Thomas Calculus 13th [Solutions]

1176 Chapter 16 Integrals and Vector Fields
x
y sin z h( z) f ( x, y, z) e ln y y sin z h( z) y cos z h ( z) y cos z h ( z) 0
x
h( z) C f ( x, y, z) e ln y y sin z C F e ln y y sin z
f
z
x
29.
P
0
N
,
M
0
P
,
N
1
M
y z y x x y
F is conservative there exists an f so that F f ;
f 2 1 3 f g 2 g 2 1 3
x 3 y y y
3
x y f ( x, y, z) x xy g( y, z) x y x y g( y, z) y h( z)
1 3 1 3
3 3
f z z z
z
f ( x , y , z ) x xy y h ( z ) h ( z ) ze h ( z ) ze e C
1 3 1 3 z z 1 3 1 3 z z
F
3 3 3 3
f ( x, y, z)
x xy y ze e C x xy y ze e
(a) work
(b) work
(c) work
B B 3 3
(1, 0,1)
1 1 z z
F dr
dt F dr
x xy y ze e
1
e e
1
A dt A 3 3 (1, 0, 0) 3 3
B 3 3
(1, 0,1)
1 1 z z
F dr
x xy y ze e
A 3 3 (1, 0, 0)
B 3 3
(1, 0,1)
1 1 z z
F dr
x xy y ze e
A 3 3 (1, 0, 0)
Note: Since F is conservative,
B
A
1
1
F dr is independent of the path from (1, 0, 0) to (1, 0, 1).
P yz yz N M yz P N yz M
y z z x x y
0 0 0 0 1 1
30. xe xyze cos y , ye , ze F is conservative there exists an f so
f yz yz f yz g yz
g
x y y y
that F f ; e f ( x, y, z) xe g( y, z) xze xze z cos y z cos y
yz f yz yz
z
yz
F
yz
g( y, z) z sin y h( z) f ( x, y, z) xe z sin y h( z) xye sin y h ( z) xye sin y
h ( z) 0 h( z) C f ( x, y, z) xe z sin y C xe z sin y
(a) work
(b) work
(c) work
B
F
A
B
F
A
B
F
A
yz
(1, /2, 0)
dr
xe z sin y (1 0) (1 0) 0
yz
(1, 0,1)
(1, /2, 0)
dr
xe z sin y 0
yz
(1, 0,1)
(1, /2, 0)
dr
xe z sin y 0
B
Note: Since F is conservative, F
A
(1, 0,1)
dr is independent of the path from (1, 0, 1) to
1, , 0 .
2
31. (a)
3 2 2 2 3
F x y F 3x y i 2 x yj ; let C be the path from ( 1, 1) to (0, 0) x t 1 and
1
2 2 3 4 4
F i j i j and
y t 1, 0 t 1 3( t 1) ( t 1) 2( t 1) ( t 1) 3( t 1) 2( t 1)
r1 i j r1 i j F r
C
1
1
1 4 4
( t 1) ( t 1) d dt dt d 3( t 1) 2( t 1) dt
1 4 5
1
0 0
5( t 1) dt ( t 1) 1; let C 2 be the path from (0, 0) to (1,1) x t and y t,
4 4
F i j and r2 i j r2 i j F r
C
2
0 t 1 3t 2t
1 5 4
t dt 1 F dr F dr 0 C C
1 F dr
C
2 2
1 2
0
2
1 4 4
t t d dt dt d 3 t 2 t dt
0
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