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1. Multiple ntegrals
.2
Double
integrals in
polar co-ordinatees
Double integral in
polar co-ordinates is the symbol
3.27
T2
fr,0) drde
(i.e) fr,0) is to be
integrated with respect tor
between the limits ri and rz (keeping 0 as
constant) which
are functions of 8 and then the resulting expression is to
be integrated with respect to 0 between the limits
e, and 2 (keepingr as constant) which are constants.
Tt/2 a cos 0
Example 3.26: Evaluate dr d 0
0
Solution:
Tt/2 a cos
cos
dr d6=
0 0 r= 0
tl2
(acos°-0)d
0
cos 0 d 0
1
Ttl2
n-1n-3 .1,nis odd)
(Since cos 0 d 0 =*
n n - 2
0

2. ntegral Ca
Multiple Integrals
3.28
3.29
a s i n 6
rd dr r/2
E x a m p l e 3 . 2 7 : E v a l u a t e
cos 0d e .(1)
Tt/2
cose is even function of 6
Solution:
a s i n e
d 8 T/2
(1) LHS= cos e d
r d r d 0 =
0
- sin'ode - 1-
Example 3.29: By
changingg into polar co-ordinates
sin'ed 2a V2ax
evaluate the integral
+ dxdy
0
1 cos 20 Solution:
de The region of
integration is shown below.
2
0
The limits are x =
0 andx =
2a
sin20 y =0 andy =
V2ax -x
(+-2ax=0)
Je =0
(t-0)- (0-0)]- +-2ax
,0)
T/2 2 cos 6
Example 3.28: Evaluate drde P (a, 0)
2a
- Tt/2 0
The strip PQ moves in the x
direction from * =0 to
= 2a and in the y direction from y = 0 to y =
N2ax-
Solution:
T/2 2 cos T/2 -2 cos 8
x+-2ax =
0 is a circle with centre(a, 0) and radius
Pdrde- d
Na +0 +0-0=a
r =0
-T/2 0 - Tt/2
T/2
8cosed8
- Tt/2

3. C
3.30
then x+y2 Multiple Integrals
Let
r
= r
cos0,y = r sin 6
d x d y = r
drd 0
Here
v a r i e s
from 0 to t/2 (since.
3.E
the integra
in the
first
q u a d r a n t )
+ y 2 a x r ° = 2 a r c o s e
r= 2a cos 0
r
varies
from 0 to 2a cose
(t,
8 )
t/2 2a cos
V2ax-x
rdrde
+ de dy =
T/2 +=a2P=r
2a cos8
i.e)rvaries from 0 to a and 0 varies from 0 to n/2
d
SS y dx dy= cosrsinrdr d
Tt/2
T/2
16a cos* 0de 0
0
Tt/2
Ttl2 cose sin 0 dr d 0
= 4a
0
cos* 0de
0
-E.cos0 sin 6d0
=o
3 na 0
4a 42 4
-
cosfesin 6de
n/2 0
cos"e d
9 = 2 - 1 n
-3
n
cos0 d (- cos 8)
over
Example 3.30: Evaluate S y dedyover t/2
quadrant of the
c i r c l e + y = a
Solution:
The region of integration
is shown
below Cos
-coo
Let x = r cose and y = r sin 6 -50-1)=
de
Then +y=a and drdy=rdrd"

4. Integral c
3.34
Multiple Integrals
TT/2 Cos 0
3.35
x =r
cos8,y =
r sins 0
Then dx dy =
rdr d 6
r/2
+y=P2
.- [?-a costey2-(a3/2da
Here r
varies from a to b
0
varies from 0 to 2 T
t/2
aa sin'e-a)de 2 T
dx dy =
rdr d
2
3 / 2
-- io-1d 2 T
cos 0sinedr de
n/2
2 T
-(0)
coso sinede
-
(3 Tt 4)
cos-0 sin 0d0
4
0
S dxdy
+
Example 3.33: Evaluate
cose(1-coso) d6
4
0
annular region bounded by +y =a,?+f
2r
4 (cos 0-cos'e) de
Solution:
integration is
region of
a2 (coße-cos'0)d 0
4
The
0
2+y=a?,+y=b (b>a)
n/2
(cose-cos0)d0
4
0
-
n(6-a
16

