r fJ R r2 sinS dr •••• (5)

where the double integral is to be taken over the

given area.

If the area is enclosed by two radii vectors at 8 = a and 8 = 13 and the curve r = /(8) is rotated about the initial

line, we have by the above result.

X

v = 2n If} sine de r= J{a>r2 dr a 0

V = ¥-J! r3 sinS de. 7. Surface of Solid Revolution :

Suppose the area bounded by the curve y = f(x), the Y ordinates at x = a , x = b and the x-axis, rotates about the

x-axis. Comsider an elementary arc ds at a point P (x, y). This arc when rotated generates a disc of radius y and the

Fig. (10.71)

•••• (6)

curved surface area of the disc is 21t y ds. By integration

over the given curve, the surface of solid of revolution is

given by

(i) Cartesian Coordinates When the curve is given by y = f (x), we have,

Fig. (10.72)

Engineering Mathematics - II (10-41) Appll. of Mutt. lnte.

When the curve is given by x = f (y), we have,

(ii) Parametric Equations

When the curve is given by x =f (t), y = \j1 (t), we have,

ds = ( )\ ( r-dt (iii) Polar Coordinates

When the curve is given by r = f (9), we have,

ds= When the curve is given by 9 =f(r), we have,

·dr ( Unfortunately we accept the above formulae without proof. )

Ex. 1 : Find the volume of the solid obtained by revolving the loop

i = � · about the x-axis.

Sol. : The above volume is given by

fo 2 fo 2 a+ x V = 1ty dx= 1tX ·-=-dx a a a X

Put a - x = z i. e. x = a - z . _ f2a(a-z)2(2a-z) .. V -1t a z dz

= 1t r:a [ 3 -5a 2 + 4az z 2 ] dz [ 3 ]2a =1t 2a3logz-5a2z+2a z2- Z3

a

= 2n a 2 (log 2 - t)

-a

(S.U. 2000,03, 07)

2a Fig. (10.72)

X

Ex. 2 : The area bounded by the curve y = ae x2tb2 and the ordinates at x = b

and x =-band x-axis is rotated about they-axis. Find the volume so generated.

Sol. : The equation can be written as

x2=-b2 lo� {�) .


V = Solid of revolution + volume of cylinder

:. V = 1t J x2 dy + nb2 � . From y = ae-x?-1b2

we get

a when X = b, y = e ; When X = 0, y = a

:. Now I= 1t J x2 dy = 1t r - b2 log (�) dy ale

Putting � = t, dy = a dt

I :. I = Jl/e (-1t b2) log t · a dt

I = -nab2 [(log t) t - t] 1/e =-1tab2 [-I- � log � +� ]

= -nab2 [- I + �] = 1t ab2 (1 - �) :. Required V= 1t ab2 (I- � + �)= 1t ab2 (I -- �) ·

Appll. of Mutt. lnte.

y (o,a)

Fig. (10.73)

Ex. 3 : The area enclosed by the lines y = 0, 2x + y = 2a and the curve

l = a2 xl(a- x) is rotated about the x-axis. Find the volume so generated (The

line intersects the curve in (a/2, a).) Sol. : The line 2x + y = 2a intersects the x-axis in A (a, 0) and the curve in

B (a/2, a). V = volume of solid of revolution + volume of cone

Ia/2 l a :. V = 1t o y2 dx + 3. na 2. 2

Now I - fa/2 a 2x dx -1t 0 a=x

= 1ta2 J;/2 (- I + a x ) dx

= na2 [- x-a log (a-x) ]�12

=na2 [ � - a log � + a log a ] = na2 [ � + a log 2]

:. V = na3 [log 2 - ! + i] = na3 [log 2 - -!]

y (0, 2a)

Fig. (10.74)

X


Ex. 4 : Show that the volume of the spindle formed by

revolving a parabolic arc about the line joining the vertex

to one extremity of the latus rectum is 2na2! 15'-l5 if the

latus rectum of the parabola is 4a. Sol. : Let (0, 0) be the vertex of the parabola i = 4ax and let L(a, 2a) be an extremity of the latus-rectum.

Let P (x, y) be any point on the arc OL. Let PM be

perpendicular to OL.

L (a, 2a)

Then the required volume is

V = n J (PM)2 · d (OM) Fig. (10.75)

Now the equation of OL is

y-0 x-0 . 0 -2a = 0 -a I. e. Y -2x = 0

2x-y 2x-y PM = + 1 = ..[5

i. e. 2x-y=O

And OM - -PM2 = - ..[5

_ x + 2y = 5 ..[5

= x+4v'i ..[5 (": l = 4ax)

:. d (OM) 1 ( 4..JO. ) 1 (v'i + 2va- ) = ..[5 1 + 2.;x dx = ..[5 v'X dx

v _ __ 1 (v'X+2va ) - n 0 -./5 ,;s v'X dx

s; (x2-3ax + 2a312 >IX) dx

_ 47t [ x2 _ 3ax2 + ± a312 . x312]a -5.. 3 2 3 0 _ 47t (112 3a3 4a3 ) = 21ta3

-5.. 3 - 2 + 3 15./5

Ex. 5 : Find the volume generated by revolving

the area between the parabola y = x2 - 2x - 8 and

the x-axis about line x =-5. Sol. : The parabola intersects the x-axis where

x2- 2x- 8 = 0 i.e. (x- 4) (x- 2) = 0 i.e. where

x = 4 and x = -2. The elementary area dx dy is at a distance

Fig. (10.76)

X

Engineering Mathematics - II (10-44) Appll. of Mult. Inte.

5 + x from the axis of rotation and genetares the ring of volume 2n (5 + x) dx dy. Hence the required voliume is given by

4 0 V = 2n f_2 t 2 -2x _ 8 (5 + x) dx dy

J4 0 = 2n _2 (5 + x) [ Y lx 2 2x 8 dx

4 =- 2n J (5 +x)(�-2-r 8) dx 2

J4 3 2 =- 2n _2 (x + 3 x - 18x-40) dx

[X 4 ]4 = - 2n 4 + x 3 - 9x 2 - 40x _2 = 4321t.

Ex. 6 : The curve r = a (1 + cos e) revolves about the initial line. Find the volume generated.

(S.U. 1998, 2004)

Sol. : The volume generated is given by

v = 231t J ? sin e de

y

9=n

(See (6) page 10·40.) Fig. (10.77) It should be noted that the volume is generated when half the area revolves

about the initial line. Hence, 2 J1t v = 3 0 ? sin e d e

v = 231t r; a3 (1 +cos e)3 . sine de

Put 1 +cos e = t :. -sin e de = dt 21t Jo 3 27ta 3 J2 :. V =- 3 a3

2 t dt =

o t3 dt

2 _ 21ta 3 [�] = 81ta 3 - 3 4 0 3

Ex. 7 : Find the area of the curved surface of the solid y formed by the revolution about its axis of the smaller part of the parabola y2 = 4ax cut off by the line x = 3 a.

