Calculus one and several variables 10E Salas solutions manual ch17
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Calculus one and several variables 10E Salas solutions manual...
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December 7, 2006
SECTION 17.1
CHAPTER 17 SECTION 17.1 1.
3 3
2i−1 3j+1 =
⎛
3
i=1 j=1
⎝
2i−1
i=1
3
⎞ 3j+1 ⎠ = (1 + 2 + 4)(9 + 27 + 81) = 819
j=1
2. 2 + 22 + 3 + 32 + 4 + 42 + 5 + 52 = 68 3.
4 3
(i2 + 3i)(j − 2) =
i=1 j=1
4.
5.
4
⎤
⎡ 3 (i2 + 3i) ⎣ (j − 2)⎦ = (4 + 10 + 18 + 28)(−1 + 0 + 1) = 0
i=1
j=1
2 2 2 2 2 2 4 4 4 4 4 4 6 6 6 6 6 6 4 + + + + + + + + + + + + + + + + + = 19 . 2 3 4 5 6 7 2 3 4 5 6 7 2 3 4 5 6 7 35 m
Δxi = Δx1 + Δx2 + · · · + Δxm = (x1 − x0 ) + (x2 − x1 ) + · · · + (xm − xm−1 )
i=1
= xm − x0 = a2 − a1
6. (y1 − y0 ) + (y2 − y1 ) + · · · + (yn − yn−1 ) = yn − y0 = b2 − b1
7.
m n
Δxi Δyj =
i=1 j=1
8.
9.
⎛
m
Δxi
q n
⎛ Δyj Δzk = ⎝
n j=1
m
m
(xi + xi−1 )Δxi =
⎞ Δyj ⎠ = (a2 − a1 ) (b2 − b1 )
j=1
j=1 k=1
i=1
⎝
i=1
n
⎞ Δyj ⎠
q
Δzk
= (b2 − b1 ) (c2 − c1 )
k=1 m
(xi + xi−1 )(xi − xi−1 ) =
i=1
(xi 2 − x2i−1 )
i=1
= xm 2 − x0 2 = a2 2 − a1 2 10.
n 1 j=1
11.
2
m n
(xi + xi−1 )Δxi Δyj =
i=1 j=1
∧ (Exercise 9)
12.
1 3 1 (yj − yj−1 3 ) = (b2 3 − b1 3 ) 2 j=1 2 n
(yj 2 + yj yj−1 + yj−1 2 )Δyj =
m n i=1 j=1
m
(xi + xi−1 )Δxi
i=1
n
Δyj
j=1
= a2 2 − a1 2 (b2 − b1 )
(yi + yj−1 )Δxi Δyj =
m i=1
⎤ ⎡ n Δxi ⎣ (yj 2 − yj−1 2 )⎦ = (a2 − a1 )(b2 2 − b1 2 ) j=1
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SECTION 17.2 13.
m n
m
(2Δxi − 3Δyj ) = 2
i=1 j=1
Δxi
867
m n n 1 −3 1 Δyj
i=1
j=1
i=1
j=1
= 2n(a2 − a1 ) − 3m(b2 − b1 ) 14.
m n
(3Δxi − 2Δyj ) = 3
i=1 j=1
15.
m n
Δxi − 2
i=1 j=1
q m n
Δxi Δyj Δzk =
i=1 j=1 k=1
m
m n
Δyj = 3n(a2 − a1 ) − 2m(b2 − b1 ).
i=1 j=1
⎞ ⎛ n q Δxi ⎝ Δyj ⎠ Δzk
i=1
j=1
k=1
= (a2 − a1 )(b2 − b1 )(c2 − c1 ) 16.
q m n
(xi + xi−1 )Δxi Δyj Δzk =
i=1 j=1 k=1
m
⎞
⎛ n q 2 2 (xi − xi−1 ) ⎝ Δyj ⎠ Δzk
i=1
j=1
k=1
= (a2 2 − a1 2 )(b2 − b1 )(c2 − c1 ) 17.
n n n
δijk aijk = a111 + a222 + · · · + annn =
appp
p=1
i=1 j=1 k=1
18. Start with
n
m n
aij .
Take all the aij
(there are only a finite number of them) and order them
i=1 j=1
in any order you chose. Call the first one b1 , the second b2 , and so on. Then m r n aij = bp where r = m × n. p=1
i=1 j=1
SECTION 17.2 1. Lf (P ) = 2 14 , 3. (a) Lf (P ) =
Uf (P ) = 5 34 n m
2. Lf (P ) = 3,
(xi−1 + 2yj−1 ) Δxi Δ yj ,
Uf (P ) =
i=1 j=1
(b) Lf (P ) ≤
n m
Uf (P ) = 5
(xi + 2yj ) Δxi Δyj
i=1 j=1
m n yj−1 + yj xi−1 + xi +2 Δxi Δyj ≤ Uf (P ). 2 2 i=1 j=1
The middle expression can be written n n m m 2 1 2 2 yj − yj−1 Δxi . xi − x2i−1 Δyj + 2 i=1 j=1 i=1 j=1
The first double sum reduces to m n 1 i=1 j=1
2
xi − 2
x2i−1
1 Δyj = 2
m
xi − 2
x2i−1
i=1
n j=1
Δyj
=
1 (4 − 0) (1 − 0) = 2. 2
In like manner the second double sum also reduces to 2. Thus, I = 4; the volume of the prism bounded above by the plane z = x + 2y and below by R.
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SECTION 17.2
4. Lf (P ) = −7/16, 6. (a) Lf (p) =
5. Lf (P ) = −7/24,
Uf (P ) = 7/16
m n
(xi−1 − yj )Δxi Δyj ,
Uf (P ) =
i=1 j=1
(b) Lf (P ) ≤
m n
Uf (P ) = 7/24
(xi − yj−1 )Δxi Δyj
i=1 j=1
m n xi + xi−1
−
2
i=1 j=1
yj + yj−1 2
Δxi Δyj ≤ Uf (P )
The middle expression can be written m n 1 i=1 j=1
2
(xi 2 − xi−1 2 )Δyj −
m n 1
2
i=1 j=1
(yj 2 − yj−1 2 )Δxi .
The first sum reduces to ⎞ ⎛ n m 1 1 1 (xi 2 − xi−1 2 ) ⎝ Δyj ⎠ = (1 − 0)(1 − 0) = . 2 i=1 2 2 j=1 In like manner the second sum also reduces to 12 . 7. (a) Lf (P ) =
m n
(4xi−1 yj−1 ) Δxi Δyj ,
Thus I = m n
Uf (P ) =
i=1 j=1
(b) Lf (P ) ≤
1 2
−
1 2
= 0.
(4xi yj ) Δxi Δyj
i=1 j=1
m n
(xi + xi−1 ) (yj + yj−1 ) Δx1 Δyj ≤ Uf (P ).
i=1 j=1
The middle expression can be written m n
xi − 2
x2i−1
yj − 2
2 yj−1
i=1 j=1
=
m
xi − 2
x2i−1
i=1
∧
n
yj − 2
2 yj−1
j=1
by (17.1.5) = b 2 − 02 d 2 − 02 = b 2 d 2 . It follows that I = b2 d2 . 8. (a) Lf (P ) =
m n
3(xi−1 2 + yj−1 2 )Δxi Δyj ,
Uf (P ) =
i=1 j=1
m n
3(xi 2 + yj 2 )Δxi Δyj
i=1 j=1
m n 2 (b) Lf (P ) ≤ (xi + xi xi−1 + xi−1 2 ) + (yj 2 + yj yj−1 + yj−1 2 ) Δxi Δyj ≤ Uf (P ) i=1 j=1
Since in general (A2 + AB + B 2 )(A − B) = A3 − B 3 , m n
(xi 3 − xi−1 3 )Δyj +
i=1 j=1
which reduces to m
m n
the middle expression can be written
(yj 3 − yj−1 3 )Δxi ,
i=1 j=1
⎛ xi 3 − xi−1 3
i=1
⎝
n
⎞ Δyj ⎠ +
j=1 3
3
m
⎛ Δxi
i=1 2
2
This can be evaluated as b d + bd = bd(b + d ).
⎝
n
⎞ yj 3 − yj−1 3 ⎠ .
j=1
It follows that I = bd(b2 + d2 ).
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SECTION 17.2 9. (a) Lf (P ) =
m n
3 x2i−1 − yj 2 Δxi Δyj ,
Uf (P ) =
i=1 j=1
(b) Lf (P ) ≤
m
n
m n
869
2 Δxi Δyj 3 xi 2 − yj−1
i=1 j=1
2 xi 2 + xi xi−1 + x2i−1 − yj 2 + yj yj−1 + yj−1 Δxi Δyj ≤ Uf (P ).
i=1 j=1
Since in general A2 + AB + B 2 (A − B) = A3 − B 3 , m n
m n 3 3 xi 3 − x3i−1 Δyj − yj − yj−1 Δxi ,
i=1 j=1
which reduces to m
the middle expression can be written
i=1 j=1
⎞ ⎞ ⎛ n ⎛ n m 3 ⎠ . xi 3 − x3i−1 ⎝ Δyj ⎠ − Δxi ⎝ yj 3 − yj−1
i=1
j=1
i=1
This can be evaluated as b3 d − bd3 = bd b2 − d
2
j=1
. It follows that I = bd (b2 − d2 ).
10. On each subrectangle, the minimum and the maximum of f are equal, so f is constant on each subrectangle and therefore (since f is continuous) on the entire rectangle R. Then f (x, y) dxdy = f (a, c)(b − a)(d − c). R
11.
dxdy =
Ω
b
φ(x) dx a
12. Suppose that there is a point (x0 , y0 ) on the boundary of Ω at which f is not zero. As (x, y) tends to (x0 , y0 ) through that part of R which is outside Ω, f (x, y), being zero, tends to zero. Since f (x0 , y0 ) is not zero, f (x, y) does not tend to f (x0 , y0 ). Thus the extended function f can not be continuous at (x0 , y0 ). 13. Suppose f (x0 , y0 ) = 0. Assume f (x0 , y0 ) > 0. Since f is continuous, there exists a disc Ω with radius centered at (x0 , y0 ) such that f (x, y) > 0 on Ω . Let R be a rectangle contained in Ω . Then f (x, y) dxdy > 0, which contradicts the hypothesis. R
14.
(x + 2y) dx dy = 4;
area(R) = (2)(1) = 2,
so average value =
4 =2 2
R
4xy dxdy = 22 32 = 36. Thus
15. By Exercise 7, Section 17.2, R
favg =
1 area (R)
4xy dxdy =
1 (36) = 6 6
R
(x2 + y 2 ) dxdy =
16. R
bd(b2 + d2 ) ; 3
area(R) = bd,
so average value =
b2 + d 2 3
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SECTION 17.3
17. By Theorem 16.2.10, there exists a point (x1 , y1 ) ∈ Dr such that
dxdy = f (x1 , y1 )πr2
f (x, y) dxdy = f (x1 , y1 ) Dr
=⇒
f (x1 , y1 ) =
1 πr2
R
f (x, y) dxdy Dr
As r → 0, (x1 , y1 ) → (x0 , y0 ) and f (x1 , y1 ) → f (x0 , y0 ) since f is continuous. The result follows. 18. 0 ≤ sin(x + y) ≤ 1 for all (x, y) ∈ R. Thus, 0 ≤
sin(x + y) dxdy ≤
R
19. z =
dxdy = 1
R
4 − x2 − y 2 on Ω : x2 + y 2 ≤ 4, x ≥ 0, y ≥ 0;
4 − x2 − y 2 dxdy is the volume V of one
Ω
quarter of a hemisphere; V = 43 π.
20. 8 − 4 x2 + y 2 on Ω is a cone with height h = 8 and radius r = 2;
V =
32π 3
x y z + + = 1; the solid is the tetrahedron bounded by the coordinate planes 3 2 6 x y z and the plane: + + = 1; V = 16 (3)(2)(6) = 6 3 2 6 2 ∼ ∼ 22. (a) Lf (P ) = 35.4603; Uf (P ) = 36.5403 (c) 3y − 2x dxdy = 36 21. z = 6 − 2x − 3y ⇒
R
SECTION 17.3 1 3 2 1. x dy dx = 0
0 3
1
x+y
e
3x2 dx = 1
0
2. 0
1
0 1
3
dx dy = 0
3
1
2
3.
xy dy dx = 0
0 1
0
x
3
4.
