Calculus one and several variables 10E Salas solutions manual
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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
SECTION 17.8 899
25. V =∫ π/2
−π/2
∫ cos θ
0
∫ r cos θ
r2r dz dr dθ =
∫ π/2
−π/2
∫ cos θ
0
(r2 cos θ − r3
)dr dθ
=∫ π/2
−π/2
112
cos4 θ dθ =132
π
26. V =∫ 2π
0
∫ 3
0
∫ √25−r2
r+1
r dz dr dθ =413π
27. V =∫ 2π
0
∫ 1/2
0
∫ √1−r2
r√
3
r dz dr dθ =∫ 2π
0
∫ 1/2
0
(r√
1 − r2 − r2√
3)dr dθ =
13π(2 −
√3)
28. V =∫ 2π
0
∫ a
0
∫ √a2+r2
√2r
r dz dr dθ =2π3a3(
√2 − 1)
29. V =∫ 2π
0
∫ 3
1
∫ √9−r2
0
r dz dr dθ =∫ 2π
0
∫ 3
1
r√
9 − r2 dr dθ = 323 π
√2
30. V =∫ 2π
0
∫ 2
1
∫ 12
√36−r2
0
r dz dr dθ =13π(35
√35 − 128
√2).
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.
M =∫ 2π
0
∫ R
0
∫ h
0
kzr dz dr dθ =12kπR2h2
32. xM = yM = 0 by symmetry
zMM =∫ 2π
0
∫ R
0
∫ h
0
kz2r dz dr dθ =13kπR2h3
M =12kπR2h2, zM =
23h
The center of mass lies on the axis of the cylinder at a distance 23h from the base of zero mass
density.
33. I = Iz = k
∫ 2π
0
∫ R
0
∫ h
0
zr3 dr dθ dz
= 14 kπR
4h2 = 12
(12 kπR
2h2)R2 = 1
2 MR2
∧from Exercise 31
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JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
900 SECTION 17.9
34. (a) I =M
πR2h
∫ 2π
0
∫ R
0
∫ h
0
r3 dz dr dθ =12MR2
(b) I =M
πR2h
∫ 2π
0
∫ R
0
∫ h
0
(r2 sin2 θ + z2)r dz dr dθ =14MR2 +
13Mh2
(c) I = 14MR2 + 1
3Mh2 −M( 12h)2 = 1
4MR2 + 112Mh2
35. Inverting the cone and placing the vertex at the origin, we have
V =∫ h
0
∫ 2π
0
∫ (R/h)z
0
r dr dθ dz =13πR2h.
36. xM = yM = 0 by symmetry
zMM =∫ h
0
∫ 2π
0
∫ (R/h)z
0
(M
V
)zr dr dθ dz =
(M
V
)πR2h2
4=⇒ zM =
πR2h2
4V=
34h
On the axis of the cone at a distance 34h from the vertex.
37. I =M
V
∫ h
0
∫ 2π
0
∫ (R/h)z
0
r3 dr dθ dz =310
MR2
38. I =M
V
∫ h
0
∫ 2π
0
∫ (R/h)z
0
z2 r dr dθ dz =35Mh2.
39. V =∫ 2π
0
∫ 1
0
∫ 1−r2
0
r dz dr dθ =12π 40. M =
∫ 2π
0
∫ 1
0
∫ 1−r2
0
kzr dz dr dθ =16πk
41. M =∫ 2π
0
∫ 1
0
∫ 1−r2
0
k(r2 + z2
)r dz dr dθ =
14kπ
SECTION 17.9
1.(√
3, 14 π, cos−1
[13
√3])
2.(
12
√6, 1
2
√2,
√2)
3. ( 34 ,
34
√3, 3
2
√3 ) 4. (2
√10, 2
3π, cos−1[ 310
√10])
5. ρ =√
22 + 22 + (2√
6/3)2 =4√
63
6.(
8, −π
4,
5π6
)
φ = cos−1
(2√
6/34√
6/3
)= cos−1(1/2) =
π
3
θ = tan−1(1) =π
4
(ρ, θ, φ) =
(4√
63
,π
4,π
3
)
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JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
SECTION 17.9 901
7. x = ρ sinφ cos θ = 3 sin 0 cos(π/2) = 0
z = ρ cosφ = 3 cos 0 = 3
y = ρ sinφ sin θ = 3 sin 0 sin(π/2) = 0
(x, y, z) = (0, 0, 3)
8. (a) (5, 12π, arccos 4
5 )
(b) (5, 32π, arccos 4
5 )