5. Multiple Integrals
3.36
the 3.37
E x a m p l e
3.34:
E v a l u a t e
J sine drd 0 ov
of
the
c i r c l e
r
= a
c o s 0
er the po
S o l u t i o n :
Here
the region of
i n t e g r a t i o n
is taken
P,0)
q u a d r a n t
of the
circle.
0 varies from
to
r
varies from 0 to a cos
T/2 acos
0 varies from 0 to t/2
:. [[ sinfedrde =
J r*sinedra
0
rvaries from 0 to 2a cose
cos68
sin0 dr de Tt/2 2a cos 0
Jsinedr d0 =
sin0dr d0e
0 R
0
cose sin'0 de
Pa cos 0
sin 0d e
T/2
(coste-cos0)de mi2
cos'e sin 0 d0
0
Ta
128
3 cos'ed(-cos 0)
t/2
Example 3.35: Evaluate sin drd 6,Ris
R
circle r 2a cos 6 above the initial line.
co-cos'o
Solution:
The region of integration is shown low
2a3
3 0-1=2a3
3

6. Multiple Integralss
3.39
3.38 a V a -
y+)dxdy
- sin 0 d 0
us
E x a m p l e 3 . 3 6 : E v a l u a t e
n/2
-
p o l a r co-ordinates.
sin0d 0=
-cos 0)6
aVa-
Solution:
Va-
y+y)
( a y + )
dxdy
= a ,
--0-1)-
x
varies
from 0 to a
Example 3.37: Evaluatee
dedy by
varies
from 0 to
V a - x 2
taking polar co-ordinates.
y = 0 , y = V a " - x 2 » 2 + y 2 = a 2
Solution:
I= + dxdy .1)
I a Let =r cos 0, y =
r
sin0
Then
+y=and dxdy =
rdr de
r=0
X
P
Take x=r cos
0,y =r sin 6
Then+y=r and dxdy =r dxde
r0
r varies from 0 to a;0 varies from 0 to 2
t/2
since x
varies from 0 to o, and y varies from 0 to
o, the
region of
integration is the 1st
quadrant.
I= r
sin0rdrd0
0 0
In the first
quadrant 0 varies from 0 to t/2
r/2
sin drd 0
-

7. Integral Calcu
Multiple Integrals
x/2 S rdrdde
3.41
T/2
(,a)
B
- -
0
T/2
de
A
,0)
n/2 (0,0
- - - d e
-
0
n/2
Cose sec0
n/2
(0-1)
de=J de rvaries from 0 to a sec0
0
e varies from 0 to
T/2 a sec 8
dy= f cos rdr de
V
mple 3.38:
Evaluate the integral
0
T/2 asec 0
J cos2e drde
hanging to polar
co-ordinates. 0
t/2
-T5
sec 6
tion: cos0 de
dxdy
+
Let 2 T/2
asece
cOse
cose de
3
0
The region of integration is bounded by1=*
Tt/2
,y =a sece d0
Let X=r cos 6, y = r sin6
Then 2+y2 =2 log (sec0+ tan 0)1
T/2
and dxdy = rdr d0
The line AB is rcos 6 a

8. 3 . 4 2
[log
(V2+1)-
(0+0)]
3
log (2+1)
Exercise 3.2
a
Tt/2.
2dr
drd Ans.
364
1 Evaluate
0 a(1-cos0)
a (1+Cos0)
rdrd e
Ans:
2. Evaluate
0 0
T/2
drd 0 Ans
3. Evaluate
0 0
4, Evaluate r dr d8 over the positive quad
the circle +y2=a2 by changing into
4
co-ordinates.
A
Tt/2
rdrd
(+022
5: Evaluate
Ans
0 0