Thomas Calculus 13th [Solutions]

1178 Chapter 16 Integrals and Vector Fields
yC g f
0 ( , ) ( )
zC
( )
zC
y
g y z h z z
h z
x y z x y z x y z
2 2 2
3/2
2 2 2
3/2
2 2 2
3/2
C
GmM
1 1 1
x y z x y z
h( z) C f ( x, y, z) C . Let C 0 f ( x, y, z )
is a potential
function for F.
(b) If s is the distance of ( x, y, z ) from the origin, then
gravitational field F is work
2 2 2
1/2
2 2 2
1/2
2 2
s x y z 2 . The work done by the
P
2
P2
d GmM GmM GmM GmM
P 2 2 2
1
x y z
s2 s1 s2 s1
P1
1 1 ,
F r as claimed.
16.4 GREENS THEOREM IN THE PLANE
M M N
y y x
1. M y a sin t , N x a cos t , dx a sin t dt , dy a cos t dt 0, 1, 1, and 0;
Equation (3):
R
Equation (4):
R
2 2
M dy N dx a t a t a t a t dt dt
C 0 0
( sin )( cos ) ( cos )( sin ) 0 0;
M N dx dy 0 dx dy 0, Flux
x y
R
2 2 2 2
M dx N dy a t a t a t a t dt a dt a
C 0 0
2 2
a
2 2 2 2
2
N M
a a x a
1
dx dy 2 dy dx 4 a x dx 4
x a x a
sin
x
x y a c a
2 a 2
a
2 2
2 2
2a 2 a , Circulation
( sin )( sin ) ( cos )( cos ) 2 ;
M M N N
x y x y
2. M y a sin t, N 0, dx a sin t dt, dy a cos t dt 0, 1, 0, and 0;
Equation (3):
Equation (4):
R
M dy
C
N dx
2 2 2 2
2
a 1
0 t cos t dt a
2
t 0; 0 dx dy
0
R
0, Flux
M dx
C
N dy
2 2 2 2 sin 2
2
2
a
0 t dt a t t a
2 4
0
2 a
2 2
2
N M dx dy 1 dx dy r dr d a d a
x y
0 0 0 2
, Circulation
R
3. M 2x 2a cos t, N 3y 3a sin t, dx a sin t dt, dy a cos t dt M 2, M 0, N 0, and
x y x
N
y
3;
2
Equation (3): M dy N dx (2a cos t)( a cos t) (3a sin t)( a sin t)
dt
C 0
2 sin 2
2
sin 2
2
2 2 2 2 2 2 2 2 2
0 2 cos 3 sin 2 t
2 4 3 t
a t a t dt a t
a t
0
2 4
2 a 3 a a ;
0
2 a
2 2
2
M N 1 dx dy r dr d a d a
x y
0 0 0 2
, Flux
R
R
2
Equation (4): M dx N dy (2a cos t)( a sin t) ( 3a sin t)( a cos t)
dt
C 0
2 2 2 2 2
2
1
0 2 a sin t cos t 3 a sin t cos t dt 5 a
2
sin t 0; 0 dx dy 0, Circulation
0
R
N
y
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