1
x y dy dx = 0
0 1
x
xy 3 dy dx =
0 1
x3
x
x2 y 2 dy dx = 0
7.
1
3
0
1
dx =
9x dx = 0
0
1 4 y 4
x2
x
dx = 0
0
9 2
1
1 1 5 x dx = 4 24
x3 1 dx = 3 18
π/2
sin (x + y) dy dx = 0
x2 1 dx = 2 12
1
x
0 π/2
1 x y3 3
0
6. 0
0
5. 0
3 (e1+y − ey ) dy = e1+y − ey 0 = e4 − e3 − e + 1
0
π/2
π/2
[− cos (x + y)]0
dx = 0
π/2
π cos x − cos x + dx = 2 2
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SECTION 17.3
π/2
π/2
8.
π/2
sin
cos(x + y) dx dy = 0
0
π/2
9.
π/2
(1 + xy) dy dx = 0
10.
0
√
−
−1
1
y
11.
(x + 3y ) dx dy =
1
y
2 3/2 x 3
1 − y 2 dy = 0
y
y(ey
1
dy = 0
y2
(integrand is odd)
2 2 2 y − y 7/2 dy = 3 27
2
y2 ey − − 1) dy = 2 2
2
0
0
6y 3
1
yex dx dy =
12. 0
√
1
xy dx dy = 0
y2
−1
1−y 2
√
π/2 π/2 1 1 1 2 1 2 1 dx = π + π x dx = π 2 + π 4 y + xy 2 2 8 4 64 0 0 1
3
y2
0
0
√1−y2
1
π π/2 =0 + y − sin y dy = cos y − cos +y 2 2 0
0
π/2
π
1 = 0
1 (e − 2) 2
2 4− 1 y2 2 2 1 2 1 2 2 2 13. dy 4 − y dx dy = 4−y 4− y − y 2 2 −2 21 y 2 −2
2
=2
0
1
x2
14. I =
x3
0
1
(x4 + y 2 ) dy dx = 0
512 16 − 8y 2 + y 4 dy = 15
x2 1 6 4x y3 x9 x4 y + dx = − x7 − dx 3 x3 3 3 0
1 4x7 9 x8 x10 = = − − 21 8 30 0 280 15. 0
by symmetry (integrand odd in y, Ω symmetric about x-axis)
1
2y
−y 2 /2
16.
e 0
2
0
x/2
2ye
2
2
0
0
−1
−y 2 /2
dy = −2e
1
0
x5 +
x8 x6 − x4 − 2 2
x6 x9 x5 x7 = + − − 6 18 5 14
1 =2 1− √ e
0
x4
dx + 0
x4 x3 1 y2 y2 xy + xy + dx + dx 2 x3 2 x4 −1 0
x3
(x + y) dydx = 0
1
2 1 x2 1 x2 1 4 = xe dx = e e −1 2 4 4 0
x4
x3
ex dy dx =
(x + y) dy dx +
−1
=
−y 2 /2
0
0
0
1
dx dy =
0
17.
18.
1
x4 +
0
x6 x8 − x5 − 2 2
x5 x7 x6 x9 + + − − 5 14 6 18 −1
1 =− 0
1 3
dx
871
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SECTION 17.3
19.
20.
1
y 1/4
f (x, y) dx dy
21.
1
√
f (x, y) dydx x
22.
0
−1
1
−x
1
1
f (x, y) dy dx +
f (x, y) dy dx 0
1/2
√ 3
x
23.
y 1/3
1
f (x, y) dxdy + 1/2
1/2
y 1/3
f (x, y) dxdy y
1/2
24.
2
y
4
y
f (x, y) dx dy + 1
f (x, y) dx dy
1
2
y/2 8
4
−2
3
1 2 x+2
dy dx =
4y−y 2
26.
3
dx dy = 0
y
27.
dx dy = 0
2y 3/2
0
−y
f (x, y) dx dy +
9 2
1 y − 2y 3/2 dy = 160
1
+
9
3
f (x, y) dx dy
−1
1
1 2 1 x + 2 − x dx = 9 2 4
(3y − y 2 ) dy =
1/4
3
y/2
0
y 1/4
4
−2
1/4x2
−3
f (x, y) dx dy 4
−1
4
+
25.
0
y 1/2
0
1
1
3
√
f (x, y) dx dy y
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SECTION 17.3
3
5−x
28.
3
dy dx = 2
6/x
(5 − x − 6/x) dx =
2
5 2 + 6 ln 2 3
29.
1
y2
sin 0
0
30.
1
y3 + 1 2
x2 dy dx = =
ln 2
32.
1
0
2
2/y
33.
dx dy = 1
y−1 1
34.
1
2
0
0
√
y
1−x
0
1
y3 + 1 y sin dx dy = dy 2 0 3 1 2 y +1 = − cos 3 2 0 1 2 cos − cos 1 = 3 2
1
−1
1
2
x2 (1 − x2 ) dx
4 15
e−x dy dx =
ln 2
e−x (2 − ex ) dx
0
ln 2 = −2e−x − x 0 = 1 − ln 2
x3 x4 + y 2
2 1 − (y − 1) dy = ln 4 − y 2
(x + y) dy dx = 0
2
ex
0
0
x2 −1
−1
31.
873
1 (1 − x)2 dx = x(1 − x) + 2 3
1 √ ( 2 − 1)y dy 0 2 1 √ = ( 2 − 1) 4
dx dy =
1
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SECTION 17.3
2
3− 32 x
35. 0
0 1
3 2− 2 y 3 4 4 4 − 2x − y dy dx = 4 − 2x − y dx dy = 4 3 3 0 0
1
36.
1
(2x + 3y) dy dx = 0
0 2
0
1
2
39.
0 √
1−x2
√ − 1−x2 √
2−2y
x3 y dx dy =
0
2x−x2
(2x + 1) dy dx =
1
1
=
−1
=6
40.
1
√ 1− 1−x2
−1
1
√ 1+ 1−x2
41.
0 1
√ − 1−x2
−1
1
1−x2
43.
(1 − x) dy dx =
2
2
0
2
44.
(1 + xy) dx dy = 1
2y−1 a
√
a2 −x2
45. a
0
b(1−x/a)
0 1
47. 0
1
1
0
1 6
dx =
11 2x3 − x4 − x6 dx = 70
2
3 55 (6 + 9y − 3y 2 − y 3 ) dy = 2 8 a
dy dx =
1
0 cos−1 y
(a2 − x2 ) dx =
2 3 a 3
a bc x 2 x y abc dy dx = 1− c 1− − dx = a b a 6 0 2
ey/x dx dy =
y
48. 0
−
x2
1
4 1 2x − x3 − x + 3 3
a2
= 3π
1 2 2 43 2 3/2 2 2 6 1 − x − x 1 − x − (1 − x ) dx = π. 2 3 16 −1
0
46.
2
(2 1 − x2 − 2x 1 − x2 ) dx = π
0
0
1
π
0
5−y
1 − y 2 dy = 6
1
(x + 3y ) dy dx = x2
1
−1
x
1−y 2
2
0 √
1
−1
(x + y ) dy dx = 0
42.
1
2
(2x + 1) dx dy 1−y 2
1−
1 (4 − y − x2 ) dy dx = 4
2
1−y 2
1+√1−y2 x2 + x √ dy
2
1−x
√
1−
1
1+
√
−1
2 15
1 π x2 1 − x2 + (1 − x2 )3/2 dx = 3 2 −1
(x2 + y 2 ) dy dx = 2
√ − 2x−x2
0
3 5 (2x + ) dx = 2 2
2
x3 y dy dx =
0
−1
0 1− 12 x
37.
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x
ey/x dy dx =
0
1
xey/x
0
esin x dx dy = 0
π/2
0
cos x
x
dx =
0
0
esin x dy dx = 0
π/2
1
x(e − 1) dx =
1 (e − 1) 2
cos xesin x dx = e − 1
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SECTION 17.3
1
1
49.
2 y4
1
y
x e dy dx = 0
y
50.
0
y2
e 0
51. favg
1
−1
4
2
yey dy =
1
−1
1
2
8x dx =
1
1−x2
2 π
0
0
ln 2
Ω=1
√
xy dy dx =
2 ln 2
1 dy = 3 0
1
4
y 3 ey dy =
0
1 (e − 1) 12
−1
x2 dx =
2 3
2 ln 2
1
1 1 dy dx = xy (ln 2)2
ln 2
x(1 − x2 ) dx =
0 2 ln 2
1 2π
1 ln 2 dx = 1 x
ln 2
(parallelogram)
1
x+1
Average value =
x+y
e 0
55.
y
(quarter circle of radius 1)
4 Average value = π 1 (ln 2)2
1 3 y4 x e 3
1 (e − 1) 2
1 x y dy dx = 8
0
0
2
52. area of Ω = 14 π
54. area of
0
0
1 = 8
53. favg =
1
dx dy =
0
1
x e dx dy =
x
1
2 y4
x−1
d
0
f (x)g(y) dxdy =
1 e2x+1 − e2x−1 dx = (e3 − 2e + e−1 ). 2
b
d
=
a d
f (x)g(y) dx c
g(y)
c
b
f (x) dx
b
f (x)g(y) dx dy = c
R
1
dy dx =
dy
a
a
b
dy =
f (x) dx
d
g(y) dy
a
c
56. Symmetry about the origin [ we want Ω to contain (−x, −y) whenever it contains (x, y)]. Ω = { (x, y) : 0 ≤ x ≤ y,
57. Note that Set
0 ≤ y ≤ 1}.
Ω = { (x, y) : 0 ≤ y ≤ x, 0 ≤ x ≤ 1}. 1 y f (x)f (y) dxdy = f (x)f (y) dx dy 0
Ω
0 1
x
=
f (y)f (x) dy dx 0
x and y are dummy variables
0
1
=
x
f (x)f (y) dy dx = 0
0
f (x)f (y) dxdy. Ω
Note that Ω and Ω don’t overlap and their union is the unit square R = { (x, y) : 0 ≤ x ≤ 1,
0 ≤ y ≤ 1}.
875
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SECTION 17.3 If 0
1
f (x) dx = 0, then 1 0= f (x) dx 0
1
f (y) dy
=
f (x)f (y) dxdy
0
R
by Exercise 55
=
f (x)f (y) dxdy +
=2
f (x)f (y) dxdy
and therefore
f (x)f (y) dxdy Ω
Ω
Ω
f (x)f (y) dxdy = 0. Ω
x
b
58. 0
a
∂f (t, y) dy dt = ∂x
b
a
x
0
b
=
∂f (t, y) dt dy ∂x
[f (x, y) − f (0, y)] dy
a
b
=
f (0, y) dy.
a
Thus
b
x
b
a
and d dx
a
b
0
d f (x, y) dy = dx =
d dx
a
x
0
b
f (0, y) dy a
b d ∂f f (0, y) dy (t, y) dy dt + ∂x dx a
b
a
x
a
∂f (t, y) dy dt + ∂x
f (x, y) dy =
b
f (x, y) dy −
H(t) dt + 0 = H(x) =
0
b
∂f (x, y) dy. ∂x
a
59. Let M be the maximum value of | f (x, y) | on Ω.
φ2 (x+h)
φ1 (x+h)
φ1 (x)
= φ1 (x+h)
| F (x + h) − F (x) | =
φ2 (x+h)
+ φ1 (x)
φ2 (x+h)
φ2 (x)
+
φ2 (x)
f (x, y) dy −
φ1 (x+h)
=
φ1 (x)
φ1 (x)
f (x, y) dy
φ2 (x+h)
f (x, y) dy +
φ1 (x+h)
≤
φ2 (x)
φ1 (x)
φ1 (x+h)
φ2 (x)
f (x, y) dy +
f (x, y) dy
φ2 (x+h)
φ2 (x)
f (x, y) dy
≤ φ1 (x) − φ1 (x + h) M + | φ2 (x + h) − φ2 (x) M. The expression on the right tends to 0 as h tends to 0 since φ1 and φ2 are continuous.