Sol. : We have s = J 2n y ds and ds = + r dx Since, l = 4ax

. !!J!. . !!i!__2a • • 2Y dx = 4a .. dx -y-

Fig. (10.78)

Engineering Mathematics - II (10-45) Appll. of Mtilt. lnte.

4a 2 :. ds = + y 2 · dr: = + 4ax · dx

I3a :. s = 21t 0 2 Wi · .. • dx

= 41t Wi J3a (x + a)112 dx = 41t Wi-13 [ (x + a)312 ]30 0 0

- 87tVa [ (4 )3/2 ( )312 l - 56 2 - 3 a -a - 3 1t a

Ex. 8 : Find the volume and the surface generated by revolving the cycloid x = a (8 + sin 8}, y = a (1 - cos 8) about the tangent at the vertex. Sol. : The volume, because of symmetry, is given Y by fl = rr

v = 2 J1ti dx

:. V=21tJ: a2 (1-cos8)2· a(l+cos8)d8

Jn: 8 8 = 2na3 0 4sin4 2 · 2cos2 2 d8

Jn: 8 8 [ 8 ] :. V =161ta3 0 sin4 2 cos2 2 de. Put 2 = t

n:/2 - 321ta3 J sin4 t cos2 t dt - 0

Fig. (10.79)

3 l i572 3 (3/2)(1/2)1172 ·(1/2) .ifi2 = 321ta . 2 14 = l61ta 3. 2. 1

= l61ta3 • i· i (11/2 ) 2= 1t2a3 • [ ·: 1112 =-/i]

Now, the surface is given by s = 21t Jy cis

= { a 2 ( 1 + cos 8)2 + a 2 sin2 8 } = +cos8) =2a cos �

:. ds =2a cos � ·de

:. s = 21t ( a (1 - cos 8) · 2a cos Q2 de -11:

e=rr

Engineering Mathematics - II (10-46) Appli. of Mult. Inte.

= 8na2 ( 2 sin2 � cos � de

[

Put sin� = t ]

I [ 3 ]I = 16na2 · 2 f0 t2 dt = 32na2 T 0

32 ') = 31ta-Ex. 9 : Find the volume and surface generated by rotating the above cycloid

about the y-axis.

Sol. : Thevolume generated is

V = 1t J :l- dy = 1t J a2 (0 + sin e)2 a sin e dO 11:

= na3 f0 (02 sin e + 2e sin2 e + sin30) dO

Now r: e2 sin 0 dO= [ e2 (-cos e)-J(-cos e) 20 de]

= [-e2 cos e + { 2e sin e + 2cos e } ]� = (n2 -2) -2 = n2 - 4

r: 20 sin2 e dO = r: e (1 -cos 2e) de= r: e dO-I: e cos 2e de

JTI: - 3 fn:/2. 3 2. _.4_ 0 sm e dO = 2 0 sm e de = 2 · 3

· 1 - 3

. V - 3[1t2-4+1t2+.4.]=1ta3[31t2 _.£] · · -1ta 2 3 2 3 Now, surface is given by

s = J 2nxds n: e

= 21t J0 a (0 + sin e) · 2a cos 2 dO

= 4na2 f: (e cos � +sine cos �) de

But f: ecos �dO= [e · (2 sin �)- {- 4 cos �)]:= 2n _ 4 n: e

And J0 sin 0 cos 2 dO fn: e e [ e ] = 2 0 sin 2 cos2 2 de. Put cos 2 = t

fo 2 .4. = 4 I t ( -dt) = 3 .

:. s = 4na2 [21t- 4 + f ]= 41ta2 { 21t-�)

r'


Ex. 10 : A circular arc of radius a making an angle 2a at the centre rotates about the chord. Find the surface area of the volume so generated.

Sol. : Let the equation of the circle be x = a cos e, y =a sin e.

The arc AB revolves about the chord AB. Let p be the perpendicular distance of any

Appli. of Mutt. Inte.

A

point P on the arc AB from the chord AB. B Now p =PM= x- a cos a= a (cos e- cos a) Fig. (10.80) ds = dx)2 + (dy

e)2 = + de- a de de d

-

But s = J 2npds = 2 J;2npds

= 4n f: a (cos e- cos a) ·a de

= 4na2 [ sin e - e cos a ]� = 4na2 [ sin a.- a cos a ]

Exercise- V

1. The segment cut off from the curve al = x3 by the line x = k revolves about the x-axis. Find the volume so generated. Prove that this volume is onefourth that of the cylinder of height k and the same base.

[ 7tk4 ] (See fig 10.38 page (10.25)) Ans.: 4a

2. Find the volume generated by rotating the parabola i = 4£U from x = 0 to x = h about the x-axis. (See fig 10.39 page (10.25)) [ Ans. 2nah2]

3. Find the volume of the solid of revolution of the loop i = 2

about the x-axis. (See fig 10.11 page (10.6)) (Ans.: 27ta3 [log 2 -logj-]]

4. F.ind the volume of the solid formed by the revolution of the curve (a - x) y 2 = a2x about its asymptote x =a. (S.U. 1997, 2001)

[ 7t2a3 ] (See fig 7.86 page (7.19)) Ans.

3 5. Find the volume of the solid formed by revolving the curve l = ax

- x

about its asymptote x = a. (See fig 7.45 page (7 .1 0)) [ Ans. : 3 ]

6. The ellipse � + � = 1 is divided into two parts by the line x = �


and the smaller part is rotated about this line. Find the volume so generated.

(Draw ellipse.) [Ans.: -j)] 7. Find the volume of the solid generated by revolving the curve a3

y = + a about the x-axis.

(See Fig 7.67 Page (7.17), 2 a =a) [Ans.: 8. Find the volume of the solid generated by revolving the hypocycloid

?3 + ..,W = Q1 about the x-axis. (See fig 8.2 page (8.2)) ·

Find also the surface of the solid.(S.U.1997)[Ans.: 1ta3, xa2] 9. The cycloid x =a (6- sin 6), y =a (1 -cos 6) rotates about its base.

Find the volume and surface so generated.

(See fig 7.116 page (7.29))

10. For the cycloid x =a (6 + sin 6), y = a (1 -cos 6), find the volume and the surface of the solid generated by the revolution of one arc about the

y-axis. (Seefig7.115 page (7.29)) [Ans.: 1ta3 (� x2 -J); 4xa2 (2 x -J)] 11. Find the volume and the surface of the solid generated when one arc

of the above cycloid is rotated about the base line.