9. The circular cylinder x2 + y2 = 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 z2 = x2 + y2.
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 + y2 + (z − 12 )2 = 1
4 of radius 12 and center (0, 0, 1
2 )
15. Sphere of radius 2 centered at the origin:∫ 2π
0
∫ π
0
∫ 2
0
ρ2 sinφdρ dφ dθ =83
∫ 2π
0
∫ π
0
sinφdφ dθ =163
∫ 2π
0
dθ =32π3
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
0
ρ2 sinφdρ dφ dθ =π
12
17. The first quadrant portion of the sphere that lies between the x, y-plane and the plane z = 32
√3.
∫ π/2
π/6
∫ π/2
0
∫ 3
0
ρ2 sinφdρ dθ dφ = 9∫ π/2
π/6
∫ π/2
0
sinφdθ dφ
= 92 π
∫ π/2
π/6
sinφdφ
= 92 π [− cosφ]π/2π/6 = 9
4π√
3
18. A cone of radius 1 and height 1;∫ π/4
0
∫ 2π
0
∫ secφ
0
ρ2 sinφdρ dθ dφ =13π
19.∫ 1
0
∫ √1−x2
0
∫ √2−x2−y2
√x2+y2
dz dy dx =∫ π/4
0
∫ π/2
0
∫ √2
0
ρ2 sinφdρ dθ dφ
= 23
√2∫ π/4
0
∫ π/2
0
sinφdθ dφ
=√
23 π
∫ π/4
0
sinφdφ =√
26
π (2 −√
2)
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902 SECTION 17.9
20.∫ π/4
0
∫ π/2
0
∫ 2
0
ρ4 sinφdρ dθ dφ =16π5
∫ π/4
0
sinφdφ =8π5
(2 −√
2)
21.∫ 3
0
∫ √9−y2
0
∫ √9−x2−y2
0
z√x2 + y2 + x2 dz dx dy
=∫ π/2
0
∫ π/2
0
∫ 3
0
ρ cosφ · ρ · ρ2 sinφdρ dθ dφ
=∫ π/2
0
12
sin 2φdφ
∫ π/2
0
dθ
∫ 3
0
ρ4 dρ =[− 1
4cos 2φ
]π/20
(π2
) [15ρ5
]30
=243π20
22.∫ π/2
0
∫ π/2
0
∫ 1
0
1ρ2
ρ2 sinφdρ dθ dφ =π
2
23. V =∫ 2π
0
∫ π
0
∫ R
0
ρ2 sinφdρ dφ dθ =43πR3
24. r = ρ sinφ, θ = θ, z = ρ cosφ
25. V =∫ α
0
∫ π
0
∫ R
0
ρ2 sinφdρ dφ dθ =23αR3
26. M =∫ 2π
0
∫ π
0
∫ R
0
k(R− ρ)ρ2 sinφdρ dφ dθ =13kπR4
27. M =∫ 2π
0
∫ tan−1(r/h)
0
∫ h secφ
0
kρ3 sinφdρ dφ dθ
=∫ 2π
0
∫ tan−1(r/h)
0
kh4
4tanφ sec3 φdφ dθ
=kh4
4
∫ 2π
0
13[sec3 φ
]tan−1(r/h)
0dθ =
kh4
4
∫ 2π
0
13
⎡⎣(√
r2 + h2
h
)3
− 1
⎤⎦ dθ
=16kπh(r2 + h2
)3/2 − h3
28. V =∫ 2π
0
∫ tan−1 r/h
0
∫ h secφ
0
ρ2 sinφdρ dφ dθ =2π3
∫ tan−1(r/h)
0
h3 tanφ sec2 φdφ
=π
3h3 tan2(tan−1
( rh
)=
13πr2h
29. center ball at origin; density =M
V=
3M4πR3
(a) I =3M
4πR3
∫ 2π
0
∫ π
0
∫ R
0
ρ4 sin3 φdρ dφ dθ =25MR2
(b) I = 25 MR2 + R2M = 7
5 MR2 (parallel axis theorem)
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SECTION 17.9 903
30. The center of mass is the centroid; V =23πR3
zV =∫ 2π
0
∫ π/2
0
∫ R
0
(ρ cosφ)ρ2 sinφdρ dφ dθ =14πR4
z =38R; (x, y, z) =
(0, 0,
38R
)
31. center balls at origin; density =M
V=
3M4π(R2
3 −R13)
(a) I =3M
4π(R2
3 −R13) ∫ 2π
0
∫ π
0
∫ R2
R1
ρ4 sin3 φdρ dφ dθ =25M
(R2
5 −R15
R23 −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.