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SECTION 17.4 60. (a)
3
2
1 3
2
(b)
xy dy dx ∼ = 4.5720 x2 + y 2
2x + y dx dy ∼ = 2.8133
√ 1+ y−1
y 2 −2y+2
1
3−y
4
1
1 dy dx =
(b)
√ 1+ x−1
62. (a) 0
7
3
x2 −2x+2
1 1
dy dx ∼ = −25.9893
2
61. (a)
−xy
xe
−1
5
2
x/2
(b)
2y
0
877
1 dx dy =
1 3
2x + y dy dx +
0
2
3
3−x
2x + y dy dx
0
∼ = 2.8133 SECTION 17.4 π/2 sin θ 1. r cos θ dr dθ = 0
0 π/4
0
cos 2θ
2.
π/4
0 π/2
0
3 sin θ
3.
2
0 2π/3
−π/3
π/2
r dr dθ = 0
π/2 1 1 1 = sin2 θ cos θ dθ = sin3 θ 2 6 6 0
cos2 2θ π dθ = 2 16
r dr dθ = 0
4.
π/2
2
9 sin θ dθ = 9 0
π/2 1 3 (1 − cos θ) sin θ dθ = 9 − cos θ + cos θ =6 3 0
π/2
3
0
2 cos θ
r sin θ dr dθ = 0
2π/3
2 cos2 θ sin θ dθ =
−π/3
5. (a) Γ : 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 1 cos r2 r drdθ =
2π
0
Γ
Γ
cos r2 r dr dθ = 2π
0
(b) Γ : 0 ≤ θ ≤ 2π, 1 ≤ r ≤ 2 2π cos r2 r drdθ = 0
1
1 6
2
Γ
cos r2 r dr dθ = 2π
1
Γ
r cos r2 dr = π sin 1
2
r cos r2 dr = π(sin 4 − sin 1)
1
0
1
r sin r dr = 2π(sin 1 − cos 1)
0
(b) Γ : 0 ≤ θ ≤ 2π, 1 ≤ r ≤ 2. 2π 2 (sin r)r dr dθ = (sin r)r dr dθ = 2π 0
1
0
6. (a) Γ : 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 1. 2π 1 (sin r)r dr dθ = (sin r)r dr dθ = 2π 0
1
2
r sin r dr = 2π[cos 1 − 2 cos 2 + sin 2 − sin 1]
1
7. (a) Γ : 0 ≤ θ ≤ π/2, 0 ≤ r ≤ 1 (r cos θ + r sin θ)r drdθ =
π/2
0
Γ
=
1
r2 (cos θ + sin θ) dr dθ
0
π/2
(cos θ + sin θ) dθ 0
1
2
r dr 0
2 1 = =2 3 3
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SECTION 17.4 (b) Γ : 0 ≤ θ ≤ π/2, 1 ≤ r ≤ 2 (r cos θ + r sin θ)r drdθ =
π/2
0
r2 (cos θ + sin θ) dr dθ
1
Γ
2
π/2
= 0
8. Γ : 0 ≤ θ ≤
0≤r≤
π 3,
+
y2
π/2
−π/2
√
1−x2
1/2 sec θ
12. 0
π/3
= 0
1
√
π/3
1
sin 0
π/2
2π
π/3
x2
+
1
0
y2
e−r r dr dθ = 2π 2
1
0
√
2x−x2
15.
π/2
π/3
sec3 θ dθ
0
π/2
1
r2 dr dθ =
1 1 − sec2 θ 2 8
dθ =
π 2
2
r2 dr =
0
4 π 3
√ 1 3 π− 6 8
r4 cos θ sin θ dr dθ
0
1
π/2
0
(sin 1 − cos 1) dθ =
0
π (sin 1 − cos 1) 2
1 2 re−r dr = π(1 − ) e
0
0
2
0
sin(r) r dr dθ =
2 cos θ
8 r cos θ r dr dθ = 3
x dy dx = 0
π/2
dy dx =
0
2
0
0
7 1 19 1 cos θ sin θ dθ = + = 5 480 40 480
0
14.
sec θ
sin θ dθ + 5(32) cos4 θ
13.
1 2
r4 cos θ sin θ dr dθ +
π/3
r dr dθ =
π/3
1−x2
0
0
π/2
10.
0
dθ 1 = 3 cos3 θ 3
0
1 π 3
r2 dr dθ =
0
π/3
π/3
r · r dr dθ =
0
dy dx =
1/ cos θ
π/3 √ 1 1 1√ 1 1 = = 3 + ln(2 + 3) sec θ tan θ + ln | sec θ + tan θ| 3 2 2 3 6 0
11. 1/2
π/3
0
0
1
1
7 14 = 3 3
=2
1
dx dy =
Γ
9.
r2 dr
1 cos θ
x2
2
(cos θ + sin θ) dθ
0
π/2
cos4 θ dθ =
0
∧
8 3 1 π π · · · = 3 4 2 2 2
(See Exercise 62, Section 8.3) 16. The region is the inside of the circle (x − 1/2)2 + y 2 = 1/4, which has polar equation r = cos θ. So the integral becomes
π/2
−π/2
π/3
17. A =
3 sin 3θ
r dr dθ = 0
0
9 2
0
π/3
cos θ
r3 dr dθ =
0
sin2 3θ dθ =
9 4
0
π/2
−π/2
π/3
1 3π cos4 θ dθ = 4 32
(1 − 6 cos θ) dθ =
3π 4
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SECTION 17.4
2π
2(1−cos θ)
18. A =
2π
2(1 − cos θ)2 dθ = 6π
r dr dθ = 0
0
0
19. First we find the points of intersection: r = 4 cos θ = 2
=⇒
π θ=± . 3
=⇒
A=
π/3
r dr dθ =
−π/3
1+2 cos θ
0
b
23.
3 cos θ
1+cos θ
π/4
−π/3
2π
π/3 1 3θ 2 2 =π 9 cos θ − (1 + cos θ) dθ = + sin 2θ − sin θ 2 2 −π/3
b 1 3 b r sin θ + r2 dθ 3 2 0 0 2π 1 1 = b3 sin θ + dθ = b3 π 3 2 0
1
24. V = 0 π/2
π/3
r2 sin θ + br dr dθ =
0
2π
1
(1 − r )r dr dθ = 2π 2
0
0
0
(r − r3 ) dr =
0
r 12 − 3r2 dr dθ = 8 2
2
25. 8
π/2
0
π/2
=8 0
2π
√
26. V = 0
2π
r 0
π/2
−π/2
π 2
3/2 2 1 dθ 12 − 3r2 − 18 0 4√ 16 √ 3 dθ = 3π 3 3 √
5 − r2 r dr dθ = 2π
5
r
6
5 − r2 dr =
0
30 1/6 (5) π 7
3/2 1 1 − 4 − r2 dθ 3 0 0 2π √ 8 √ 2 = − 3 dθ = (8 − 3 3 )π 3 3 0
4 − r2 dr dθ =
0
28. V =
6
0 1
27.
5
0
√ 4π +2 3 3
0
r dr dθ =
−π/3
−π/3
(2 + 4 cos 2θ) dθ =
cos 2θ dθ = 4
0
π/3
π/3
1 (1 + 2 cos θ)2 dθ = 3π 2
r dr dθ = 8
2π
√ 2 cos 2θ
π/4
0
0
(8 cos2 θ − 2) dθ =
0
21. A = 4
2π
π/3
r dr dθ = 0
22. A =
1 2
−π/3
2
2π
4 cos θ
20. A =
cos θ =
cos θ
2π
(1 − r2 )r dr dθ =
π/2
−π/2
cos2 θ cos4 θ − 2 4
dθ =
5π 32
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SECTION 17.5
29.
December 7, 2006
π/2
−π/2
2 cos θ
2r2 cos θ dr dθ =
π/2
2 3 r cos θ 3
−π/2
0
=
π/2
2 cos θ dθ 0
16 32 cos4 θ dθ = 3 3
−π/2
π/2
cos4 θ dθ =
0
∧
32 3
3 π 16
= 2π
Ex. 46, Sect. 8.3 30.
31.
π/2
−π/2
b a
2a cos θ
r2 dr dθ =
a sin θ
r 0
a2
0
2π
r
32. (a) V = 2 0
(b)
−
a sin θ 1 2 2 3/2 − a −r dθ 3 0 0 π 1 1 = a2 b 1 − cos3 θ dθ = πa2 b 3 3 0
b dr dθ = a
π
R2
−
s2
s ds dθ = 4π
0
r
S
R2 − S 2 dS =
0
4π 3 R − (R2 − r2 )3/2 . 3
4 3 4 πR − V = π(R2 − r2 )3/2 3 3
π/4
2
ex
0
1 2
0
+y 2
2π
dxdy = 0
Ω
4
π/4
r dr dθ = 2 0
2 cos 2θ
33. A = 2
34.
r2
32 3 8a3 cos3 θdθ = a 3 9
−π/2
0
π
π/2
π/4
(2 cos 2θ)2 dθ = 2
(1 + cos 4θ) dθ = 0
2 rer dr dθ = π e16 − e4
2
SECTION 17.5 1 1 2 1. M = x2 dy dx = 3 −1 0 xM M =
1
−1
yM M =
1
√
x3 dy dx = 0
=⇒
1
1
−1
1 1 2 x dx = 2 3
2
x y dy dx = 0 x
2. M =
1
(x + y) dy dx = 0
0
xM = 0
0
−1
1
1
√
x
xM M =
1
5/2
x(x + y) dy dx = 0
1
x
0
yM M =
√
0
x
y(x + y) dy dx = 0
0
0
1
1/3 1 = 2/3 2
x 13 dx = 2 20
x3/2 +
0
1
yM =
=⇒
x2 + 2
x2 x3/2 + 2 3
dx =
190 19 =⇒ xM = 42 273
dx =
3 6 =⇒ yM = 10 13
π 2
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SECTION 17.5
1
1
π
sin x
1 1 1 3. M = xy dy dx = (x − x5 ) dx = 2 6 0 x2 0 1 1 1 4 1 2 2/21 xM M = =⇒ xM = = x2 y dy dx = (x2 − x6 ) dx = 2 21 1/6 7 2 0 x 0 1 1 1 1 1 1/8 3 yM M = xy 2 dy dx = (x − x7 ) dx = =⇒ yM = = 3 8 1/6 4 2 0 x 0
π
sin2 x π dx = 2 4 0 0 0 π sin x π 2 sin x π π2 xM M = xy dy dx = x dx = =⇒ xM = 2 8 2 0 0 0 π sin x π 3 sin x 16 4 yM M = y 2 dy dx = dx = =⇒ yM = 3 9 9π 0 0 0
4. M =
y dy dx =
8
x1/3
5. M = 0
0
y 2 dy dx =
1 3
8
x dx = 0
32 3
1 8 2 512 xM M = xy dy dx = x dx = 3 9 0 0 0 8 x1/3 1 8 4/3 96 yM M = y 3 dy dx = x dx = 4 0 7 0 0
a
8
√
x1/3
2
a2 −x2
=⇒ =⇒
xM = yM =
16 512/9 = 32/3 3
96/7 9 = 32/3 7
a
x 2 a4 (a − x2 ) dx = 8 0 0 0 2 a √a2 −x2 a 2 x 2 a5 8 xM M = x2 y dy dx = (a − x2 ) dx = =⇒ xM = a 2 15 15 0 0 0 a √a2 −x2 a x 2 a5 8 2 yM M = xy dy dx = (a − x2 )3/2 dx = =⇒ yM = a 3 15 15 0 0 0
6. M =
xy dy dx =
1
3x
7. M =
xy dy dx = 0
2x
1
3x
xM M =
0 1
2x 3x
yM M = 0
2x
5 2
1
x3 dx =
0
5 x y dy dx = 2
1
2
xy 2 dy dx =
x4 dx =
0
19 3
1
0
5 8 1 =⇒ 2
x4 dx =
19 15
xM = =⇒
1/2 4 = 5/8 5
yM =
152 19/15 = 5/8 75
1 3 3 x(3 − x) + (3 − x)2 dx = 5 2 2 2 0 0 0 2 3− 3 x 2 2 7 x 3 2 3 7 2 xM M = x (3 − x) + (3 − x) dx = =⇒ xM = x(x + y) dy dx = 2 2 2 2 10 0 0 0 2 3− 3 x 2 2 (3 − 32 x)3 6 3 x yM M = y(x + y) dy dx = (3 − x)2 + dx = 6 =⇒ yM = 2 2 3 5 0 0 0
8. M =
2
3− 32 x
(x + y) dy dx =
2
881
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December 7, 2006
SECTION 17.5
2π
1+cos θ
9. M = 0
0
2π
1 r dr dθ = 3
1+cos θ
xM M = 0
2π
2
0
(1 + 3 cos θ + 3 cos2 θ + cos3 θ) dθ =
0
1 r cos θ dr dθ = 4
2π
3
1 = 4
5π 3
(1 + cos θ)4 cos θ dθ
0 2π
cos θ + 4 cos2 θ + 6 cos3 θ + 4 cos4 θ + cos5 θ dθ
0
7π = 4 7π/4 21 = . 5π/3 20
Therefore, xM =
2π
1+cos θ
yM M = 0
0
1 r sin θ dr dθ = 4
2π
3
0
2π 1 1 5 (1 + cos θ) sin θ dθ = =0 (1 + cos θ) 4 5 0 4
Therefore, yM = 0.