[Ans.: 51t2a3; 634 1ta2]

12. Find the volume and the surface area of the solid gnerated by revolving

the loop of the curve X = r' y = t- t r about x-axis. (S.U. 2001, 2003)

(See fig 7.60 page (7.15))

(Hint :The loop lies between t = 0 and t = ..f3 ) [Ans. : t 1t; 3 1t] 13. The part of the parabola J2 = 4ax cut off by the latus-rectrum revolves

about the tangent at the vertex. Find the surface of revolution. (See fig 10. 39 page (10.25)) IAns.: 1ta2 [3V2 +log (Vl-1)])

14. The curve r =a (1-cos 6) revolves about line. Find the volume and the surface of the solid so generated.

(See fig 7.89 page (7 .20))

15. Find the area of the surface of revolution formed by revolving the circle r = 2a cos 6 about the initial line.

(See fig 9.9pa�e (9.20)) [ Ans.: 4na2]

Engineering Mathematics - II (10-49) Appll. of Mutt. Inte.

16. A quadrant of a circle of radius a revolves round its chord. Find he

surface area of the solid so formed. (S.U. 2002) [Ans. : 2na 2 ..ti (I -*)] (See fig 10.80 page (10.46)) (Hint: In solved Ex. 10 put a.= 1t/4 )

17. Prove that the surface generated by the revolution of the tractrix

x = a cost+ ..!. a log tan2 �, y = a sin t about it asymptote i.e. about the x-axis 2 2

is equal to the surface of the sphere of the radius a.

(See fig 7.118 page (7.31)) ( • J1r./2 ds ) Hmt S = 2 0 2ny dt

· dt

18. The region enclosed by the curves y = sin x, y = cos x and the x-axis

is rotated about the x-axis. Find the volume of the solid so generated.

(Fig. left to you.) [ Ans. : * (n- 2)]

19. Find the volume of the solid generated by revolving the loop of the

curve i = � (2- x) about the x-axis.

(See fig 7.63 page (7.15)) [Ans.: t n] 20. Find the volume of the solid of revolution of the curve r = 5 + 4 cos e

revolving about the initial line. (S.U. 2000, 2005)

(Fig. similar to fig 7.93 page (7.22)) [Ans.:

8. Moment of Inertia

Definition : Let a mass m be situated at a point P which is at a distance r

from a line then the product mr2 is called the moment of inertia of the mass m about the line or the axis.

(a) Moment of Inertia of a plane lamina

Consider a plane lamina. Let p be the density at any point P. Consider a small area dx dy at a point P (x, y). Then the mass of the elementary area is p dx dy.

If p is distance of this elementany mass from the axis, then the moment of inertia about this axis is given by

pp2 X

Fig. (10.81)

The moment of inertia of the elementary area about the x-axis is p y2 dx dy.

:. Moment of inertia about the x-axis is

M.I.=fJ py2dxdy

R


where R is region of integration.

Similarly moment of inertia about they-axis is

I. =lj px2 dx If we consider a small area r dr de at a point P (r, e). If pis the distance of

the point P from the axis then.

M. I. = Jf p p 2 r dr de R

If the line is the x-axis then since p = y = r sin e, the moment of in tertia is

given by

M. I. = fJ p (r 2 sin 2 e) r dr de R

If the line is they-axis then since p = x = r cos e, the moment of in tertia is

given by

M. I.= fJ p (r 2 cos 2 e) r dr de R

(b) Moment of Inertia of solid

Consider a solid of volume V and let p be the density at a point P(x, y, z). We consider a small volume dx dy dz. Its mass is p dx dy dz. Since its distance

from the x-axis Y 2 + z 2 , the moment of inetia about the x-axis is

M. I. =ffJ p (y2 +z 2)dx dy dz v

Similarly moment of inertia about they-axis is given by

M. I. =Jff p(x2 +z 2)dxdydz v

And moment of inertia about the z-axis is given by

M.l. =Jff p(x2 + y2) dx dy dz v

(c) Theorem of perpendicular axes

If lx, ly are the moments of inertia of a plane lamina about two

perpenc:\icular axes ox and oy respectively then the moment of inertia Iz about

the axis perpendicular to the plane of the lamina through 0 i.e. the axis oz is

given by

(d) Theorem of parallel axes

If 18 is the moment of inertia of a mass M about an axis through its centre


of gravity (C.G.) then its moment of inertia JP about a line parallel to the above axis at a distance dis given by

==lg We accept these theorems without proof.

(e) Radius of Gyration lf the moment of inertia M.I. of a body of mass M isM k2 then k is called

the radius of gyration. In other words the radius of gyration is given by

or Radius of gyration == .

Ex. 1 : Find the moment of inertia about the x-axis of the area enclosed by the lines x = 0, y = 0, (xla) + (ylb) = 1. Sol. : Consider a small area dx dy at a distance y from the x-axis, then

M. I. = JJ p y 2 dx dy

On the strip x varies from 0 to a( 1- f) and

the strip varies from y = 0 toy = 6.

JbJa(b-y)lb 2 . . M.l. = p 0 0 y dx dy

== pJ� Y2 [x]�(b y)lb dy

y

(0, b)

Fig. (10.82)

== pfb y 2 a(b- y) dy == p� Jb (by 2- y 3) dy 0 b b 0

== p�[bL - L]b = p�- � = pab 3 b 3 4 b 12 12 0

Ex. 2 : Prove that the moment of inertia of the area included between the parabolas y2 == 4ax and

x2 = 4ay about hex-axis is 144

M a2 where M is the 35

mass of the area included between the curves.

Sol. : Consider a small area dx dy at a distance y from the x-axis, then M.I. is

X

Fig. (10.83)

Engineering Mathematics - II (10-52) Appll. of Mult. inte.

I=Jfpy2dKdy On the strip x varies from x = y 2f a to x = 2�y and the strip varies from

y=O toy =4a.

f4a JJi; 2 :. M.I.= 2 p y dKdy 0 y 14a

= py2 [x]Ji; dy y214a = J:a py2 [ ]dy = pf:a[ 2� · y512- dy

= p[2� .�.y 112 _J_L]4a =p[2sa4(�) --1 . 45a5]

7 4a 5 0 7 4a 5

= p(4a)4 (�-.!.) = �(4a)4 ·p. 7 5 35 Now, the mass M of the area included between the curves is given by

f4a JJi; J4a J4;zy M = 0 y2t4apdKdy=p 0 [x]y2/4a dy

=Pta _X.]dy= p[.J4; ·�y3/2 __ 1 .L]4a 0 4a 3 4a 3 0

= p[�·(4a)2 _ (4a)2 ]= p (4a)2 :. p= 3M 3 3 3 16a2

3 4 3 4 3M But M.I.= 35

·(4a) p= 35

(4a) . 16a2

144 2 .:. M.l.= 35

a M.

Ex. 3 : Find the polar moment of inertia of the area bounded by the parabola y2 = 2x and the line y = x. Sol. : ''Polar'' moment of inertia means the moment of inertia about the axis through the origin and perpendicular to the plane of the area. Since polar M.l. is required we shall transf9rm the problem to polar coordinates.

