Thomas Calculus 13th [Solutions]

814 Chapter 11 Parametric Equations and Polar Coordinates
31.
2 2
dx
dy
dy
2 2
sin t and cos t dx
( sin t) (cos t) 1
dt dt dt dt
Area
2 2 2 2
y ds
0
2 (2 sin t )(1) dt 2 2 t cos t
0
2 (4 1) (0 1) 8
32.
2
1/2 1/2
2
1
2
dx t and dy t dx dy t t t 1 Area
dt dt dt dt t
3 3/2
2
1
2 x ds 2
0
2 t t dt
3 t
4
3 2
2
t t 1 dt;
u t 1 du 2 t dt; t 0 u 1, t 3 u 4
3 0
4 3/2
4
2 u du 4 u 28
1 3 9 1 9
3 3/2
2
Note: 2 2 t t 1 dt is an improper integral but lim f ( t ) exists and is equal to 0, where
0 3 t
t 0
2
3/2
f ( t) 2 2 t t 1.
Thus the discontinuity is removable: define F( t) f ( t ) for t 0 and
3 t
3
F (0) 0 F 28
0
( t ) dt
9
.
33.
2 2
dx dy
dy 2
2
2
1 and t 2 dx
1 t 2 t 2 2t
3
dt dt dt dt
Area
2 2
2 x ds 2 t 2 t 2 2t 3 dt;
2
2
u t 2 2t 3 du 2t 2 2 dt; t 2 u 1, t 2 u 9
9 3/2
9
u du 2 u 2 (27 1) 52
1 3 1 3 3
2 2
34. From Exercise 30, dx dy
tan t Area
dt dt
/3
2 cos t 2 1 ( 1)
0 2
/3 /3
2 y ds 2 cos t tan t dt 2 sin t dt
0 0
35. dx 2
dt
and
dy
2 2
2 2
1 dx dy
2 1 5
dt dt dt
2 1
Area
1
2 y ds 2 ( t 1) 5 dt
0
2 5 t t
2
3 5. Check: slant height is 5 Area is (1 2) 5 3 5.
0
36.
2 2
dx
2 2
h and dy r dx dy h r Area
dt dt dt dt
2 2 2 1
2 2 2 2
r h r t dt
0 2 r h r t r h r
2
.
0
Check: slant height is
2 1
2 2
h r Area is
2 2 . r h r
1 2 2
2 y ds 2 rt h r dt
0
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