10. M =
5π/6
2 sin θ
y dx dy =
5π/6
r sin θ r dr dθ = π/6
Ω
1
π/6
8 1 sin4 θ − sin θ 3 3
√ 3 3 + 8π dθ = 12
xM = 0
by symmetry 5π/6 2 sin θ yM M = r2 sin2 θ r dr dθ = π/6
1
Therefore, yM
5π/6
1 4 sin θ − sin2 θ 4
6
π/6
√ 11 3 + 12π dθ = 16
√ 33 3 + 36π = √ 12 3 + 32π
11. Ω : −L/2 ≤ x ≤ L/2, −W/2 ≤ y ≤ W/2 M 2 4M W/2 L/2 2 1 Ix = y dx dy = y dxdy = MW 2 LW LW 0 12 0 Ω
∧ symmetry
M 2 1 x dxdy = ML2 , LW 12
Iy = Ω
Iz =
M 2 1 x + y 2 dxdy = M L2 + W 2 LW 12
Ω
√
√ L 3 Ky = Iy /M = 6
W 3 , 6 √ √ 3 L2 + W 2 Kz = Iz /M = 6
Kx =
Ix /M =
L 12. λ(x, y) = k x + 2 Ix =
−W/2
=
W/2
W3 12
M=
L/2
L k x+ 2 −L/2
M W
=
W/2
−W/2
L/2
L k x+ 2 −L/2
y 2 dx dy =
1 MW2 12
W/2
−W/2
dx dy =
y 2 dy
L/2
kW L2 2
L k x+ 2 −L/2
dx
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SECTION 17.5 Iy =
L/2
L/2
L k x+ 2 −L/2
−W/2
Iz =
W/2
W/2
L k x+ 2 −L/2
−W/2
L k x+ 2
13. M =
x2 dx dy =
kL4 W 1 = M L2 24 12
(x2 + y 2 ) dx dy = Ix + Iy =
1 1 1 kL dxdy = kL( area of Ω) = kL2 W 2 2 2
dxdy =
Ω
1 M (L2 + W 2 ). 12
Ω
∧ symmetry L 1 kx2 + Lx dxdy x k x+ dxdy = 2 2
xM M = Ω
kx2 dxdy = 4k
=
0
Ω
Ω
W/2
L/2
x2 dx dy =
0
∧
∧
symmetry =
1
1 6
2 2 kL W
yM M =
L=
1 kWL3 12
symmetry 1 6
ML;
xM =
L y k x+ dxdy = 0; 2
1 6
L
yM = 0
Ω
∧ by symmetry
λ(x, y)[x2 + y 2 ] dx dy =
14. Iz = Ω
Ω
15. Ix = Ω
4M = πR2
4M 2 4M y dxdy = 2 πR πR2
Iy = 14 MR2 ,
π/2
Ω
π/2
0
2
sin θ dθ 0
λ(x, y)y 2 dx dy = Ix + Iy .
M Kz 2 = M Kx 2 + M Ky 2
Since Iz = Ix + Iy , we have
λ(x, y)x2 dx dy +
R
Kz 2 = Kx 2 + Ky 2 .
r3 sin2 θ dr dθ
0 R
3
r dr 0
therefore
1 4M π 1 4 = R = MR2 2 πR 4 4 4
Iz = 12 MR2
Kx = Ky = 12 R, 16. Iz = IM + d2 M.
√ Kz = R/ 2 Rotation doesn’t change d,
doesn’t change M , and doesn’t change IM .
883
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December 7, 2006
SECTION 17.5
17. IM , the moment of inertia about the vertical line through the center of mass, is M 2 x + y 2 dxdy πR2 Ω
where Ω is the disc of radius R centered at the origin. Therefore 2π R M 1 IM = r3 dr dθ = MR2 . πR2 0 2 0 1 2
MR2 + d2 M where d is the distance from the center of the disc to the origin. Solving √ this equation for d, we have d = I0 − 12 MR2 M. We need I0 =
b
f (x)
18. Ix = a
0
b
f (x)
Iy =
λ λy dy dx = 3
b
2
b
2
λx dy dx = λ a
[f (x)]3 dx
a
0
x2 f (x) dx.
a
19. Ω : 0 ≤ x ≤ a, 0 ≤ y ≤ b b√ 2 2 4M 2 4M a a a −x 2 1 Ix = y dy dx = M b2 y dxdy = πab πab 0 0 4 Ω
Iy =
4M 2 4M x dxdy = πab πab
a
0
Ω
b a
√
a2 −x2
x2 dy dx =
0
1 M a2 4
Iz = 14 M a2 + b2
1
√
x
20. Ix =
1
2
(x + y)y dy dx = 0
0 1
0 √
x
Iy =
1
2
0 1
0 1
21. Ix = x2
0
1
1
Iy = 1
1
Iz = 8
0
1
8
0
x3 + 2
(x − x9 ) dx =
0
1
√ 3
x
(x3 − x7 ) dx =
y 2 · y 2 dy dx =
x
y 2 · x2 dy dx = 0
dx =
5 28
dx =
25 ; 72
Iz = Ix + Iy .
1 16
13 80
8
x5/3 96 dx = 5 5
8
x3 1024 dx = ; 3 3
0 √ 3
1 10
0
0
Iy = 0
1 2
7/2
xy(x2 + y 2 ) dy dx = Ix + Iy =
22. Ix =
1 4
x2
0
x3 y dy dx =
x2
0
xy 3 dy dx =
x
(x + y)x dy dx = 0
x2 x5/2 + 3 4
Iz = Ix + Iy
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SECTION 17.5
2π
1+cos θ
23. Ix =
0
2π
0 1+cos θ
Iy =
0 2π
0
1+cos θ
(1 + cos θ)5 sin2 θ dθ =
0
1 5
r4 cos2 θ dr dθ =
2π
(1 + cos θ)5 cos2 θ dθ =
0
33π 40 93π 40
63π 20
r4 dr dθ = Ix + Iy =
0
x1 M1 + x2 M2 , M 1 + M2
24. xM =
2π
2
0
Iz =
1 r sin θ dr dθ = 5 4
y 1 M1 + y 2 M 2 M1 + M 2 A = π r22 − r12
yM =
25. Ω : r12 ≤ x2 + y 2 ≤ r22 ,
(a) Place the diameter on the x-axis. M 2 M 2π r2 2 2 1 Ix = r sin θ r dr dθ = M r22 + r12 y dxdy = A A 0 4 r1 Ω (b) 14 M r22 + r12 + M r12 = 14 M r22 + 5r12 (parallel axis theorem) (c) 14 M r22 + r12 + M r22 = 14 M 5r22 + r12 26. Set r1 = r2 = r in the proceeding problem. Then the required moments of inertia are 1 2 2Mr
(a)
(b)
3 2 2Mr .
27. Ω : r12 ≤ x2 + y 2 ≤ r22 , A = π r22 − r12 M 2 1 M 2π r2 3 2 I= r dr dθ = M (r22 + r12 ) x + y dxdy = A A 0 2 r1 Ω
28. Let l be the x-axis and let the plane of the plate be the xy-plane. Then 2 I − IM = λ(x, y)y dx dy − λ(x, y)(y − yM )2 dx dy Ω
Ω
2 λ(x, y)[2yM y − yM ] dx dy
= Ω
yλ(x, y) dx dy −
= 2yM
2 yM
λ(x, y) dx dy
Ω
= 29. M =
2 2yM M
Ω
−
2 yM M
k R − x2 + y 2 dxdy = k 0
Ω
π
= R
0
2 yM M
by symmetry yM M = y k R − x2 + y 2 dxdy = k
yM = R/π
= d M.
1 Rr − r2 dr dθ = kπR3 6
xM = 0
Ω
2
0
π
0
R
1 Rr2 − r3 sin θ dr dθ = kR4 6
885
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SECTION 17.5 k(R −
30. Ix =
x2
+
y 2 )y 2
π
0
Ω
π
R
Iy = k 0
R
dx dy = k
(R − r) r2 sin2 θ r dr dθ =
0
3M R2 kπR5 = . 40 20
kπR5 3M R2 = 40 20
(R − r)r2 cos2 θr dr dθ =
0
3M R2 . 10
Iz = Ix + Iy =
31. Place P at the origin. M= k x2 + y 2 dxdy Ω
π
2R sin θ
=k 0
r2 dr dθ =
0
32 3 kR 9
xM = 0 by symmetry yM M = y k x2 + y 2 dxdy = k 0
Ω
π
2R sin θ
r3 sin θ dr dθ =
0
64 4 kR 15
yM = 6R/5 Answer: the center of mass lies on the diameter through P at a distance 6R/5 from P . 32. Putting the right angle at the origin, we have λ(x, y) = k(x2 + y 2 ). b h− h x b 1 M= k(x2 + y 2 ) dy dx = kbh(b2 + h2 ) 12 0 0 b h− h x b kb2 h(3b2 + h2 ) b(3b2 + h2 ) xM M = kx(x2 + y 2 ) dy dx = =⇒ xM = 60 5(b2 + h2 ) 0 0
b
yM M = 0
h− h bx
ky(x2 + y 2 ) dy dx =
0
kbh2 (b2 + 3h2 ) h(b2 + 3h2 ) =⇒ yM = 60 5(b2 + h2 )
33. Suppose Ω, a basic region of area A, is broken up into n basic regions Ω1 , · · · , Ωn with areas A1 , · · · , An . Then
xA =
x dxdy = Ω
n i=1
⎛ ⎝
Ωi
⎞ x dxdy ⎠ =
n
xi Ai = x1 A1 + · · · + xn An .
i=1
The second formula can be derived in a similar manner. 2 1 4 34. (a) M = (x + y) dy dx = 3 0 x/2 2 1 7 7 xM M = x(x + y) dy dx = ; xM = 6 8 0 x/2 2 1 3 yM M = y(x + y) dy dx = 1; yM = 4 0 x/2 2 1 2 1 4 4 (b) Ix = y 2 (x + y) dy dx = x2 (x + y) dy dx = Iy = 5 3 0 x/2 0 x/2
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SECTION 17.6
887
SECTION 17.6 1. They are equal; they both give the volume of T . 2. (a) Lf (P ) =
q n m
xi−1 yj−1 zk−1 Δxi Δyj Δzk ,
Uf (P ) =
i=1 j=1 k=1
(b) xi−1 yj−1 zk−1 ≤
xi yj zk Δxi Δyj Δzk
i=1 j=1 k=1
xi + xi−1 2
xi−1 yj−1 zk−1 Δxi Δyj Δzk ≤
yj + yj−1 2
zk + zk−1 2
≤ xi yj zk
1 2 xi − xi−1 2 yj 2 − yj−1 2 zk 2 − zk−1 2 ≤ xi yj zk Δxi Δyj Δzk 8
1 2 xi − xi−1 2 yj 2 − yj−1 2 zk 2 − zk−1 2 ≤ Uf (P ). 8 i=1 j=1 m
Lf (P ) ≤
q n m
q
n
k=1
The middle term can be written ⎞ ⎛ n m q 1 2 1 1 xi − xi−1 2 ⎝ yj 2 − yj−1 2 ⎠ zk 2 − zk−1 2 = (1)(1)(1) = . 8 i=1 8 8 j=1 k=1
Therefore
I=
3.