M2 =M
VV2 =
3M4π(R2
3 −R13) (4
3πR2
3
)=
MR23
R23 −R1
3 ; similarly M1 =MR1
3
R23 −R1
3 .
Then
I = I2 − I1 = 25 M2R2
2 − 25 M1R1
2 =25
(MR2
3
R23 −R1
3
)R2
2 − 25
(MR1
3
R23 −R1
3
)R1
2
=25M
(R2
5 −R15
R23 −R1
3
).
(b) Outer radius R and inner radius R1 gives
moment of inertia =25M
(R5 −R1
5
R3 −R13
). [part (a) ]
As R1 → R,
R5 −R15
R3 −R13 =
R4 + R3R1 + R2R12 + RR1
3 + R14
R2 + RR1 + R12 −→ 5R4
3R2=
53R2.
Thus the moment of inertia of spherical shell of radius R is
25M
(53R2
)=
23MR2.
(c) I = 23MR2 + R2M = 5
3 MR2 (parallel axis theorem)
32. (a) The center of mass is the centroid; using the result of Exercise 30,
x = y = 0
z =z2V2 − z1V1
V=
38R2
43πR2
3 − 38R1
43πR1
3
43π(R2
3 −R13)
=3(R2
2 + R12)(R2 + R1)
8(R22 + R2R1 + R1
2)
(b) Setting R1 = R2 = R in (a), we get x = y = 0, z = 12R
33. V =∫ 2π
0
∫ α
0
∫ a
0
ρ2 sinφdρ dφ dθ =23π (1 − cosα) a3
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JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
904 SECTION 17.9
34.∫ 2π
0
∫ π/4
0
∫ 1
0
eρ3ρ2 sinφdρ dφ dθ =
13π(e− 1)
(2 −
√2)
35. (a) Substituting x = ρ sinφ cos θ, y = ρ sinφ sin θ, z = ρ cosφ
into x2 + y2 + (z −R)2 = R2
we have ρ2 sin2 φ + (ρ cosφ−R)2 = R2,
which simplifies to ρ = 2R cosφ.
(b) 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π/4, R secφ ≤ ρ ≤ 2R cosφ
36. (a) M =∫ 2π
0
∫ π/2
0
∫ 2R cosφ
0
kρ3 sinφdρ dφ dθ =85kπR4
(b) M =∫ 2π
0
∫ π/2
0
∫ 2R cosφ
0
kρ3 sin2 φdρ dφ dθ =14kπ2R4
(c) M =∫ 2π
0
∫ π/2
0
∫ 2R cosφ
0
kρ3 cos2 θ sin2 φdρ dφ dθ =18kπ2R4
37. V =∫ 2π
0
∫ π/4
0
∫ 2
0
ρ2 sinφdρ dφ dθ +∫ 2π
0
∫ π/2
π/4
∫ 2√
2 cosφ
0
ρ2 sinφdρ dφ dθ
=13
(16 − 6
√2)π
38. V =∫ 2π
0
∫ π
0
∫ 1−cosφ
0
ρ2 sinφdρ dφ dθ =83π
39. Encase T in a spherical wedge W . W has spherical coordinates in a box Π that contains S. Define
f to be zero outside of T . Then
F (ρ, θ, φ) = f (ρ sinφ cos θ, ρ sinφ sin θ, ρ cosφ)
is zero outside of S and
∫∫T
∫f(x, y, z) dxdydz =
∫∫W
∫f(x, y, z) dxdydz
=∫∫
Π
∫F (ρ, θ, φ) ρ2 sinφdρdθdφ
=∫∫S
∫F (ρ, θ, φ) ρ2 sinφdρdθdφ.
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SECTION 17.9 905
40. Break up T into little basic solids T1, . . . , TN . Choose a point (x∗i , y
∗i , z
∗i ) from each Ti and view
all the mass as concentrated there. Now Ti attracts m with a force
Fi∼= −Gmλ(x∗
i , y∗i , z
∗i )(Volume of Ti)ri3
ri
where ri is the vector from (x∗i , y
∗i , z
∗i ) to (a, b, c). We therefore have
Fi∼= Gmλ(x∗
i , y∗i , z
∗i )[(x
∗i − a)i + (y∗i − b)j + (z∗i − c)k]
[(x∗i − a)2 + (y∗i − b)2 + (z∗i − c)2]3/2
(Volume of Ti).