Consider a small area r d6 dr at a distance r from the axis. Then

Fig. (10.84)

-··t�••• ..... •••t� ..... -...... ,. •• _ •. \ •v 'tVJ

I=Jfpp2 rdrd8=Jfpr3drd8

-1'1'••• -• •••w••• ••••w•

Now the line y = X in polar coordinates becomes r sin e = r cos e :. sin 8 =cos 8 i.e. 8 = n/4. Also the distance p = r.

The parabola y2 = 2x becomes r2 sin2 8 = 2r cos 8 i.e. r = 2 cos 8/ sin2 e. Hence, r varies from 0 to 2 cos 8 I sin2 e and 9 varies from n/4 to n/2.

Jn/2 J2cosatsin 2 a 3 p Jn/2( 2 cosO) 4 :. I = p r dr dO=- dO

n/4 0 4 n/4 sin e

J7tl2 cos 4 e Jn/2 4 4 =4p cot e cosec Ode n/4 sin e n/4

= 4pfn12 cot 4 e (1 +cot

2 8) cosec 2e dO n/4 Put cot 8 = t :. - cosec2 8 dO= dt

:. I= 4pf1°

(t4 +t6

)(-l)dt = 4pf� (t4 +t6

) dt

= 4p[c

+�] I

= 4. g

p = 48

p 5 7 0 35 35

Now, the mass M of the area is

2 [ 2 ]2cosa/sin 2 a Jn/2 I

2cosa/sin a e

J1tl2 r M= prdrd =p - dO n/4 0 n/4 2 0

Jn/2 4 cos 2 e Jn/2 2 2 = p 4 dO = 2p cot e cosec e de n/4 2 sin e n/4

=2p[- cot3e ]n'2

=�

p :.p= 3M

3 . 3 2 n/4

48 48 3M 72 But M.I.=-p=-·-=-M.

35 35 2 35

Ex. 4 :An area is bounded by the curve y = c cosh (xlc), the axes and the ordinate x = c. Find the radius of gyration about they-axis.

Sol.: The curve is a catenary as shown in the figure. OAPB is the given area. Let p be the density. If we consider a small area 4x dy at P (x, y) then its M.I. about they-axis is p x2 dx dy.

On the stripy varies from y = 0 toy = c cos h (xlc) and the strip varies from x = 0 to x = c.


. -Jcfccosh(xlc) 2 .. M.I.- 0 0 p X dx dy

= J� p X 2 [y] �cos h(x/c) dx

= p f�· x 2ccos h (xI c) dx

Integrating by parts, we get,

Fig. (10.85)

M.I.= pc[ x2 ( csin h �) � ( 2 x{ c2 cosh �) + ( 2{ c 3

sinh �) J:

= pc [c3 sinh 1 - 2c3 cosh I + 2c3 sinh 1]

. e-e e+e [ -1 -1 ] = pc

4[3sm hl-2cos hl] = pc

4

c4( 5) pc

4 2

=p- e-- =- (e -5) 2 e 2e

fc

Jccos h(xlc)

Now area OAPB = 0 0 dy dx

= J�[y]�cosh(xlc)dx= J�ccosh(xlc)dx

= c 2 [sin h(x I c)] � = c 2 sin hI = c

2

2 ( e-;)

Mass M of the area is given by

c2( I) pc

2 2 M =p- e-- =-(e -1)

2 e 2e

Hence the radius of gyration k is given by

4 k 2 _ M.I. _ c ( 2 5) 2e

--- -p.- e - . Mass 2e

Ex. 5 : Find the moment of inertia of a circular disc of radius a about the axis through the centre and perpendicular to the plane of the disc.

Sol. : Consider a small area r dr de situated at P (r, e) of the disc. Let us consider the axes as shown in the figure. Hence, the equation of the circle is r =a. If p is the density then M.l. of this elementary mass is p · ,:2 · r dr de

Now, r varies from 0 to a and e varies from 0 to 21t. X

:. The required M.l. is given by Fig. (10.86)

engineering Mathematics - II (10-55) Appli. of Mult. lnte.

f2n fa 3 J2n a 4 M.I. = p r dr de= p de 0 0 0 4

a 4 [ ] 2n a 4 na 4 =p4 eo

u M is the mass of the disc then

2 M M = pn a :.p= --2 na M na4 Ma2

:. M.l.=-2 ·- =--na 2 2

Ex. 6 : Show that the M.I. of a rectangle of sides a, b about its diagonal is

M ( a2b2 ) - 2 2 where M is the mass of the rectangle. Y 6 a +b Sol. : Consider the rectangle and the axes as shown in

the figure. Consider a small area dx dy at a point

P (x, y). Let p be the length of the perpendicular from

P to the diagonal OB. Hence, the M.l. of elem�ntary a x

is Fig. (10.87)

M.l.= fJ p2pdxdy To findp consider the equation of the line OB. It passes through the origin

and its slope is bla. Hence, its equation is y = ( �) x i.e. bx- ay = 0.

bx-ay The length of the perpendicular P = 2 Hence, the M.I. is given by

M.l.= a b (bx-ay)2 ·pdxdy fx=ofy=O (a2 +b2) _

p a (bx-ay)3 dx

[

l

b

(a2 +b2)jo -3a o

r[(bx-ab)3 -b 3x3J dx 3(a2 +b2)a 0

3 ? r[(x-a)3-x3]dx

3(a +b-)a 0


pb3 x4]a ==-

3a(a2+b2) 4 -4 0

pb3 (-a4-a4) == - ==

3a(a 2 + b 2) 4 6(a + b M

But the mass M of the rectangle M == p ab :. P == ab

M a3b3 Ma2b2 :. M.I.==

ab· 6(a2+b2) ==

6(a2+b 2).

Ex. 7 : Find the M.l. of the semi-circle about the line

joining one end of the bounding diameter to the

midpoint of the arc.

Sol. : Let one end of the semi-circle be the origin, the

bounding diameter OA be the x-axis, the perpendicular

to it through 0 be the y-axis. Let a be the radius.

Now the equation of the circle is (x - a)2 + y2 == a2 i.e. x2 + y2

== 2ax i.e. r == 2a cos e. If B is the mid point of the arc then the equation of OBis y ==X.

If P (x, y) is any point of the semi-circle then since the equation of the line

OBis y == x i.e. x- y == 0 the length of the perpendicular p from it to the line OB

is p ==

JJ 2 JJ (x-y) 2 :. M.I. == p p dx dy == p

2 dx dy

== � fJ (x 2 + y 2 -2xy) dx dy

Changing to polar coordinates, since r varies from 0 to 2a cos e and e

varies from 0 to 1t/2, we get,

p J7t/2 J2acos6 2 2 M.I. ==- ( r -2r sin ecose) r dr de 2 0 0

� � J:" (1-2M OcosO) [ rm• dO

== � J1t12 (1-2sinecose) ·16a 4

cos 4

e de 8 0

== 2a 4pf:12 (cos 4 e-2cos 5 esin e) de

r.ngmeermg Matnematics - II (10-57) Applications of Differential .....