1 . 8 dx dy dz = α (volume of Π) = α(a2 − a1 )(b2 − b1 )(c2 − c1 )
α dx dy dz = α Π
Π
4. Since the volume is 1, the average value is
xyz dx dy dz =
1 . 8
Ω
5. Let P1 = {x0 , · · · , xm },
P2 = {y0 , · · · , yn },
P3 = {z0 , · · · , zq } be partitions of [ 0, a ], [ 0, b ], [ 0, c ]
respectively and let P = P1 × P2 × P3 . Note that xi + xi−1 yj + yj−1 xi−1 yj−1 ≤ ≤ xi yj 2 2 and therefore xi−1 yj−1 Δxi Δyj Δzk ≤
1 4
xi 2 − x2i−1
2 yj 2 − yj−1 Δzk ≤ xi yj Δxi Δyj Δzk .
It follows that Lf (P ) ≤
q m n 1 2 2 xi − x2i−1 yj 2 − yj−1 Δzk ≤ Uf (P ). 4 i=1 j=1 k=1
The middle term can be written ⎞ ⎛ n m q 1 2 1 2 2 2 xi − xi−1 ⎝ yj − yj−1 ⎠ Δzk = a2 b2 c. 4 i=1 4 j=1 k=1
6. Ix = Ixy + Ixz ,
Iy = Ixy + Iyz ,
Iz = Ixz + Iyz
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SECTION 17.6
7. x1 = a,
y 1 = b,
z 1 = c;
x0 = A,
x1 V1 + xV = x0 V0
y 0 = B,
z0 = C
=⇒
a2 bc + (ABC − abc) x = A2 BC
=⇒
x=
A2 BC − a2 bc ABC − abc
similarly y= 8. Encase T in a box Π.
AB 2 C − ab2 c , ABC − abc
z=
ABC 2 − abc2 ABC − abc
A partition P of Π breaks up II into little boxes Πijk .
Since f is
nonnegative on Π, all the mijk are nonnegative. Therefore 0 ≤ Lf (P ) ≤ f (x, y, z) dx dy dz. T
9. M =
Kz dxdydz Π
Let P1 = {x0 , · · · , xm }, P = P1 × P2 × P3 .
P2 = {y0 , · · · , yn },
P3 = {z0 , · · · , zq } be partitions of [ 0, a ] and let
Note that zk−1 ≤
1 2
(zk + zk−1 ) ≤ zk
and therefore 2 Kzk−1 Δxi Δyj Δzk ≤ 12 K Δxi Δyj zk 2 − zk−1 ≤ Kzk Δxi Δyj Δzk . It follows that Lf (P ) ≤
q m n 1 2 Δxi Δyj zk 2 − zk−1 ≤ Uf (P ). K 2 i=1 j=1 k=1
The middle term can be written ⎞ ⎛ n m q 1 1 1 2 2 Δxi ⎝ Δyj ⎠ zk − zk−1 = K(a) (a) (a2 ) = Ka4 . K 2 2 2 i=1 j=1 k=1
M = 12 Ka4 where K is the constant of proportionality for the density function. 10. xM M =
Kzx dx dy dz =
Ka5 1 =⇒ xM = a 4 2
Kzy dx dy dz =
Ka5 1 =⇒ yM = a 4 2
Kz 2 dx dy dz =
Ka5 2 =⇒ zM = a. 3 3
Π
yM M = Π
zM M = Π
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SECTION 17.7 11.
Kz x2 + y 2 dxdydz
Iz = Π
Kx2 z dxdydz +
= !
Π
"#
Ky 2 z dxdydz . !
$
Π
"#
I1
$
I2
We will calculate I1 using the partitions we used in doing Exercise 9. Note that 2 xi + xi xi−1 + x2i−1 zk + zk−1 x2i−1 zk−1 ≤ ≤ xi 2 zk 3 2 and therefore 2 ≤ Kxi 2 zk 2 Δxi Δyj Δzk . Kx2i−1 zk−1 Δxi Δyj Δzk ≤ 16 K xi 3 − x3i−1 Δyj zk 2 − zk−1 It follows that Lf (P ) ≤
q m n 1 3 2 xi − x3i−1 Δyj zk 2 − zk−1 ≤ Uf (P ). K 6 i=1 j=1 k=1
The middle term can be written ⎞ ⎛ n m q 1 1 1 3 2 3 2 ⎠ ⎝ K xi − xi−1 Δyj zk − zk−1 = Ka3 (a)(a2 ) = Ka6 . 6 6 6 i=1 j=1 k=1 Similarly I2 = 16 Ka6 and therefore Iz = 13 Ka6 = 23 12 Ka4 a2 = 23 M a2 . ∧ by Exercise 9 2 12. (a) Lf (P ) ∼ 3y − 2x dxdy = 57 (c) = 56.4803 Uf (P ) ∼ = 57.5603 R
SECTION 17.7 a b c 1. dx dy dz = 0
0
1
0 x
a
0
y
1
0
0
0
1
2y
3.
bc dz = abc 0
x
y dz dy dx = 0
a
c dy dz =
0
2.
b
1
y 2 dy dx =
0
0
x
1
2y
(x + 2z) dz dx dy = 0
1
0
0
1
1
= 0
1
1+x
4.
1
xy
0
1−x
2
5. 0
= 0
1
3
−1 0 2 1 −1
2 3 x 3
2y
0
(z − xy) dz dy dx =
(4 − 2xy) dy dx = 0
0 2
2
1 1
2 2
0 1
−1
1 2 z − xyz 2
0
1−x
1
dy =
2x y dy dx =
0
x xz + z 2 0 dx dy =
1+x
4z dz dy dx =
x3 1 dx = . 3 12
dy dx 1
2
8 dy = 16 0
0
1
1
16 3 2 y − 3 3
2y
2x2 dx dy
dy =
2 3
2x2 11 (1 + x)3 − (1 − x)3 dx = 3 9
3
1 2y − xy 2 −1 dx =
889
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SECTION 17.7
2
6.
−1
0
1
√
0
π/2
√
1
0
2
−1
y−2
e2
e
y2
x+y dz dx dy = z
ln x
9.
y
2
0
y2
y2
= π/2
2
y(x − 1) dx dy =
y
1
1
π/2
10.
1 2 x y − xy 2
y2
0
0
1
0
2
dy = y
π/2
z
π/2
e cos x sin y dz dy dx = 0
(y − 2)2 − 1 3 + y(y − 3) dy = 2 2
x [yez ]ln dx dy 0
1
2
−1
1
y
1
y−2
(x + y) dx dy =
−1
2
z
2
ye dz dx dy = 1
0
1
2
2
[xy cos z]0 1−x dx dz
1 1 π/2 1 1 π/2 1 − (1 − x2 )3/2 cos z dz = x 1 − x2 cos z dx dz = cos z dz = 3 3 3 0 0 0 0
π/2
=
4z dz = 8 0
0
0
2
(2z − 4x) dx dz =
x cos z dy dx dz = 0
8.
1−x2
7.
1
−1
0
1
2
(z − xy) dy dx dz =
1
π/2
3
1 5 3 3 47 2 y − y + y dy = 2 2 24
(e − 1) cos x sin y dy dx
0 π/2
=
(e − 1) cos x dx = e − 1
0
11.
c2
b2
a2
f (x)g(y)h(z) dxdydz = c1
Π
b1
c2
f (x)g(y)h(z) dx dy dz
a1
b2
= c1
b1
c2
=
a1
h(z)
c1
a2
a2
1
x3 dx
0
2
z dz
0 1
13.
x2 dx
0
3
y 2 dy
0
kxyz dx dy dz = k
=
c
y dy
z dz
0
a
kx yz dxdydz = k
2
x dx 1 3
a3
1 2 2 2 ka b c 8
b
y dy
0
=k
=
0
Π
1 8
h(z) dz c1
b
2
By Exercise 14, M =
c2
g(y) dy
x dx
xM M =
b2
8 27 1 =8 3 3 3
0
Π
15.
a
8 9 1 =3 4 3 2
z 2 dz
0
14. M =
=
0 2
g(y) dy dz
b1
3
y 2 dy
b2
b1
f (x) dx a1
f (x) dx
a1
=
12.
f (x) dx dy dz
a2
g(y)h(z)
z dz
0
1
ka2 b2 c2 . Therefore x =
2 2 3
b2
1 2
c2 =
c
0 1 12
ka3 b2 c2 .
a. Similarly, y =
2 3
b and z =
2 3
c.
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SECTION 17.7
891
16. (a) I= kxyz (x − a)2 + (y − b)2 dx dy dz Π
a
x(x − a) dx
=k
c
y dy
0
=
b
2
z dz
0
x dx
a
+k
0
0
b
y(y − b) dy 2
0
1 2 2 2 2 ka b c (a + b2 ). 48 M = 18 ka2 b2 c2 ,
Since IM −
(b)
1 2 (a + b2 ) 18
I = 16 M (a2 + b2 ).
by parallel axis theorem
17.
18. V =
1
1
1−y
dx dy dz =
dz dy dx = 0
T
0
0
1 2
19. center of mass is the centroid 1 2
x=
by symmetry yV = y dxdydz =
1
1
1−y
0
0
1
= V =
1 2
1 2
0
1
0
1
0
0
0
1 1 − 2y + y 2 dy dx = 2
(by Exercise 18 );
20. Ix =
1
y − y 2 dy dx
0
0
T
0
1 1 1 2 1 3 1 1 = y − y dx = dx = 2 3 6 0 0 6 0 1 1 1−y zV = z dxdydz = z dz dy dx =
1
y dz dy dx =
0
T
y = 13 ,
z=
0
1
1
0
1
1 (1 − y)2 dy dx 2
1 1 11 1 1 y − y 2 y 3 dx = dx = 3 2 0 3 6 0
1 3
1 M 2 (y + z 2 ) dx dy dz = M V 3
T
Iy = T
M 2 1 (x + z 2 ) dx dy dz = M V 2
Iz = T
M 2 1 (x + y 2 ) dx dy dz = M V 2
c
z dz 0
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r
−φ(x)
−r
√
r2 − (x2 + y 2 ) ,
26.
−
−1 √
k
2
dz dy dx 1−x2 /2
2+y 2
27.
k 2−y 2
1 k z − 3x2 − y 2 dz dy dx 4
4−x2 −y 2 /4
4−z 2
x2 + y 2 + z 2 dz dy dx
4−x2 −y 2
√2−y2 −
k the constant of proportionality
dz dy dx
3x2 +y 2 /4
√
x2 + y 2 dx dz dy
z 2 +2y 2
2
0
2
3
= 0
2
1
1
3
y
1 z+y 3
1
2
1
x2 y 2 z 2 dx dy dz =
29. T
=
=
=
=
0
−1
1 3 1 3 1 9
y+1
0 0
−1
0
0
−1
1 3 2 2 x y z 3
y+1
−1
y+1
0
dy +
2
dz = 0
1
1
0
0
1
1
3
1
1 3 x z + xy 3
1 dy dz 0
2 28 z + 4 dz = 3 3 1
(4y 2 ey − 2yey + 2y − 2y 2 ) dy = 4e −
x2 y 2 z 2 dx dz dy +
1
1
0
1−y
0
1−y
0
0
0
1 3
0
0
y+1 0
0
1 3
3
x
dz dy +
y 2 z 2 dz dy +
1 2 3 y z 3
0
1
0
−1
0
0
2y(x + y)e dx dy =
0
2
x z + y dx dy dz =
y
2ye dz dx dy = 0
2
1 1 zy + y 2 3 2
0
x
0
dy dz =
x+y
28. 0
1
x z + y dx dy dz =
T
r2 − x2 ,
1−y 2
√ −2 2−2x2
√ − 2
with φ(x) =
x2 +y 2
−2x−3y−10
√ 2 2−2x2
1
1
√
x−x2
√
√ − 2
1−x2
√1−x2 /2
2
k r − x2 + y 2 + z 2 dz dy dx
−ψ(x,y)
√ − x−x2
0
25.
