Thomas Calculus 13th [Solutions]

Section 11.5 Areas and Lengths in Polar Coordinates 837
26. 2
2sin
1 cos , 2 ;
2
r dr
therefore Length 2
2sin
d
2
2
(1 cos )
/2 1 cos (1 cos )
2 2
2
4 (1 cos ) sin
1 sin d 2
d
/2
2 2
/2 1 cos
2
(1 cos ) (1 cos ) (1 cos )
2 2
2 1 1 2cos cos sin d
/2 1 cos (since 1 cos 0 on 2
(1 cos )
2
)
2
d
2 1 2 2cos
3 3
d
/2 1 cos 2 2 d
(1 cos ) /2 2 2 d
(1 cos ) /2 /2 csc d
2 /2
csc d
2
2sin
(sincecsc 0 on )
2 2
2 3/2
2
3/2
2
/2 3 csc cot
/2
1
/2
csc u du 2 u u
csc u du (use tables)
/4 2 /4 2 /4
2 1 1
/2
1 1
2 2 ln | csc u cot u | 2 /4 2 2
ln 2 1 2 ln 1 2
27.
28
3 2
r cos dr sin cos ; therefore Length
3 d 3 3
/4 3
2
2
2
cos sin cos d
0
3 3 3
/4 6 2 4 /4 2 2 2 /4 2
cos sin cos d cos cos sin d cos d
0 3 3 3 0 3 3 3 0 3
2
3
/4 1 cos
3 2
/4
d 1 sin
3
0 2 2 2 3 0 8 8
1
1/2 1/2
r 1 sin 2 ,0 2 dr (1 sin 2 ) (2cos 2 ) (cos 2 )(1 sin 2 ) ; therefore
d 2
Length
2 2
2 2 2
2
cos 2
1 2sin 2 sin 2 cos 2 2 2sin 2
0 (1 sin 2 ) (1 sin 2 ) d 0 1 sin 2 d 0 1 sin 2
d
2
2
2 d 2 2
0 0
29. Let r f ( ). Then
x f 2 2
( ) cos dx ( ) cos ( )sin dx ( ) cos ( )sin
d
f f d
f f
2 2 2 2
f ( ) cos 2 f ( ) f ( )sin cos f ( ) sin ; y f ( )sin dy f ( )sin f ( ) cos
d
dy
2
2 2 2 2 2
f ( )sin f ( ) cos f ( ) sin 2 f ( ) f ( )sin cos f ( ) cos . Therefore
d
2 2
dx dy
2 2 2 2 2 2 2 2 2 2
f ( ) cos sin f ( ) cos sin dr f ( ) f ( ) r .
d d d
Thus,
2 2
dx dy
2 dr
2
L d r d
d d d
.
30. (a) dr
2 2 2 2 2
r a 0; Length a
d
0 0 d
0
| a | d a
0
2 a
(b) cos dr
2 2 2 2 2
r a a sin ; Length ( a cos ) ( a sin ) d a cos sin d
d
0 0
0 | a | d a 0
a
(c) r asin dr a cos ; Length
d
2 2 2 2 2
( a cos ) ( a sin ) d a cos sin d
0 0
0 | a | d a 0
a
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