The sum of these approximations is a Riemann sum for the triple integral given and tends to thattriple 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
S : 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π/4, R secφ ≤ ρ ≤ 2R cosφ.
T has volume V = 23 πR
3. By symmetry the i, j components of force are zero and
F =
⎧⎨⎩3GmM
2πR3
∫∫T
∫z
(x2 + y2 + z2)3/2dxdydz
⎫⎬⎭k
=
⎧⎨⎩3GmM
2πR3
∫∫S
∫ (ρ cosφρ3
)ρ2 sinφdρdθdφ
⎫⎬⎭k
=
{3GmM
2πR3
∫ 2π
0
∫ π/4
0
∫ 2R cosφ
R secφ
cosφ sinφdρ dφ dθ
}k
=GmM
R2
(√2 − 1
)k.
42. With the coordinate system shown in the figure, T
is the set of all points (x, y, z) with cylindrical coor-dinates (r, θ, z) in the set
S : 0 ≤ r ≤ R, 0 ≤ θ ≤ 2π, α ≤ z ≤ α + h.
The gravitational force is
F =
⎡⎣∫∫
T
∫ (GmM
V
)z
(x2 + y2 + z2)3/2dxdydz
⎤⎦k
=
⎡⎣GmM
πR2h
∫∫S
zr
(r2 + z2)3/2dr dθ dz
⎤⎦k
=[GmM
πR2h
∫ 2π
0
∫ R
0
∫ α+h
α
zr
(r2 + z2)3/2dzdrdθ
]k
=2GmM
R2h
(√R2 + α2 −
√R2 + (α + h)2 + h
)k
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906 SECTION 17.10
SECTION 17.10
1. ad− bc 2. 1 3. 2(v2 − u2
)4. u ln v − u 5. −3u2v2 6. 1 +
1uv
7. abc 8. 2 9. ρ2 sinφ
10. |J(r, θ, z)| =
∣∣∣∣∣∣∣∣cos θ sin θ 0
−r sin θ r cos θ 0
0 0 1
∣∣∣∣∣∣∣∣ = r.
11. J(ρ, θ, φ) =
∣∣∣∣∣∣∣∣sin φ cos θ sin φ sin θ cos θ
−ρ sin φ sin θ ρ sin φ cos θ 0
ρ cos φ cos θ ρ cos φ sin θ −ρ sin φ
∣∣∣∣∣∣∣∣ = −ρ2 sin φ; |J(ρ, θ, φ)| = ρ2 sin φ.
12. (a) dx− by = (ad− bc)u0 (b) cx− ay = (bc− ad)v0
13. Set u = x + y, v = x− y. Then
x =u + v
2, y =
u− v
2and J (u, v) = −1
2.
Ω is the set of all (x, y) with uv-coordinates in
Γ : 0 ≤ u ≤ 1, 0 ≤ v ≤ 2.
Then∫ ∫Ω
(x2 − y2
)dxdy =
∫ ∫Γ
12uv dudv =
12
∫ 1
0
∫ 2
0
uv dv du
=12
(∫ 1
0
u du
)(∫ 2
0
v dv
)=
12
(12
)(2) =
12.
14. Using the changes of variables from Exercise 13,∫∫Ω
4xy dx dy =∫ 1
0
∫ 2
0
4(u2 − v2
4
)12dv du =
12
∫ 1
0
∫ 2
0
(u2 − v2
)dv du = −1
15.12
∫ 1
0
∫ 2
0
u cos (πv) dv du =12
(∫ 1
0
u du
)(∫ 2
0
cos (πv) dv)
=12
(12
)(0) = 0
16. Set u = x− y, v = x + 2y. Then
x =2u + v
3, y =
v − u
3, and J(u, v) =
13
Ω is the set of all (x, y) with uv-coordinates in the set
Γ : 0 ≤ u ≤ π, 0 ≤ v ≤ π/2.
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SECTION 17.10 907
Therefore∫∫Ω
(x + y) dxdy =∫∫Γ
19(u + 2v) du dv =
19
∫ π
0
∫ π/2
0
(u + 2v) dv du =118
π3.
17. Set u = x− y, v = x + 2y. Then
x =2u + v
3, y =
v − u
3, and J(u, v) =
13.
Ω is the set of all (x, y) with uv-coordinates in the set
Γ : 0 ≤ u ≤ π, 0 ≤ v ≤ π/2.