√
1
23.
24.
December 7, 2006
ψ(x,y)
√ − 1−x2
−1
√
1
φ(x)
ψ(x, y) =
22.
T1: PBU
SECTION 17.7
21.
QC: PBU/OVY
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1
x2 y 2 z 2 dx dz dy
0
1 dz dy 0
2 2 1 y z 0 dz dy
1 2 3 y z 3
1 y 5 + 3y 4 + 3y 3 + y 2 dy + 9
1 3 2 2 x y z 3
0
1−y
1−y
0
dy 0 1
1 y 2 − 3y 3 + 3y 4 − y 5 dy = 270
29 3
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SECTION 17.7
2
√
√4−x2 −y2
4−x2
30.
√
2
4−x2
xy dz dy dx = 0
0
xy 0
0
π/2
0
1 2
3
y 2 dx dy dz =
31.
0
T
3
0
2
r2 cos θ sin θ
2−2x/3
4 − r2 r dr dθ
0
2
r3
0
6−2x−3y
1 4 √ 32 4 − r2 dr = (4 u − u3/2 ) du = 4 0 15
3
y 2 dz dy dx =
0
= 0
2−2x/3
4 − x2 − y 2 dy dx
0
= =
0
0
2−2x/3
2 6−2x−3y dy dx y z 0
(6y 2 − 2xy 2 − 3y 3 ) dy dx
0
2−2x/3 2 3 2y 3 − xy 3 − y 4 dx 3 4 0 0 1 3 2 12 = 2 − x dx = 4 0 3 5
3
=
1
1−x2
√
1−y
1
2
32.
1−x2
y dz dy dx = 0
0
0
2
x+2
x
V = x2
0 2
0
x
2
x2
2
x2
2
x+2
2
x
x2
11 , 10
x=
0
x2 2
1
0
9 , 4
y=
1
z=
kz dx dy dz = 0
0
1
0
1
M=
2
−1
3
0
0
4−x2
1
2−x
2
1 3 x + 4x2 + 4x − x5 dx = 6 2
x+2
x2
0
1 2 x dy dx = 2
0
1 k 2
k(x2 + y 2 + z 2 ) dz dy dx = k
27 2 ;
(x, y, z) =
1
3 12 2, 2, 5
44 x3 + 2x2 − x4 dx = 15
2
0
dz dy dx = 0
11 20
1
M=
2
0
x+2
z dz dy dx = 0
xy dy dx = 0
zV =
8 x2 + 2x − x3 dx = 3
2
x2
0
0
x+2
x dy dx =
x
2
0
y dz dy dx = 0
35. V =
0
0
x+2
yV =
(b)
x+2
x dz dy dx = x2
1 − y dy dx
x dy dx =
0
x+2
xV =
34. (a)
2
dz dy dx =
2 4 2 16 1 − x3 + x5 − x7 + dx = 3 5 7 105 12
1
0
y2
0
0
=
33.
1 3 22 x + 2x2 − x4 dx = 2 15
893
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December 7, 2006
SECTION 17.7
(x − x) dx dy dz =
36. T
x dx dy dz − x
T
dx dy dz = xV − xV = 0. T
Similarly, the other two integrals are zero.
a
φ(x)
x 1 abc with φ(x) = b 1 − , 6 a
ψ(x,y)
37. V =
dz dy dx = 0
0
0
1
(x, y, z) =
1 4
a,
4
b,
1 4
c
M 2 1 (x + y 2 ) dx dy dz = V 30
38. Iz =
M V
=
x y ψ(x, y) = c 1 − − a b
1 M 5
T
39. Π : 0 ≤ x ≤ a, 0 ≤ y ≤ b, 0 ≤ z ≤ c a b c M 2 1 (a) Iz = x + y 2 dz dy dx = M a2 + b2 3 0 0 0 abc 1 (b) IM = Iz − d2 M = 13 M a2 + b2 − 14 a2 + b2 M = 12 M a2 + b2 ∧ parallel axis theorem (17.5.7) 1 1 (c) I = IM + d2 M = 12 M a2 + b2 + 14 a2 M = 13 M a2 + 12 M b2 ∧ parallel axis theorem (17.5.7)
2
2
40. V = 1
2
2
1
2
2
zV = 1
1
1
y
41. M =
2
0
z dz dy dx =
73 73 =⇒ z = . 12 72
2
0
(xM , yM , zM ) =
42. T is symmetric
1
109 109 =⇒ x = =y 12 72
k x +y +z 0
2
x dz dy dx =
1+x+y
−2
1
1+x+y
−2
1
(3 + x + y) dy dx = 6 1
xV =
2
dz dy dx =
−2
1
1+x+y
7 34 37 12 , 45 , 90
2
1 3 k x y + y + y dy dx dz dy dx = 3 0 0 1 1 1 2 1 = x + dx = k k 2 3 2 0
(a) about the yz-plane,
(d) about the origin.
by symmetry
1
1
2
3
(b) about the xz-plane,
(c) about the xy-plane,
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SECTION 17.7 43. (a)
0 by symmetry (b) (a1 x + a2 y + a3 z + a4 ) dxdydz = a4 dxdydz = a4 (volume of ball) = T
4 3
T
∧ by symmetry
2
2
4−y 2
x2 y 2 dz dy dx =
44. 0
2−y
0
a
√
a2 −x2
352 45
√a2 −x2 −y2
45. V = 8
a
√
a2 −x2
dz dy dx = 8 0
0
0
∧
π/2
a
a2 − r2 r dr dθ
=8 0
0
π/2
= −4 =
a
8 3
46. 8 0
47.
M=
2
√
4−x2 /2
√ − 4−x2 /2
−2
2
√
0
4x − x − 4xy 3
2π
V =2 0
49. M =
2
−1
3
0
2
4−x2
2
2−z
2
√
1
2
2−x
(xM , yM , zM ) =
dy dz dx 0
5
0
2
0
(c) V =
2−x
9
√
9−y
dz dx dy + 0
0
0
2−x
dz dx dy 5
3/2 128 k x 4 − x2 dx = 15
9−x2
(b) V =
0
2
(r − r3 ) dr dθ = π
0
0
4−y 2
0
dy dx dz 0
x2 +3y 2
9−x2
50. (a) V =
4−x2 /2
0
4 dy dx = k 3
135 k; k(1 + y) dz dy dx = 4
2−x
4 π a3 3
kx dz dy dx 0
dθ 0
k|x| dz dy dx = 4 x2 +3y 2
0
48. using polar coordinates
a
4 πabc. 3
4−y 2
4−x2 /2
= 4k
π/2
0
0
2 2 (a − r2 )3/2 3
dθ =
dz dy dx = 0
0
b√1−x2 /a2 c√1−x2 /a2 −y2 /b2
a2 − x2 − y 2 dy dx
0
0
polar coordinates
0
0
1 9 12 , , 2 5 5
πa4
895
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SECTION 17.8
6
3
6−x
51. (a) V =
dy dx dz 0
z/2
3
2x
x
6−x
(b) V =
dy dz dx 0
(c)
0
6
x
3
y
V =
6
(12−z)/2
6−y
dx dy dz + 0
z/2
Ωxy
(c) V =
dx dy dz 0
3
z/2
2 4 − y dx dy
52. (a) V =
z/2
4
−4
4
√
V =
dz dy dx
(d)
Ωxy
4
(c) V =
√
(b) V =
4−y
√ − 4−y
0
dx dz dy
−y
(d) V =
(4 − z − |x|) dx dz
54. (a) V =
−2
(d) V =
4
0
5
√
3
4+x
√ − 4+x 2
4−z
1
4−z 2
2
dy dz dx +
|x|
0
√
4−x
2
√ − 4−x
56. Let f (x, z) = 36 − 9x2 − 4z 2 and g(x, z) = 1 −
4−y
√ − 4−y
dx
dydz
y
−y
dx dy dz
4−z 2
|x|
dz dx dy
y
4−z 2
dx dy
dy
dx dz
dy dx dz
|x|
dz
2
3−(3/2)x
4−z 2
dy dz dx
|x|
ln xy dz dy dx ∼ = 6.80703 z
55. (a) 2
z 2 −4
−2
4−z
2
√
0
Ωxy
2
−2
(b) V =
Ωxy
(c) V =
2
2
−y
Ωxy y
y
−y
0
2y dydz
4
V =
53. (a) V =
4−y
√ − 4−y
Ωxy 4−y
√ − 4−y
|x|
(b)
√
4
2
(b) 0 1 2
x−
1 3
1
0
3
√ 16 3 √ √ 4 2−2 x yz dz dy dx = 3
z. Then
f (x,z)
V =
1 dy dz dx = 71 0
0
g(x,z)
SECTION 17.8 1. r2 + z 2 = 9
2. r = 2
3. z = 2r
4. r cos θ = 4z
5. 4r2 = z 2
6. r2 sin2 θ + z 2 = 8
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SECTION 17.8
π/2
2
4−r 2
7. 0
0
π/2
0
2
= 0
2
1
0
r dz dr dθ
4
4r − r2 dr dθ
0 π/2
=
2
4 dθ = 2π 0
0 1
0 2
π/4
1
√
1−r 2
8.
r dz dr dθ = 0
0
0
2π
2
π 12
r2
9.
r dz dr dθ 0
0
0
2π
2
= 0
4
r3 dr dθ
0 2π
=
-2 -
4 dθ = 8π 0
2 0 0
2
3
2π
0 2
3
10.
r dz dθ dr = 9π 0
0
r
1
√
1−x2
√4−(x2 +y2 )
11.
π/2
1
√
4−r 2
dz dy dx = 0
0
0
π/2
1
=
r 0
= 0
π
1
12. 0
0
r
1
z 3 r dz dr dθ = 0
r dz dr dθ 0
π
0
1
0
0
4 − r2 dr dθ
0 π/2
8 √ − 3 3
1 π (1 − r4 ) r dr dθ = 4 12
dθ =
√ 1 8−3 3 π 6
897
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December 7, 2006
SECTION 17.8
3
√9−y2 √9−x2 −y2
13. 0
0
x2
0
1 +
π/2
3
√
9−r 2
dz dx dy =
y2
0
0 π/2
0 3
= 0
1 · r dz dr dθ r
9 − r2 dr dθ
0
3 π/2 r 9 −1 r 2 = 9 − r + sin dθ 2 2 3 0 0
9π = 4
π/2
1
√
1−r 2
14.
π/2
1
0 1
√
0
0
1−x2
0
0
0
2
π/2
sin(x2 + y 2 ) dz dy dx =
15. 0
0
1
0
2π
1
2−r 2
0
0
r2 dz dr dθ = 4π
r2
1
dθ = 0
0
π/2
9 2 π 8
0
sin(r2 )r dz dr dθ
2r sin(r2 ) dr dθ = 12 π(1 − cos 1) ∼ = 0.7221
(1 − r2 )r2 dr =
0
2
0 1
=
16.
