Thomas Calculus 13th [Solutions]

Section 16.4 Greens Theorem in the Plane 1183
b
34. f ( x)
dx Area of R y dx , from Exercise 33
a C
x ( x, y)
dA x dA x dA
M y R R R
35. Let ( x, y) 1 x Ax x dA ( x 0) dx dy
M ( x, y)
dA dA A
R
R
R R
2
x dy, Ax x dA (0 x) dx dy xy dx , and Ax x dA 2 x 1 x dx dy
C 2
C
3 3
R R
R R
1 2 1 1 2 1 2
C 3 x dy 3 xy dx 2 C x dy C xy dx 3 C
x dy xy dx Ax
2 2 2 3
36. If ( x, y ) 1 then I ( , ) 0 1
y x x y dA x dA x dy dx x dy,
3 C
R R R
2 2 2
2 2 2
x dA 0 x dy dx x y dx , and x dA 3 x 1 x dy dx
C
4 4
R R
R R
1 3 1 2 1 3 2 1 3 2 1 3 2
x dy x y dx x dy x y dx x dy x y dx x dy x y dx I
C 4 4 4 C 3 C C 4 C
y
37.
38.
2 2 2 2
f f f f f f f f
M , N M , N
dx dy dx dy 0 for such curves C
y x y 2 x 2
C y x
2 2
y x x y
R
M 1 2 1 3 1 2 2
4 x y , M
, N 1
3 y N x y 4
x y x
Curl N M 1 2 2
1 x y 0 in the interior of
x y 4
2 2
the ellipse 1
2 2
x y 1 work F dr 1 1 x y dx dy will be maximized on the region
4
C
4
R
2 2
R {( x, y)|curl F } 0 or over the region enclosed by 1 1 x y
4
39. (a) 2 2 y
2 2 y
f x i j M x , N ; since M, N are discontinuous at (0, 0), we
2 2 2 2 2 2 2 2
x y x y x y x y
compute
C f n ds directly since Greens Theorem does not apply. Let x a cos t , y a sin t
dx a sin t dt,
dy a cos t dt , M 2 cos , 2 sin , 0 2 ,
a
t N a
t t so f n ds M dy N dx
C
C
2x
2 2
2 2 2
cos t ( a cos t) sin t ( a sin t) dt 2 cos t sin t dt 4 . Note that this holds for
0 a
a
0
any a 0, so 4
C f n ds for any circle C centered at (0, 0) traversed counterclockwise and
4
C f n ds if C is traversed clockwise.
(b) If K does not enclose the point (0, 0) we may apply Greens Theorem: f n ds M dy N dx
C
C
2 2 2 2
2 y x 2 x y
M N dx dy dx dy 0 dx dy 0. If K does enclose the point (0, 0)
x y
2 2
2
2 2
2
R R x y x y
R
we proceed as follows:
Choose a small enough so that the circle C centered at (0, 0) of radius a lies entirely within K. Greens
Theorem applies to the region R that lies between K and C. Thus, as before, 0 M N dx dy
x y
R
M dy N dx M dy N dx where K is traversed counterclockwise and C is traversed clockwise.
K
C
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