Therefore∫∫Ω
sin (x− y) cos (x + 2y) dxdy =∫∫Γ
13
sinu cos v dudv =13
∫ π
0
∫ π/2
0
sinu cos v dv du
=13
(∫ π
0
sinu du
)(∫ π/2
0
cos v dv)
=13(2)(1) =
23.
18. Using the change of variables from Exercise 16,∫∫Ω
sin 3x dx dy =∫ π
0
∫ π/2
0
sin(2u + v)13du dv = 0.
19. Set u = xy, v = y. Thenx = u/v, y = v and J (u, v) = 1/v.
xy = 1, xy = 4 =⇒ u = 1, u = 4
y = x, y = 4x =⇒ u/v = v, 4u/v = v =⇒ v2 = u, v2 = 4u
Ω is the set of all (x, y) with uv-coordinates in the set
Γ : 1 ≤ u ≤ 4,√u ≤ v ≤ 2
√u.
(a) A =∫∫Γ
1vdudv =
∫ 4
1
∫ 2√u
√u
1vdv du =
∫1
4
ln 2 du = 3 ln 2
(b) xA =∫ 4
1
∫ 2√u
√u
u
v2dv du =
73
; x =7
9 ln 2
yA =∫ 4
1
∫ 2√u
√u
dv du =143
; y =14
9 ln 2
20. J(r, θ) = abr, Γ : 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π
A =∫∫
Γ
abr dr dθ = ab
∫ 2π
0
∫ 1
0
r dr dθ = πab
21. Set u = x + y, v = 3x− 2y. Then
x =2u + v
5, y =
3u− v
5and J (u, v) = −1
5.
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908 SECTION 17.10
With Γ : 0 ≤ u ≤ 1, 0 ≤ v ≤ 2
M =∫ 1
0
∫ 2
0
15λ dv du =
25λ where λ is the density.
Then
Ix =∫ 1
0
∫ 2
0
(3u− v
5
)2 15λ dv du =
8λ375
=475
(25λ
)=
475
M,
Iy =∫ 1
0
∫ 2
0
(2u + v
5
)2 15λ dv du =
28λ375
=1475
(25λ
)=
1475
M,
Iz = Ix + Iy =1875
M.
22. x =u + v
2, y =
v − u
2, J(u, v) =
12
Γ : −2 ≤ u ≤ 2, −4 ≤ v ≤ −u2
A =∫∫Γ
12du dv =
12
∫ 2
−2
∫ −u2
−4
dv du =163
23. Set u = x− 2y, v = 2x + y. Then
x =u + 2v
5, y =
v − 2u5
and J (u, v) =15.
Γ is the region between the parabola v = u2 − 1 and the line v = 2u + 2. A sketch of the curves showsthat
Γ : −1 ≤ u ≤ 3, u2 − 1 ≤ v ≤ 2u + 2.Then
A =15
(area of Γ) =15
∫ 3
−1
[(2u + 2) −
(u2 − 1
) ]du =
3215
.
24. xA =12
∫ 2
−2
∫ −u2
−4
u + v
2dv du = −32
5yA =
12
∫ 2
−2
∫ −u2
−4
v − u
2dv du = −32
5
A =163
=⇒ x = y = −65
25. The choice θ = π/6 reduces the equation to 13u2 + 5v2 = 1. This is an ellipse in the uv-planewith area πab = π/
√65. Since J(u, v) = 1, the area of Ω is also π/
√65.
26.∫∫Sa
e−(x−y)2
1 + (x + y)2dx dy =
12
∫∫Γ
e−u2
1 + v2du dv
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
12
∫ a
−a
∫ a
−a
e−u2
1 + v2du dv ≤ 1
2
∫∫Γ
e−u2
1 + v2dudv ≤ 1
2
∫ 2a
−2a
∫ 2a
−2a
e−u2
1 + v2du dv.
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JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38
SECTION 17.10 909
The two extremes can be written
12
(∫ a
−a
e−u2du
) (∫ a
−a
11 + v2
dv
)and
12
(∫ 2a
−2a
e−u2du
) (∫ 2a
−2a
11 + v2
dv
).
As a → ∞ both expressions tend to 12 (
√π) (π) = 1
2π3/2. It follows that∫ ∞
−∞
∫ ∞
−∞
e−(x−y)2
1 + (x + y)2dx dy =
12π3/2.