= 2a 4p[l:! · �- 2 ·.!.]::a 4p[31t-3.]

4·2 2 6 8 3 Since the area of the semi circle 1ta2f2 its mass M is given by

1ta 2 2M M==p- :.p::-2 1ta2

:. M.l.= a4. 2M [31t _3_]== Ma2 (� -.i_) 1ta 2 8 3 4 31t Its radius of gyration k is given by

k2:: M.l. ==a2(�-.i.). Mass 4 31t

Ex. 8 : Find the M.l. of the area bounded by one arc of the cycloid X= a (e-sin e), y =a (1 -cose) from e = 0 toe= 21t and the x-axis about the x-axis. Sol. : Consider a small area dx dy at a point P (x, y ) . If p is the density then M.I. of this area about the x-axis is p y2 dx dy.

X = a (e -sin e), dx = a ( 1 -COS e) y = a ce -cos e), dy = a sin e de

M.l.= ff py 2dxdy= f f�py 2 dx dy

3 0

X

== pfLdx=£ fy3 dx de 3 3 de

= £J2n a 3( 1-cos e) 3 ·a (1-cos e)-de 3 0

Fig. (10.89)

a4pJ21t 4 a4pJ21t 8 e =- (1-cose) -de=- 16sin -·de 3 0 3 0 2

16a 4 p f 1t • 8 [ . e J 0 sm $·2·d$ Puttmg 2"=<1>

J2a Ja Ifj(2a-x)=f( x)then 0 f(x)dx=2 0J(x)dx ... (A)

6 4 4 :. M.l.= �Jn/2 sin 8$·d(j> = 64a p. 7 ·5 · 3·1 .�:: 35 1tpa4 3 0 3 8·6·4·2 2 1 2

If M i s the mass of the area, M == J J p dx dy =.J J� p dx dy == f p y dx = pf ·de

f21t ==p 0 a(1-cose)·a(l-cose)·de

Engineering Mathematics - II (10-58) Applications of Differential •....

=a 2pJ 2n (1-cose}2 ·de= a 4 pi 2n 4 cos 4 �·de 0 0 2

= 4a 2pi; cos 4�·2 ·d� [Putting � = � J

2 I n/2 4 2 3 · l 7t 2 M = 16 a p cos �·d�= 16a p ·-= 3na p [By (A)] 0 4·2 2

:. M.l.= 35 ·7t . _l!_ a4 =35M a 2. 12 3na2 36 Ex. 9 : Find the moment of inertia of the area included between the arc of the cycloid x = a (e + sin e), y = a (1 - cos e), the straight line x = a 1t and the tangent at the vertex about the x-axis.

Sol.: Consider a small area-dx dy at a point P (x, y). If pis the density then M.l. of this area about the x-axis is p · y2 dx dy.

·: x =a (e +sin e), dx =a (1 +cos e)· d8 . . y =a (I-cos e), dy =a sin e· de

M.l. = H p y 2 dx dy = I I� p y 2 dx dy

I y3 P I 3dx =p 3 dx=3 Y de

·de

From 0 to B, e varies form 0 to 1t

:. M.I.=� I: a3(1-cose)3a(l+cose)·de

Fig. (10.90)

= a34 PI;[ 2sin2(e/2)r { 2cos2(e/2)} de

Put e 12 = �

16a4 Jn . 6 2 = -- p sm (e/2)·COS (e/2)·de 3 0

l6a 4 . 6 2 :. sm �cos �·2·d�

32a4 (5·3·1)( 1) 1t 5 4 =--·P· ·-=-7t·pa 3 8·6·4·2 2 24

If M is the mass of the area,

M = H p dx dy = J J� p dx dy

= J pydx= J ·de

8=7t

Engineering Mathematics - II (10-59) AppUcations of Differential •....

:. M = J;pa(l-cos8)·a(l+cos8)·d8

= pa 2 J; 2sin 2(8/2) · 2cos 2(8/2) ·d8

=a 2pJ; 4sin 2(8/2)·cos 2(8/2)d8

Put 8/2 = cj)

. . M =a 2p J;12 4 sin 2 cj) ·cos 2 q, · 2dcj)

= Sa 2p (1) · (1) . � = a 2pn 4·2 2 2

5 4 2M 5 2 :. M.I.=-·na --=-Ma . 24 a2n 12

2M :.p= a21t

Ex. 10 : Find the moment of inertia of one loop of lemniscate r 2 = a 2 cos 28 about the initial line.

Sol. : Consider a small element r dr d8 at a distance p = y = r sin 8 from the

initial line.

Its M.I. about the initial line is Y

Jn/4 2 d8 d :. I= p p r r -n/4 0

fn/4 2 • 2 e 8 d = 2p r sm ·rd r 0 0

n/4 r . 2 [ 4

= 2pf0 4 0 sm 8·d8

Fig. (10.91)

a4 Jn/4 2 2 a4 Jn/4 2 ( 1- cos 28 ) = p- cos 28·sin 8·d8 = p- cos 28 d8

2 0 2 0 2

a4 Jn/2 2 dt =p- cos t(l-cost)·-

4 0 2

a4 fn/2 2 3 =p- [cos t-cos t)dt 8 0

a4 [ 11t 2 ] a4 =ps 22-3 =p96(31t-S)

Now, mas M of the loop is

[Put 28 = t]

M = JJ P r dr d8 = pf1t14 r dr d8 -n/4 0

Engineering Mathematics - II (10-60) Applications of Differential .....

e

J1t/4[r2

de = 2p r dr d = 2p -0 0 0 2

0

Jn/4 2 a 2 n/4 a 2 2M =p a cos2e·de=p- [sin2e] 0 =-p :. p=-2 o 2 2 a

a 4

2M a 4 3n -8 2 :. M.l.=p% (3n-8)= a 2

·%

(3n-8)=� Ma

Ex. 11 : Find the moment of inertia of a circular plate about a tangent .

r=2a cosa

Sol. : Consider a circle with centre at (a, 0) and radius a, so that y-axis is the tangent to the circle at the origin.

Now the equation of the circle is (x-a )2 + y2 = a2

i.e. x2 + y2 = 2ax i.e. r2 = 2ar cos e i.e. r = 2a cos e.