π/2
r π (1 − r2 ) dr dθ = 2 16
zr dz dr dθ = 0
8π 15
17. (0, 1, 2) → (1, 12 π, 2)
18. (0, 1, −2) → (1, 12 π, −2)
19. (0, −1, 2) → (1, 32 π, 2)
20. (0, 0, 0) → (0, arbitrary, 0)
21. V =
π/2
−π/2
22. V =
π/2
−π/2
23. V =
π/2
π/2
−π/2
0
r
π/2
−π/2
0
2a cos θ
r2 dr dθ
0
8 3 32 3 a cos3 θ dθ = a 3 9
2a cos θ
r 2 /a
r dz dr dθ = 0
0
a cos θ
0
0
3
a
0
1 1 cos2 θ − cos3 θ 2 3
2 cos θ
0
3 3 πa 2
a−r
r dz dr dθ =
π/2
−π/2
2a cos θ
r dz dr dθ =
−π/2
24. V =
−π/2
=
=
π/2
2+ 12 r cos θ
π/2
−π/2
dθ =
r dz dr dθ =
a cos θ
r(a − r) dr dθ
0
1 3 a (9π − 16) 36
5 π 2
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SECTION 17.8 25. V =
π/2
3
√
25−r 2
26. V =
0
2π
r+1 √
1/2
27. V = 0
a
√
√
0
2π
3
1
2π
2r
9−r 2
√ √ 1 r 1 − r2 − r2 3 dr dθ = π 2 − 3 3
2
0
2π
0
1 2
1
√
1/2
3
r
0
30. V = 0
r2 cos θ − r3 dr dθ
2π 3 √ a ( 2 − 1) 3
a2 +r 2
r dz dr dθ =
√
2π
r dz dr dθ =
0
0
29. V = 0
cos θ
−π/2
r dz dr dθ =
r 3
28. V = 0
1−r 2
√
0
2π
41 π 3
r dz dr dθ = 0
π/2
1 1 cos4 θ dθ = π 12 32
−π/2
2π
r2
0
π/2
r cos θ
r dz dr dθ =
−π/2
=
cos θ
899
36−r 2
r dz dr dθ =
0
9 − r2 dr dθ =
1
32 3 π
√
2
√ √ 1 π(35 35 − 128 2). 3
31. Set the lower base of the cylinder on the xy-plane so that the axis of the cylinder coincides with the z-axis. Assume that the density varies directly as the distance from the lower base.
2π
R
M=
h
kzr dz dr dθ = 0
0
0
1 kπR2 h2 2
32. xM = yM = 0 by symmetry 2π R h 1 zM M = kz 2 r dz dr dθ = kπR2 h3 3 0 0 0 M=
1 kπR2 h2 , 2
zM =
2 h 3
The center of mass lies on the axis of the cylinder at a distance density.
2π
R
h
33. I = Iz = k =
1 4
0
0
4 2
1 2
kπR h =
1 2
zr3 dr dθ dz
0
kπR2 h2 R2 =
1 2
∧ from Exercise 31
MR2
2 3h
from the base of zero mass
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SECTION 17.9
34. (a)
M I= πR2 h
(b)
0
M πR2 h
I=
2π
R
0 2π
0
h
r3 dz dr dθ =
0 R
0
h
1 M R2 2
0
I = 14 M R2 + 13 M h2 − M ( 12 h)2 = 14 M R2 +
(c)
1 1 M R 2 + M h2 4 3
(r2 sin2 θ + z 2 )r dz dr dθ = 1 2 12 M h
35. Inverting the cone and placing the vertex at the origin, we have
h
2π
(R/h)z
V =
r dr dθ dz = 0
36. xM = yM = 0
2π
(R/h)z
zM M = 0
0
0
M V
zr dr dθ dz =
On the axis of the cone at a distance
M 37. I = V
h
0
M 38. I = V
0
2π
2π
0
h
1
(R/h)z
r3 dr dθ dz =
0
2π
0
(R/h)z
1−r 2
0
2π
0
1
1−r 2
41. M = 0
0
0
M V
πR2 h2 3 πR2 h2 =⇒ zM = = h 4 4V 4
from the vertex.
3 MR2 10
z 2 r dr dθ dz =
r dz dr dθ =
3 4h
0
39. V = 0
0
by symmetry
h
0
1 2 πR h. 3
3 M h2 . 5
1 π 2
40.
2π
1
1−r 2
kzr dz dr dθ =
M= 0
0
0
1 k r2 + z 2 r dz dr dθ = kπ 4
SECTION 17.9 1. 3. 5.
√
3,
( 34 ,
3 4
1 4
√
π, cos−1 3,
3 2
√
1 √ 3 3
3)
√ √ 4 6 2 2 2 ρ = 2 + 2 + (2 6/3) = 3 √ 2 6/3 π √ φ = cos−1 = cos−1 (1/2) = 3 4 6/3 π θ = tan−1 (1) = 4 √ 4 6 π π (ρ, θ, φ) = , , 3 4 3
2. 4.
1√ 2
1 2
√ (2 10,
6.
6,
√
√ 2
2,
2 3 π,
√ 3 cos−1 [ 10 10])
π 5π 8, − , 4 6
1 πk 6
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SECTION 17.9
901
8. (a) (5, 12 π, arccos 45 )
7. x = ρ sin φ cos θ = 3 sin 0 cos(π/2) = 0
(b) (5, 32 π, arccos 45 )
z = ρ cos φ = 3 cos 0 = 3 y = ρ sin φ sin θ = 3 sin 0 sin(π/2) = 0 (x, y, z) = (0, 0, 3) 9. The circular cylinder x2 + y 2 = 1;
the radius of the cylinder is 1 and the axis is the z-axis.
10. The xy-plane. 11. The lower nappe of the circular cone z 2 = x2 + y 2 . 12. Vertical plane which bisects the first and third quadrants of the xy-plane. 13. Horizontal plane one unit above the xy-plane. 14. sphere
x2 + y 2 + (z − 12 )2 =
1 4
of radius
1 2
and center (0, 0, 12 )
15. Sphere of radius 2 centered at the origin: 2π π 2 8 2π π 16 2π 32π ρ2 sin φ dρ dφ dθ = sin φ dφ dθ = dθ = 3 0 3 0 3 0 0 0 0 16. That part of the sphere of radius 1 that lies in the first quadrant between the x, z-plane and the plane y=x
π/4
0
π/2
0
1
π 12
ρ2 sin φ dρ dφ dθ =
0
17. The first quadrant portion of the sphere that lies between the x, y-plane and the plane z =
π/2
π/2
3
π/2
2
π/2
ρ sin φ dρ dθ dφ = 9 π/6
0
sin φ dθ dφ
0
π/6
π/4
2π
π
=
9 2
√ π/2 π [− cos φ]π/6 = 94 π 3
sec φ
1
19. 0
0
√
1−x2
√2−x2 −y2 √
0 π/4
0
π/2
√
2
=
2 3
√
2 3
0
π/4
π/2
2
ρ2 sin φ dρ dθ dφ
sin φ dθ dφ
√
=
0
ρ2 sin φ dρ dθ dφ =
0
dz dy dx = x2 +y 2
sin φ dφ π/6
18. A cone of radius 1 and height 1; 0
π/2
9 2
=
0
0
π
0
π/4
sin φ dφ = 0
√
√ 2 π (2 − 2) 6
1 π 3
3 2
√
3.
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SECTION 17.9
π/4
π/2
2
0
21.
0 3
16π 5
ρ4 sin φ dρ dθ dφ =
20. 0
π/4
sin φ dφ = 0
√ 8π (2 − 2) 5
√9−y2 √9−x2 −y2 z x2 + y 2 + x2 dz dx dy 0
0
0
π/2
π/2
3
= 0
0 π/2
ρ cos φ · ρ · ρ2 sin φ dρ dθ dφ
0
1 sin 2φ dφ 2
= 0
π/2
3 π/2 π 1 1 5 ρ dρ = − cos 2φ ρ 4 2 5 0 0
3
4
dθ 0
0
243π = 20
π/2
π/2
1
1 2 π ρ sin φ dρ dθ dφ = 2 ρ 2
22. 0
0
0
2π
π
R
23. V = 0
0
0
24. r = ρ sin φ,
α
θ = θ,
π
R
25. V = 0
0
2π
ρ2 sin φ dρ dφ dθ =
0
0
π
0 2π
z = ρ cos φ
ρ2 sin φ dρ dφ dθ = R
26. M =
2π
4
=
tan−1 (r/h)
0
h sec φ
tan−1 (r/h)
=
2π
0
kh4 tan φ sec3 φ dφ dθ 4
1 3 tan−1 (r/h) kh4 sec φ 0 dθ = 3 4
2π
0
⎡ √
1⎣ 3
r2
+ h
h2
3
⎤ − 1⎦ dθ
3/2 1 − h3 kπh r2 + h2 6
2π
tan−1 r/h
h sec φ
28. V = 0
=
kρ3 sin φ dρ dφ dθ
0
0
kh 4
1 kπR4 3
0
= 0
2 αR3 3
k(R − ρ)ρ2 sin φ dρ dφ dθ =
27. M = 0
4 πR3 3
0
ρ2 sin φ dρ dφ dθ =
0
r 1 π 3 h tan2 (tan−1 = πr2 h 3 h 3
29. center ball at origin; density = (a) I =
3M 4πR3
(b) I =
2 5
2π
0 2
0 2
π
R
7 5
tan−1 (r/h)
0
M 3M = V 4πR3
ρ4 sin3 φ dρ dφ dθ =
0
MR + R M =
2π 3
MR2
2 MR2 5
(parallel axis theorem)
h3 tan φ sec2 φ dφ
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SECTION 17.9 30. The center of mass is the centroid;
2π
π/2
R
zV = 0
0
3 z = R; 8
(a) I =
(x, y, z) =
31. center balls at origin; 3M 4π R2 3 − R1 3
density
2π
0
2 3 πR 3
V =
(ρ cos φ)ρ2 sin φ dρ dφ dθ =
0
3 0, 0, R 8
=
π
1 4 πR 4
M 3M = V 4π R2 3 − R1 3
R2
ρ4 sin3 φ dρ dφ dθ =
R1
0
903
5 2 R2 − R1 5 M 5 R2 3 − R1 3
This result can be derived from Exercise 29 without further integration. View the solid as a ball of mass M2 from which is cut out a core of mass M1 . MR2 3 M 3M 4 3 3 M2 = = ; V2 = πR 2 V 3 4π R2 − R1 3 R2 3 − R1 3 Then I = I2 − I1 =
2 5
M2 R 2 2 −
2 5
similarly
M1 =
MR1 3 . R2 3 − R1 3
2 MR2 3 MR1 3 2 − R R1 2 2 5 R2 3 − R1 3 R2 3 − R1 3 5 R2 − R1 5 2 . = M 5 R2 3 − R1 3
M1 R 1 2 =
2 5
(b) Outer radius R and inner radius R1 gives
5 2 R − R1 5 moment of inertia = M . 5 R 3 − R1 3
[part (a) ]
As R1 → R, R 5 − R1 5 R4 + R3 R1 + R2 R1 2 + RR1 3 + R1 4 5R4 5 = −→ = R2 . 3 2 3R2 3 R 3 − R1 R2 + RR1 + R1 Thus the moment of inertia of spherical shell of radius R is 2 2 5 2 M R = MR2 . 5 3 3 (c) I = 23 MR2 + R2 M =
5 3
MR2
(parallel axis theorem)
32. (a) The center of mass is the centroid;
using the result of Exercise 30,
x=y=0
3 4 3 4 R2 πR2 3 − R1 πR1 3 z 2 V2 − z 1 V1 3(R2 2 + R1 2 )(R2 + R1 ) 8 3 8 3 z= = = 4 V 8(R2 2 + R2 R1 + R1 2 ) π(R2 3 − R1 3 ) 3 (b) Setting
2π
R1 = R 2 = R
α
33. V = 0
0
0
a
in (a), we get
ρ2 sin φ dρ dφ dθ =
x = y = 0,
2 π (1 − cos α) a3 3
z = 12 R
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SECTION 17.9
2π
π/4
1
34. 0
0
3
eρ ρ2 sin φ dρ dφ dθ =
0
35. (a) Substituting
x = ρ sin φ cos θ,
z = ρ cos φ
ρ2 sin2 φ + (ρ cos φ − R)2 = R2 ,
we have
which simplifies to (b) 0 ≤ θ ≤ 2 π,
2π
ρ = 2R cos φ.
