Thomas Calculus 13th [Solutions]

Section 15.1 Double and Iterated Integrals Over Rectangles 1085
27.
28.
29.
1 1 1 2
1 1
2
1
V f ( x , y ) dA 1 3 3 1
0 0 (2 x y ) dy dx
0 2 y xy y dx x dx x x
2 0 0 2 2 2
1 0
R
( , ) 4 2 y
4 y 4 4
V f x y dA dy dx dx
0 0 2 0 4 0
1 dx x
0
4
R
0
2 2
/2 /4 /2 /4 /2
V f ( x , y ) dA
0 0 2 sin x cos y dy dx
0 2 sin x sin y dx
0 0
2 sin x dx
R
/2
2 cos x 2
0
30.
31.
1 2 2
2 1 3 1
16 16
1
V f ( x, y) dA 4 y dy dx 4y 1 y dx dx x 16
0 0 0 3 0 0 3 3 0 3
R
2
2 x 3 2
3 2 k 3 2 9 2 27
kx y dx dy x y dx 9ky dy ky k
1 0 3 1
1
x 0 2 1 2
Thus we choose k 2/27.
32.
/2
1 /2 1
sin y dy is some number, say a . Then xsin y dy dx a x dx 0 since the integral of
0
1 0 1
the odd function x over an interval symmetric to 0 is equal to 0.
33. By Fubinis Theorem,
2 1 1 2
x
x
dx dy
dy dx
1 xy
1 xy
0 0 0 0
1
y 2 1
0
y 0 0
34. By Fubinis Theorem,
1 2x
3
ln(1 xy) dx ln(1 2 x) dx ln(1 2 x) 1 ln 3 1
2 2
1 3 3
xy
1 xy
xe dx dy xe dy dx
0 0 0 0
3
y 1 3 3
xy x x
3
e dx e dx e x e
y 0 0
0
0
1 4 16.086
1
0
1 2
y x 1
y x 2
35. (a) MAPLE gives
dx dy and
dy dx
3
0 0 ( x y)
3
3 . This does not contradict
0 0 ( x y)
3
Fubinis Theorem since the integrand is not continuous on the region R : 0 x 2, 0 y 1.
2 1
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