Consider now a small element r dr de at a distance p =X= r cos e from they-axis. Its M.l. about they-axis is

Fig. (10.92)

JJ 2 Jn/2 J2acos

a 2 2 I= P · p rdr de= p r cos e r dr de -1tl2 0

= pf1t12 [c]

2acosa cos 2 e de= 4pf

1t12

a 4

cos 6 e de -1tl2 4 -1t/2

0

4J1t/2 6 4 5 · 3· 1 1t 5 4

=p8a cot e·de=p8a · --·-= p-na 0 6·4·2 2 4

If M is the mass of the plate then

M = Jn/2

J2acosa r dr de=

Jn/2 [c]2

acosa

de -n/2 0 P P

-n/2 2 0

= p. 2 J;12 2. a 2 cos 2 e. de= p. 4a 2 J:1

2 cos 2 e. de

2 1 1t 2 =p·4a · 2'·2 =pn a .

X

[Or since the area of the circle of radius a is 1t a2 and the density is p its mass M = p 1t a2.]

5 4 M 5 4 5 2 :. M .l.=p· -1ta =--·-na =-M a 4 na 2 4 4

Ex. 12 : Find the moment of inertia about the line e = 1t /2 of the area enclosed by r = a (1 + cos e).

Engineering Mathematics • II (10-61) Applications of Differential .....

Sol. : Consider a s mall el ement r dr de at a distance p = X = r cos e from the y-ax is i.e. the line e = 11: I 2.

:. I = fJ x 2

r dr de

fn Ja(l+cos6) 2 2 =2 0 0 · r cos er drde

fn Ja(l+cos6) 3 2 = 2

0 0 r cos 8 r dr de

[ 4 ]a(1+cos6) =2J: r

4 0 cos

28·d8

a4 n

= -f (l+ cos e)4

cos 2

e ·de 2 0

Fig. (10.93)

a4 n

= -J 16cos8

(e/2)(2cos2

(e/2)-1 )2 · de 2 0

=Sa 4 J;[ 4 cos 12

(e/2)-4cos 10(8/2)+ cos

8(8/2) J de

= 8a4 J: 12

(4cos 12 t-4cos 10 t +cos 8 t) 2dt [Put 8/2 = t)

4 [ 11 ·9·7·5·3·1 1t =16a 4· ·-

12·10·8·6·4·2 2

_ 4 . 9 · 7 · 5 · 3 · 1 . !: + 7 · 5 · 3 ·1 . !:] = 4n a

4 .

10·8·6·4·2 2 8·6·4·2 2 32

Ex. 13 :Find the moment of inertia about the (i) initial line (ii) polar axis of the area of r =a (1 +cos e).

Sol. : Consider an element r dr de (i) M.I. about the initial line is given by

I = fJ y 2

r dr de

2JnJ

a(l+cos6) 2 . 2 8 d de = r sm r r

0 0

Fig. (10.94)

JnJa(l+cos6) 3 . 2 Jn a4 4

• 2 =2 r sm edrde=2 -(1+cos8) sm e de 0 0 0 4

a4 n

= -J (2cos 2 e/2) 4

(2 sin8/2cose/2)2de

4 0

= 32a4 J: cos

10(e/2)sin 2

(e/2)de [Put 8/2 = t]


= 64a 4 J :12 cos 10 t sin 2 t dt

=64a4 . (9·7·5·3·1)(1) ·�=3.1ta4. 12·10·8·6·4·2 2 32

(ii) Moment of Inertia about the polar axis is given by

JJ 2 J7t Ja(l+cos6) 3

I = r r dr de = 2 0 0 r dr de

[ 4 ]a(l+cos6)

4 =2J: r

4 0 de=a

2 J;a4(1 + cose)4de

= 8a4 J: cos 8(8/2)·de [Pute/2 = t]

= 16a 4 J n/2 cos 8 t dt = 16a 4 . 7 . 5 . 3 . 1 . � = 35 1ta 4.

0 8·6·4·2 2 16 Ex. 14 : Prove that the M.l. of the area included between the smaller arcs of r = 2a cos e and r = 2a sin e about the axis perpendicular to the plane of the

curves through the pole is a 4 ( - 2) . Y

Sol. : The curve r = 2a cos e i.e. ,.2 = 2ar cos e i.e. x2 + y2 = 2ax i.e. (x- a)2 + y2 = a2 is a circle with the centre at (a, 0) and radius a. Similarly the curve r = 2a sin e i.e. ,.2 = 2ar sin e i.e. x2 + y2

= 2ay i.e. x2 + (y- a)2 = a2 is a circle with centre at (0, a) and radius a. The area included between the smaller arcs of these circles r=2a cosa is divided equally by the line e = 1t/4. Fig. (10.95)

Consider a small area r d8 dr at a point P in the common area. Its M.L about the axis throu gh 0 perpendi cular to the plane of the circles is ,.2 ·rde dr= r3 drd8. The required M.l. is obtained by integrating r3 dr· d8 over the common area. But since e = 1t/4 divides the area equally, we have,

[ 4 ] 2asin9 M.I.=2J;'4 J:asine

r3drde=2J;'4 r

4 o de

= 8a4 J;'4 sin48·d8= 8a4 ·d6

Putting 2e = •• 2dQ = d •• M.I.= a4 J;12[1-2 cos •+ cos 2 •]·d•

-2] .


Ex. 15 : Find the moment of inertia of the quadrant of the ellipse 2x2 + y2 = 1 in which x andy are positive about an axis perpendicular to the plane where mass per unit area of the ellipse varies as the abscissa of the point at which it is situated. Sol. : The equation of the ellipse can be written as

2 2 .: + L = 1 . The ellipse is shown in the figure. 1/2 1

Consider a small area dx dy at P (x, y). The density p = "Ax. Hence M.l. is given by

Fig. (10.96)

I = fJ p p 2 dx dy = fJ A x p 2 dx dy where p 2 = x 2 + y 2

ff 2 2 2 2 :. M.I. =A (x + y ) x dx dy =A 0 0 (x + y ) xdx dy

[ 3 ]..[;.2x2 3 y x y+x- dx 0 3 0

+� (1-2x 2)3/2 ]d,-c Putting X = sin e. dx = cos e de

fn/2 ( 1 3 1 I 3 ) :. M.I.=A 0 2-fisin ecose+)· -fisine·cos e · -ficosede

[frt/2 1 3 2 frt/2 1 4 . J = A 0 4 sin e cos e de+ 0 (j ·cos e sm e de

= _2 +.!. · (-cos 5 e) = "- [__!_ + __!_] = A

4 5 . 3 . 1 6 5 0 30 30 15

Ex. 16 : Find the moment of inertia about the x-axis of the portion of the parahola y2 = 4ax bounded by the x-axis and latus rectum if density at each point varies as the cube of the abscissa. Sol. : Consider a small element dx dy at P (x, y). Since density p varies as the cube of the abscissia p = J.3.