0 ≤ φ ≤ π/4,
π/2
2R cos φ
0
0 2π
0 π/2
2R cos φ
(b) M = 0
0
2π
2R cos φ
0 2π
π/4
π
38. V = 0
2
2
2π
π/2
ρ sin φ dρ dφ dθ + 0
2π
1 2 4 kπ R 8
kρ3 cos2 θ sin2 φ dρ dφ dθ =
V =
=
1 2 4 kπ R 4
0
37.
8 kπR4 5
kρ3 sin2 φ dρ dφ dθ =
0
π/2
(c) M = 0
R sec φ ≤ ρ ≤ 2R cos φ
kρ3 sin φ dρ dφ dθ =
36. (a) M =
0
y = ρ sin φ sin θ,
x2 + y 2 + (z − R)2 = R2
into
√ 1 π(e − 1) 2 − 2 3
0
0
0
π/4
√ 2 2 cos φ
ρ2 sin φ dρ dφ dθ
0
√ 1 16 − 6 2 π 3
1−cos φ
ρ2 sin φ dρ dφ dθ =
0
8 π 3
39. Encase T in a spherical wedge W . W has spherical coordinates in a box Π that contains S. f to be zero outside of T .
Then F (ρ, θ, φ) = f (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ)
is zero outside of S and
f (x, y, z) dxdydz =
T
f (x, y, z) dxdydz W
F (ρ, θ, φ) ρ2 sin φ dρdθdφ
= Π
F (ρ, θ, φ) ρ2 sin φ dρdθdφ.
= S
Define
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SECTION 17.9 40. Break up T into little basic solids T1 , . . . , TN . all the mass as concentrated there.
905
Choose a point (x∗i , yi∗ , zi∗ ) from each Ti and view
Now Ti attracts m with a force
Gmλ(x∗i , yi∗ , zi∗ )(Volume of Ti ) Fi ∼ ri =− ri 3 where ri is the vector from (x∗i , yi∗ , zi∗ ) to (a, b, c).
We therefore have
Gmλ(x∗i , yi∗ , zi∗ )[(x∗i − a)i + (yi∗ − b)j + (zi∗ − c)k] Fi ∼ = [(x∗i − a)2 + (yi∗ − b)2 + (zi∗ − c)2 ]3/2
(Volume of Ti ).
The sum of these approximations is a Riemann sum for the triple integral given and tends to that triple integral as the maximum diameter of the Ti tends to zero. 41. T is the set of all (x, y, z) with spherical coordinates (ρ, θ, φ) in the set 0 ≤ θ ≤ 2π,
S: T has volume V =
2 3
0 ≤ φ ≤ π/4,
R sec φ ≤ ρ ≤ 2R cos φ.
πR3 . By symmetry the i, j components of force are zero and ⎧ ⎫ ⎨ 3GmM ⎬ z F= dxdydz k ⎩ 2πR3 ⎭ (x2 + y 2 + z 2 )3/2 T
⎧ ⎫ ⎨ 3GmM ρ cos φ ⎬ = ρ2 sin φ dρdθdφ k 3 3 ⎩ 2πR ⎭ ρ + = =
S
3GmM 2πR3
cos φ sin φ dρ dφ dθ 0
0
R sec φ
GmM √ 2 − 1 k. R2
42. With the coordinate system shown in the figure, T is the set of all points (x, y, z) with cylindrical coordinates (r, θ, z) in the set S:
0 ≤ r ≤ R,
0 ≤ θ ≤ 2π,
α ≤ z ≤ α + h.
The gravitational force is ⎡ ⎤ GmM z F =⎣ dxdydz ⎦ k V (x2 + y 2 + z 2 )3/2 ⎡
T
GmM =⎣ πR2 h
⎤ zr dr dθ dz ⎦ k (r2 + z 2 )3/2
S
=
GmM πR2 h
0
2π R α+h 0
α
,
2π π/4 2R cos φ
zr dzdrdθ k (r2 + z 2 )3/2
2GmM 2 2− 2 + (α + h)2 + h k = R + α R R2 h
k
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SECTION 17.10
SECTION 17.10 1.
ad − bc
2.
1
3.
2 v 2 − u2
4.
u ln v − u
5.
−3u2 v 2
6.
1+
7.
abc
8.
2
9.
ρ2 sin φ
cos θ 10. |J(r, θ, z)| = −r sin θ 0
0 0 = r. 1
sin θ r cos θ 0
sin φ cos θ 11. J(ρ, θ, φ) = −ρ sin φ sin θ ρ cos φ cos θ
sin φ sin θ
= −ρ2 sin φ; 0 −ρ sin φ cos θ
ρ sin φ cos θ ρ cos φ sin θ
dx − by = (ad − bc)u0
12. (a)
(b)
v = x − y.
13. Set u = x + y,
1 uv
|J(ρ, θ, φ)| = ρ2 sin φ.
cx − ay = (bc − ad)v0
Then
x=
u+v , 2
y=
u−v 2
1 and J (u, v) = − . 2
Ω is the set of all (x, y) with uv-coordinates in 0 ≤ u ≤ 1,
Γ: Then
x2 − y 2 dxdy =
Ω
1 = 2
1
u du 0
1 1 uv dudv = 2 2
Γ
2
v dv 0
1 = 2
1
0 ≤ v ≤ 2. 2
uv dv du 0
0
1 1 (2) = . 2 2
14. Using the changes of variables from Exercise 13, 1 2 2 1 1 2 2 u − v2 1 dv du = 4xy dx dy = 4 u − v 2 dv du = −1 4 2 2 0 0 0 0 Ω
15.
1 2
1
2
u cos (πv) dv du = 0
0
16. Set u = x − y,
1 2
v = x + 2y.
1
u du 0
2
cos (πv) dv 0
=
1 2
1 (0) = 0 2
Then
2u + v v−u 1 , y= , and J(u, v) = 3 3 3 Ω is the set of all (x, y) with uv-coordinates in the set x=
Γ:
0 ≤ u ≤ π,
0 ≤ v ≤ π/2.
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SECTION 17.10 Therefore
1 1 (u + 2v) du dv = 9 9
(x + y) dxdy = Ω
Γ
17. Set u = x − y,
π/2
(u + 2v) dv du = 0
0
1 3 π . 18
v = x + 2y. Then x=
Ω
π
2u + v , 3
y=
v−u , 3
and J(u, v) =
1 . 3
is the set of all (x, y) with uv-coordinates in the set Γ : 0 ≤ u ≤ π,
0 ≤ v ≤ π/2.
Therefore 1 1 π π/2 sin u cos v dudv = sin (x − y) cos (x + 2y) dxdy = sin u cos v dv du 3 3 0 0 Ω Γ π π/2 1 2 1 = sin u du cos v dv = (2)(1) = . 3 3 3 0 0 18. Using the change of variables from Exercise 16, π π/2 1 sin 3x dx dy = sin(2u + v) du dv = 0. 3 0 0 Ω
19. Set u = xy,
v = y.
xy = 1,
xy = 4
y = x,
y = 4x
Then x = u/v,
y=v
u = 1,
u=4
=⇒ =⇒
and
J (u, v) = 1/v.
4u/v = v =⇒ v 2 = u,
u/v = v,
v 2 = 4u
Ω is the set of all (x, y) with uv-coordinates in the set
(a) A =
1 dudv = v
Γ
4 (b) xA = 1
√ 2 u
√
4 yA = 1
u
4
√
1
u
1 dv du = v
u 7 dv du = ; 2 v 3
√ 2 u
√
√ 2 u
dv du = u
√
1 ≤ u ≤ 4,
Γ:
14 ; 3
x=
y=
u≤v≤2
√
u.
4
ln 2 du = 3 ln 2 1
7 9 ln 2
14 9 ln 2
Γ : 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π 2π 1 A= abr dr dθ = ab r dr dθ = πab
20. J(r, θ) = abr,
0
Γ
21. Set u = x + y,
0
v = 3x − 2y. Then x=
2u + v , 5
y=
3u − v 5
and
1 J (u, v) = − . 5
907
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SECTION 17.10 0 ≤ u ≤ 1,
0≤v≤2 1 2 1 2 M= λ dv du = λ 5 5 0 0
With Γ :
Then
1
2
Ix = 0
0 1
2
Iy = 0
0
3u − v 5 2u + v 5
Iz = Ix + Iy =
2
2
where
λ is the density.
1 8λ 4 λ dv du = = 5 375 75 1 28λ 14 λ dv du = = 5 375 75
2 λ 5
2 λ 5
=
4 M, 75
=
14 M, 75
18 M. 75
u+v v−u 1 , y= , J(u, v) = Γ: 2 2 2 2 1 16 1 2 −u A= dv du = du dv = 2 2 −2 −4 3
−2 ≤ u ≤ 2,
22. x =
−4 ≤ v ≤ −u2
Γ
23. Set u = x − 2y,
v = 2x + y. Then x=
u + 2v , 5
v − 2u 5
y=
and
J (u, v) =
1 . 5
Γ is the region between the parabola v = u2 − 1 and the line v = 2u + 2. A sketch of the curves shows that Γ:
−1 ≤ u ≤ 3,
Then A=
24. xA = A=
1 2
2
−2
−u2
−4
1 1 (area of Γ) = 5 5
32 u+v dv du = − 2 5
3
u2 − 1 ≤ v ≤ 2u + 2.
−1
yA =
1 2
32 (2u + 2) − u2 − 1 du = . 15
2
−2
−u2
−4
32 v−u dv du = − 2 5
16 6 =⇒ x = y = − 3 5
25. The choice θ = π/6 reduces the equation to 13u2 + 5v 2 = 1. This is an ellipse in the uv-plane √ √ with area πab = π/ 65. Since J(u, v) = 1, the area of Ω is also π/ 65. 26.
e−(x−y) 1 dx dy = 2 1 + (x + y) 2 2
Sa
e−u du dv 1 + v2 2
Γ
where Γ is the square in the uv-plane with vertices (−2a, 0), (0, −2a), (2a, 0), (0, 2a). Γ contains the square −a ≤ u ≤ a, −a ≤ v ≤ a and is contained in the square −2a ≤ u ≤ 2a, −2a ≤ v ≤ 2a. Therefore 1 2
a
−a
a
−a
e−u 1 du dv ≤ 2 1+v 2 2
Γ
e−u 1 dudv ≤ 2 1+v 2 2
2a
−2a
2a
−2a
e−u du dv. 1 + v2 2
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SECTION 17.10 The two extremes can be written a a 1 1 2 e−u du dv 2 2 −a −a 1 + v
1 2
and
2a
−2a
e−u du 2
2a
−2a
1 dv . 1 + v2
√ As a → ∞ both expressions tend to 12 ( π) (π) = 12 π 3/2 . It follows that ∞ ∞ 2 1 e−(x−y) dx dy = π 3/2 . 2 1 + (x + y) 2 −∞ −∞ 27. J = abcρ2 sin φ;
2π
0
28. x = y = 0 2π zV = 0
π/2
0
π
1
V =
1
0
abcρ2 sin φ dρ dφ dθ =
0
(cρ cos φ)abcρ2 sin φ dρ dφ dθ =
0
4 πabc 3
πabc2 3 =⇒ z = c . 4 8
2 M 3M V = πabc, λ = = 3 V 2πabc 2π π/2 1 2 2 3M Ix = b ρ sin2 φ sin2 θ + c2 ρ2 cos2 φ abcρ2 sin φ dρ dφ dθ 2πabc 0 0 0 2 1 2 = 5M b +c Iy = 15 M a2 + c2 , Iz = 15 M a2 + b2
29.
2π
1
π
30. I = 0
0
0
PROJECT 17.10 1. (a)
ρ2 (abcρ2 sin φ) dφ dρ dθ =
−1
θ = tan
ay 1/α bx
,
r=
4 πabc 5
x 2/α a
+
y 2/α α/2 b
⎫ ar1 (cos θ1 )α = ar2 (cos θ2 )α ⎪ ⎪ ⎬
(b)
br1 (sin θ1 )α = br2 (sin θ2 )α r1 > 0,
0
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