Thomas Calculus 13th [Solutions]

1164 Chapter 16 Integrals and Vector Fields
16. C1 : x t, y 3 t, 0 t 1 dx dt; C2 : x 1 t, y 3, 0 t 1 dx dt; C3
: x 0, y 3 t, 0 t 3
dx 0 x y dx x y dx x y dx x y dx
C C C C
1 2 3
1 1 3 1 1
t 3 t dt (1 t) 3( 1) dt 0 (3 t) 0 2 t dt 4 t dt
0 0 0 0 0
2/3
1
2/3
1
4
t
2
(4 t) 4
2 3
16
2 3 4
3 0 3 0 3 3
17.
2
r i j k
( t) t t , 0 t 1 dx dt, dy 0, dz 2t dt
(a)
(b)
(c)
C
1 2 1 2 1 3
1
5
0 2 3 0 6
( x y z) dx t 1 t dt t t t
1 2
( ) 1 0 0
C x y z dy 0
t t
1 2 1 2 3 3 2 4
1
( x y z) dz t 1 t 2t dt 2t 2t 2t dt
2
t t
1
t
5
C 0 0 3 2 0 6
18. r( t) (cos t) i (sin t) j (cos t) k, 0 t dx sin t dt, dy cos t dt, dz sin t dt
(a)
(b)
(c)
C
C
C
2 1 3
(cos )( cos )( sin ) cos sin (cos )
2
0 0 3 0 3
x z dx t t t dt t tdt t
3 2 1 3
0 0 0 3
0
x z dy (cos t )( cos t )(cos t ) dt cos tdt 1 sin t cos t dt (sin t ) sin t 0
2 2 1 2 1 1 cos 4t
(cos )(sin )( cos )(sin ) cos sin sin 2
0 0 4 0 4 0 2
x y z dz t t t t dt t t dt t dt dt
1
(1 cos 4 t) dt
1
t
1
sin 4t
8 0 8 32 0 8
19.
20.
2 3 2 3
r ti t j tk, 0 t 1, and F xyi yj yzk F t i t j t k and i 2tj k
F
dr
dt
3 1 3 1
0 2
2t work 2t dt
t
6
F (2 sin t) i (3 cos t) j (cos t sin t) k and r
( sin t) i (cos t)
j
1
k
dt
6
F
dr
2 2 1 1
2 2 2 1 1
dt
6 6 0
6 6
sin 2
2
3 3 t
t sin 2t t
1
sin t
1
cos t
2 4 2 6 6
0
r (cos t) i (sin t) j k, 0 t 2 , and F 2yi 3 xj ( x y)
k
3 cos t 2sin t cos t sin t work 3 cos t 2 sin t cos t sin t dt
dr
dt
21. r (sin t) i (cos t) j tk, 0 t 2 , and F zi xj yk F ti (sin t) j (cos t) k and
dr
dt
dr
dt
2
2 2
2
0
(cos t) i (sin t) j k F t cos t sin t cos t work t cos t sin t cos t dt
t sin 2t
2 4
cos t t sin t sin t
0
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