Sf 2 JaJ2.,J; 3 2 :. M.I.= py dxdv= Ax y dxdy . 0 ()

y

X

Fig. (10.97)


a3/2 Ja x912dx =SA. a3f2[X 1112 ]a= 16 A. a 7

3 0 3 11/2 0 33

Now mass M = Jf pdxdy

= J� s:-ra; A X 3dx dy = J� A X 3 [y] �..fai dx

9M :.A.=--4 as

:. M.I.= 16 .2..!!:!..a 7 =12M a2

33 4 as 11

Ex. 17 : Prove that the moment of inertia about an axis through the centre, perpendicular to the plane of a circular ring whose outer and inner radii are

bandais ..!.M(a2+b2) whereMis the mass of 2

the ring.

Sol. : Consider a small area r dr d8 at a point P (r, 8). Then the moment of inertia about the axis perpendicular to the plane through 0 is Fig. (10.98)

M.I.= Jf p·r2 ·rdrda

2n b 2n [r4 ] b =So fap·r3drd8= So P 4

o da

= £S2n (b4 -a4)·d8 = £(b4 -a4)[8]2n 4 0 4 0

= p(b 4 -a 4). �. 2

But the mass of the ring is given by

M =1t(b2 -a2)·P :. p= M

1t(b2 -a2)

Engineering Mathematics - II (10-65) Applications of Differential .••..

M 4 41t M z 2 :. M. I .= 2 " ·1t(b a )·-= (b +a ). 1t(b a�) 2 2

Ex. 18 :The density of a circular lamina varies Y as the square of the distance from a point 0 on the e=1[/2 circumference. Find the moment of inertia of the area about an axis through 0 perpendicular to the plane of the circle.

Sol. :Let the point 0 be the origin. Consider the circle passing through the origin of radius a and centre on the x-axis. Its equation is (x- a)2 + y2 = a2 i.e. x2 + y2 = 2ax cos e i.e. r2 = 2ar cos e i.e. p = 2a cos e. Fig. (10.99)

Consider a small element r dr de at P (r, 8). Then density p =). r2. Hence, the M.I. is given by

ff 2 2 J7t/2J2acose 5

M.l. = r ·I. r · r de dr = 2 0 0 /. r dr de n/2 r 64 rc/2 [ 6 ]2acose

=2A fo 6

o d0=3t.Jo a6cos6e-de

= 64/. a6 .�.!:=.!.QI.a61t 3 6·4·2 2 3 Sf f

n/2 J

2acose 2 Now mass M = p dx dy = 2 0 0 ). r r dr de n/2 r n/2 [ 4 ]2acose =2Af0 4

0 de=8/.a4J0 cos4e - de

M=SI.a4.l:..!:=3/.a4n :./.= 2M 4·2 2 2 3a4n

10 2M 6 20 2 :. 1t=-M a 3 3a 1t 9 Ex. 19 :Find the rdius of gyration for the area of the cardioide r =a ( 1 +cos e) about the axis perpendicular to its plane through the pole when the density at any point varies as the distance of point from the pole.

Sol. : Consider a small elementary area r de dr at a point P (r, e). Since the density p varies as the distance of the point from the pole p = ). r.

:. M.I. = fJ r 2 ·). r · r de dr

f1[ Ja(l+cose) 4 = 2 dr-de 0 0

Fig. (10.100)


[Put e/2 = cj>]

Now the mass of the area = JJ p r dr ·de = Sf A. r · r dr de

j1tja(l+cos6) j1t a3 3 :.M=2 A.-(l+cose) de 0 0 0 3 2A. a 3 j1t 2 3 =-- [2cos (e/2)] -de [Pute/2=cj>]

3 0

16A. a 3 j7t/2 6 =-- cos c1>·2dcj> 3 0

= 32 A.a3·�·�=�A.a31t 3 6·4·2 2 3

:. Radius of gyration k is given by

k2 = M.l. = 63 A.a51t· 3 = 189 a2. Mass 20 5A.a 3n 100

Ex. 20 : Find the M.l. of a cone of base radius a and height h about its axis.

Sol. : Consider a small disc of the cone of thickness dy at a distance y from the vertex 0. Let the radius of the disc be x. If pis the density then the mass m of the disc ism= p 1t x2 dy.

The moment of in tertia of this elementary disc of mass m about the axis oy is by the result obtained in the solved Example 5 above page (10.54) is

x2 2 x2 1t 4 m-= pn x dy·-=p·- x dy

2 2 2

Fig. (10.101)


By considering the similarity of triangles, we get,

x y ay -=- :.x= -

a h h Hence, the M.l. of elementary disc is

1t a4y4 p·-·--dy 2 h4

:. M.I. of the cone about the axis

=J"p-�-a4y4 dy=p-�-�[L]h 0 2 h4 2 h4 5 0

1t a4 ·h5 na4h = p ·-·-·-= P ·--2 Jz4 5 lO

But the mass M of the cone is

1 2 M =p·-1ta h 3

3M :. p=�h 1ta

:. M.l.= 3M _na4h=]_Ma 2 na 2h 10 10

Ex. 21 : Find the moment of inertia of a circular

cylinder of radius a and height h about its axis.

Sol.: Consider a small disc of thickness dy. Its radius

is a. If p is the density then the mass m of the disc

m=pna2dy. The moment of inertia of this elementary disc

of mass m about the axis oy is by the result obtained

in the solved Example 5 above page (10.54) is

ma2 1t 4 --=p-a dy 2 2

Hence M.I. of the cylinder is

Ih 1t 4 1t 4 h 1t 4 M.l.= 0 p2a dy= p2 a [y]0 =p2a h

But the mass of M of the cylinder is M = p 1t a2 h

M 1t 4 Ma 2 :. h=--. 1ta h 2 2

X

Fig. (10.102)

M :. p=�h 1ta

Ex. 22 : Find the moment of inertia about z-axis of the region bounded by

z =x2 + y2, z = O,x =a,x= -a, y =-a,y =a.

Engineering Mathematics - II (10-68) Applications of Differential ..•.•

Sol. : Consider a small volume dx dy dz at a P (x, y, z) on the surface z = x2 + y2.

The distance p of the point P from the z-axis is p = x 2 + y 2

:. M.I. = J J J p p 2 dx dy dz

JJ x2+y2

= J0 p(x 2+y2) dxdydz

= JJp(x2+y2)2dxdy =4J:J:p(x2+y2)2dxdy

= 4p J: J: (x 4 + 2x 2y 2 + y 4) dx dy

Fig. (10.103) [ x 5 2a 3 x 3 a 5 ]" =4p o

[a6 2a6 a6 ] 112 6 =4p 5+

9+5 = 45

pa ·

x2+y2 Mass = M = J J J p dx dy dz = p J J J0 dx dy dz

:. M = 4pJ� J� (x2 + y 2)dx dy = 4pJ� [ x 2y+ I ·dx

=4p[ a34 + a

34 ]=�pa 4

:.M.I.= 112. 3M ·a6 = 14 Ma2. 45 8a4 15

3M :. p=--4 Sa

Ex. 23 : Find the moment of inertia of a sphere about a diameter.

