Thomas Calculus 13th [Solutions]

Section 15.3 Area by Double Integration 1103
18. 2 0 4 x
2 dy dx dy dx
0 x 4 0 0
2 2 4 1/2
4 x dx x dx
0 0
3 2
3/2
4
4x
x 2 x 8 8 16 32
3
0
3 0 3 3 3
19. (a) average 1 1 1
2
0 0 sin ( x y ) dy dx
2
0 cos( x y ) dx
0
2
0
cos( x ) cos x dx
1 sin( x ) sin x 1 ( sin 2 sin ) ( sin sin 0) 0
2 0 2
(b) average 1
/2
2 /2 2
2 0 0 sin ( x y ) dy dx
2
0 cos( x y ) dx
0 2
0
cos x
2
cos x dx
2
2 sin x sin x 2 sin 3 sin sin sin 0 4
2 2 2 2 2
2
0
20. average value over the square
average value over the quarter circle
2 4 1
1 1 1 xy 1
xy dy dx dx x dx 1
0 0 0 2 0 2 4
0
2 1
0.25;
2 2 1 x
1
1 1 x
4
1 xy
2
1 3
xy dy dx dx x x dx
0 0 0 2
0
4
0
2 x x 1 0.159. The average value over the square is larger.
2 4
0
2
2
21. average height
3 2
2 2 2 2
3 2
1 2 2 1 2 y 2
x y dy dx x y dx 1 2x 8 dx 1 x 4x
8
4 0 0 4 0 3 4 0 3 2 3 3
0
3
0
2 ln 2
22. average 1
2 ln 2 2 ln 2
1 1
2 ln 2 ln y
2 ln 2
dy dx dx 1 1 ln 2 ln ln 2 ln ln 2 dx
2
ln 2 ln 2
2
ln 2
2
(ln 2) xy (ln 2) x
ln 2 (ln 2) ln 2 x
1
2 ln 2
dx 1 2 ln 2
ln x
1 (ln 2 ln ln 2 ln ln 2) 1
ln 2 ln 2 x ln 2 ln 2 ln 2
23. The region R is shaded in the following figure.
y
2
1
0 1 2
x
R
2 2
2
0
4 x
2
2 2 1
x x x
dA 1 dy dx 4 x (2 x) dx 4 x 2sin 2x 2, where
2 x
2 2 2
0
2
we use integration by parts with u 4 x and dv 1/2 to find
is a quarter of a circle of radius 2 with a triangle of area 2 removed, giving area 2.
2
4 x dx . Geometrically, the region R
2
0
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