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 . 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 : xd y y Fig. (10.2) ....(1) X (See Ex. 5) .... (2)
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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
Engineering Mathematics - II (1 0-5) Appli. of Mult. lnte.
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)
+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)
Engineering Mathematics - II (10-12) Appll. of Mult. lnte.
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
Engineering Mathematics - II (10-13) Appli. of Mult. lnte.
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)]:
Engineering Mathematics - II (10-14) Appli. of Mult. lnte.
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).
= ( 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)
Engineering Mathematics - II (10-16)
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)
Engineering Mathematics - II (10-18)
= _!_ 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
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
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)
Engineering Mathematics - II (10-23) Appli. of Mult. lnte.
[ 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
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
Engineering Mathematics - II (10-25) Appli. of Mult. lnte.
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
Engineering Mathematics - II (10-26) Appli. of Mult. lnte.
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
Engineering Mathematics - II (10-29) Appll. of Mult. lnte.
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
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
Engineering Mathematics - II (10-32)
= 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
Appli. of Mult. lnte.
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)
Engineering Mathematics - II (10-33) Appli. of Mult. lnte.
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
Engineering Mathematics - II (10-34) Appli. of Mult. lnte.
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
Engineering Mathematics - II (10-35) Appli. of Mult. lnte.
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
Engineering Mathematics - II (10-36) Appll. of Mult. lnte.
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
Engineering Mathematics - II (10-37)
= 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
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.
- 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
= { 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'
Engineering Mathematics - II (10-47)
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 = �
Engineering Mathematics - II (10-48) Appll. of Mult. lnte.
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
Engineering Mathematics - II (10-50) Appll. of Mult. lnte.
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
Engineering Mathematics - II (10-51) Appli. of Mult. lnte.
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
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
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.
Engineering Mathematics - II (10-54) Appli. of Mult. lnte.
. -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
Engineering Mathematics - II (10-56) Appll. of Mult. lnte.
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
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
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] .
Engineering Mathematics - II (10-63) Applications of Differential .....
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)
Engineering Mathematics - II (10-64) Applications of Differential .....
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)
Engineering Mathematics - II (10-66) Applications of Differential .....
[Put e/2 = cj>]
Now the mass of the area = JJ p r dr ·de = Sf A. r · r dr de
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)
Engineering Mathematics - II (10-67) Applications of Differential .....
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 ·
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.
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. .