Sol. : Let the equation of the sphere be x2 + y2 + z2 = a2. Let the z-axis be the

diameter about which M.I. is to be obtained

Engineering Mathematics - II {10-69) Applications of Differential .....

:. M.l. = 8 J J J (x 2 + y 2) p dx dy dz We change the cartesian coordinate system to spherical polar system

i.e. We put X= r sin 8 COS cjJ, )' = r sin 8 sin cp, Z = r COS 6 and replace dx dy dz by r2 sin e d8 el dr.

• . x2 + y2 = r2 sin2 e

J'lt/2 J1t/2 Ja 2 · 2 2 :. M.I. = 8 0 0 0

p · r sm 6 · r sin 6 d8 dcp dr

J'lt/2 3 J'lt/2 Ja 4 = 8p 0 sin e d6

0 dq>

0 r dr

2 1t a5 81ta5 =Sp·-·-·-=p· --

3 2 5 15 Now the mass of the sphere M is

4 3 3M M =pV=p·-1t a :. p=--3 3 41ta

5 3M 81ta 2 2 Hence M.I.=--· -- =-1ta 41ta3 15 5

Ex. 24 : Show that the M.l. of the positive octant of the ellipsoid + + � = 1 about the x-axis is M (b 2 + c 2) where M is the mass of

c 6 the ellipsoid.

Sol. : Consider a small volume dx dy dz at a point P (x, y, z) on the surface of the ellipsoid. The distance

p of the point P from the x-axis is p = y 2 + z 2 y :. M.I. = J J J p p 2 dx dy dz

= J J J p(y2 +z2) dxdy dz Fig. (10.104)

To evaluate the integral we put X = ar sin e sin cp, y = br sin e cos cp, z = cr cos e, dx dy dz = abc r2 sin e d6 dq> dr.

:. M.l. = P J:12 J:12 J � (b 2 sin 2 8 sin 2 q, + c 2 cos 2 8)r 2 ·

abc r2 sin e d8 dcp dr 1t/2 1t/2 . . . r 5

[ ]1 =pf0 J0 (b2sm28sm2cp+c2cos28)abcsm 8d9dcp·

S 0

J'lt/2 J1t/2 2 . 2 . 2 2 2 . 1 =p (b sm 9sm Cl+c cos 6)abc sm9d9d�Jl·o 0 5

Engineering Mathematics - II (10·70) Applications of Differential .....

abc J

n/2 J

n/2 2 . 2 . 2 7 2 = p- sinS d8 (b sm 8sm <!>+ c- cos 8) d<!> 5 0 0

abc J

n/2 . [ 2 . 2 1 1t 2 2 1t J = p- sm8 d8 b sm 8·-·-+c cos 8·-5 0 2 2 2

abcn J

n/2 ( b 2 3 2 2 . ) = p-- -sin 8+ c cos 8sm8 d8 10 0 2

= p abcn [!C. 3. + c] = p abcn (b 2 + c 2) 10 2 3 3 30

2 2 2 Since the vo1ume of the ellipsoid �+�+�= 1 is �nabc, the mass

a b c 3 M of the octant is given by.

1 4 M =-·- nabcp 8 3

6M :.p=-nabc

2 2 :.M.I.= 6M +c)

nabc 30 5

Exercise .: VI

1. Find the moment of inertia of the bounded by y 2 = x and x2 = y about

the x-axis. (See fig. 10.43 page ( 10.27) , a= 1) [ Ans. : 9/35 M] 2. Find the moment of inertia of the area under the curve y = sin x

(i) from x = 0 to x = 1t (ii) from x = 0 to x = 2 n about the x-axis.

(Fig. left to you.) [ Ans. : (i) 4/9 (ii) 8/9 ] 3. Prove that the moment of inertia of the area included between the

parabolas y 2 = 4 ax and x2 = 4 ay about they-axis is ( 144/35) M a2 where M is

the mass of the area. (See fig. 10.23 page (10.13)) 4. Find the polar moment of inertia of the area bounded by the parabola

y2 = 4ax and the line y = x. (Fig. left to you.) [ Ans. : (768/35) p a4] 5. Find the moment of inertia of the of the parabola y2 = 4ax bounded by

the double ordinate at a distance h from the vertex about (i) the tangent at the

vertex (ii) the axis of the parabpla.

(Fig. left to you.) [ Ans. : (i) (317) M h2 (ii) (4/5 ) M a h ] 6. Find the moment of inertia and the radius of gyration of the area

bounded by y = 4x (1 - x) and the x-axis about (i) the x-axis (ii) they-axis. (See fig. 7. 160 page (7.39 ))

[ Ans. : (i) (8/35) M; (ii) (3/10) M; ] 7. Prove that the moment of inertia of a lamina of mass M in the form of

· Enginering Mathematics - II · (1D-71) Applications ot Diferential .....

·· a right-angled triangle having hypotenuse of teng!tt c about the axis through . vertex c:ontailing the right angle perpendicular to the plane of the lamina is (M /6) c2. (Fig. left to you.)

(Hint: Jr a, bare the other sides lhen , . 2 ., M.l.= 0 p(x + .v�)dxdy And Mass M = p tlb/2 and a2 + /)!. '= c2 )

8. Find the moment of inertia qf a circular ring whose inner and outer radii are a and b aboul the x-axis.

(Se tig.10.98 page (10.64)) ( Ans.: M (til+ Il) /4]

( Hint : M.l. = JJ p y 2 dx dy . Changing to polar

M.l. = 12"

J6

P· r2 sin 2 8· r dr dB. Mas M = 1t (al- Il) p.) 0 • . 9. Find the moment of inertia in the above eunsple about they-axis.

(a2 +b2) (Hint: M.l.= 0p·r2cos28·rdrd8) [Ans.: M J - 2 2

10. Find the moment of inertia of the quadlwit o( an ellipse x 2 + Zy • I a b

of mas M about the x-axis, if the density at any point is proportional to xy •

(lfint: M.l.= J: XJ•J12dxdy)

Mas M. ==I: Mb2 [Ans.: - J 3 11. Find the moment of inertia in the above problem about the y-uis.

. . �

2 (Hiat: M.l. =I: I:·" b _, A .l"J• ..( 2dx dy ) [ Ans. : ]

12. The surface density of a circular lamina varies as the square of the dis�� 1 �i.-:t 0 on the ci::::mfere:t�. �:.nd t!-.e ;r.::nu:t:! fnerJa of the IIU aboUt the axis throuch 0 perpendicular to. the lamina.

(Se fig. 10.40 pa� (10.26)) (Hint : Take the circle with center (a, 0) and radius a. .

1 rt/2 f241CIOI8 2 2 ltse uationisr=2acos8. M.l.= . (A.r )·r ·rdrd8 -•12 0 .

M M IK/2

I2GCOI8 2) d d8, ass = r · r r ' -K/2 0

20 [Ans.: -Ma2J 9 •••

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