January 27, 2005 11:55
L24-CH15
Sheet number 1 Page number 655
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CHAPTER 15
Multiple Integrals EXERCISE SET 15.1
1
2
1.
1
(x + 3)dy dx = 0
(2x + 6)dx = 7
0 4
0
1
4
2
0
2
ln 3
ln 2
0
2
1
0
2
0
5
1
1
0
π
1
2
0 4
2
ln 2
1 1
2
1
−2
1
0 1
−3
10dx = 20 4
dx = 1 − ln 2
3
1 1 − x+1 x+2
dx = ln(25/24)
0 dx = 0 −1
2 π/2
2 x 1 − x dy dx =
3
15. 0
4
xy dy dx = x2 + y 2 + 1
14. 0
6
dy dx =
1
4xy 3 dy dx =
13. −1
1 1− x+1
7
1 x (e − 1)dx = (1 − ln 2)/2 2
0
1 dy dx = (x + y)2
12. 3
π
xy ey x dy dx =
(3 + 3y 2 )dy = 14 −2
π/2
11. 0
0
(sin 2x − sin x)dx = −2
1
ln 2
1
x cos xy dy dx =
6
8.
0
2
10. π/2
−1
4
x dy dx = (xy + 1)2
9. 0
3 dy = 3
−1
2
2
(x2 + y 2 )dx dy = −2
0
dx dy = −1
4.
1 sin x dx = (1 − cos 2)/2 2
0
7.
0
4x dx = 16 1
0
y sin x dy dx = 0
−1
3
(2x − 4y)dy dx =
ex dx = 2
0
6.
1
ln 3
ex+y dy dx =
5.
1
1 y dy = 2 3
x2 y dx dy =
3.
3
2.
1
√ √ [x(x2 + 2)1/2 − x(x2 + 1)1/2 ]dx = (3 3 − 4 2 + 1)/3
0
1
x(1 − x2 )1/2 dx = 1/3
0
π/3
(x sin y − y sin x)dy dx =
16. 0
0
0
π/2
x π2 − sin x dx = π 2 /144 2 18
17. (a) x∗k = k/2 − 1/4, k = 1, 2, 3, 4; yl∗ = l/2 − 1/4, l = 1, 2, 3, 4, 4 4 4 4 f (x, y) dxdy ≈ f (x∗k , yl∗ )∆Akl = [(k/2−1/4)2 +(l/2−1/4)](1/2)2 = 37/4
R
2
k=1 l=1
k=1 l=1
2
(x2 + y) dxdy = 28/3; the error is |37/4 − 28/3| = 1/12
(b) 0
0
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