27. J = abcρ2 sinφ; V =∫ 2π
0
∫ π
0
∫ 1
0
abcρ2 sinφdρ dφ dθ =43πabc
28. x = y = 0
zV =∫ 2π
0
∫ π/2
0
∫ 1
0
(cρ cosφ)abcρ2 sinφdρ dφ dθ =πabc2
4=⇒ z =
38c .
29. V =23πabc, λ =
M
V=
3M2πabc
Ix =3M
2πabc
∫ 2π
0
∫ π/2
0
∫ 1
0
(b2ρ2 sin2 φ sin2 θ + c2ρ2 cos2 φ
)abcρ2 sinφdρ dφ dθ
= 15 M(b2 + c2
)Iy = 1
5 M(a2 + c2
), Iz = 1
5 M(a2 + b2
)30. I =
∫ 2π
0
∫ 1
0
∫ π
0
ρ2(abcρ2 sinφ) dφ dρ dθ =45πabc
PROJECT 17.10
1. (a) θ = tan−1
[(aybx
)1/α], r =
[(xa
)2/α+(yb
)2/α]α/2(b) ar1(cos θ1)α = ar2(cos θ2)α
br1(sin θ1)α = br2(sin θ2)α
r1 > 0, 0 < θ < 12π
⎫⎪⎪⎬⎪⎪⎭ =⇒ r1 = r2, θ1 = θ2
2. J = abαr cosα−1 θ sinα−1 θ
3. (a)
x
y
a
a
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910 REVIEW EXERCISES
(b) x = ar cos3 θ, y = ar sin3 θ; x23 + y
23 = a
23 =⇒ r = 1 and x = a cos3 θ, y = a sin3 θ
A =∫ 0
π2
y(θ)x′(θ) dθ =∫ 0
π2
a sin3 θ(3a cos2 θ[− sin θ]) dθ
= 3a2
∫ π2
0
sin4 θ cos2 θ dθ = 3a2
∫ π2
0
(sin4 θ − sin6 θ) dθ
= 3a2
[3 · 14 · 2
π
2− 5 · 3 · 1
6 · 4 · 2π
2
](See Exercise 62(b) in 8.3)
=3a2π
32
(c) Entire area enclosed: 4 · 3a2π
32=
3a2π
8
4. (a)
a = 3, b = 2 a = 2, b = 3
(b) From Problem 2, Jacobian J = 8abr cos7 θ sin7 θ
A =∫ π
2
0
∫ 1
0
8abr cos7 θ sin7 θ dθ = 4ab∫ π
2
0
cos7 θ sin7 θ dθ =ab
70
REVIEW EXERCISES
1.∫ 1
0
∫ √y
y
xy2 dx dy =∫ 1
0
[12x2y2]√y
ydy =
∫ 1
0
(12y3 − 1
2y4
)dy =
[18y4 − 1
10y5]10
=140
2.∫ 1
0
∫ y
−y
ex+y dx dy =∫ 1
0
[ex+y]y−y
dy =∫ 1
0
(e2y − 1) dy =[12e2y − y
]10
=e2
2− 3
2
3.∫ 1
0
∫ 3x
x
2yex3dy dx =
∫ 1
0
[y2ex
3]3xx
dx =∫ 1
0
(9x2ex3 − x2ex
3) dx =
[3ex
3 − 13ex
3]10
=83e− 8
3
4.∫ 2
1
∫ lnx
0
xey dy dx =∫ 2
1
[xey]lnx
0dx =
∫ 2
1
x(x− 1) dx =[13x3 − 1
2x2]21
=56
5.∫ π/4
0
∫ 2 sin θ
0
r cos θ dr dθ =∫ π/4
0
[12r2 cos θ
]2 sin θ
0dθ =
∫ π/4
0
2 sin2 θ cos θ dθ =[23
sin3 θ]π/40
=√
26
6.∫ 2
−1
∫ 4
0
∫ 1
0
xyz dx dy dz =∫ 2
−1
∫ 4
0
[12x2yz]10dy dz =
∫ 2
−1
∫ 4
0
12yz dy dz =
∫ 2
−1
4z dz = 6
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REVIEW EXERCISES 911
7.∫ 2
0
∫ 2−3x
0
∫ x+y
0
x dz dy dx =∫ 2
0
∫ 2−3x
0
[xz]x+y
0dy dx =
∫ 2
0
∫ 2−3x
0
(x2 + xy) dy dx
=∫ 2
0
[x2y +
12xy2
]2−3x
0
dx =∫ 2
0
(32x3 − 4x2 + 2x
)dx
=[38x4 − 4
3x3 + x2
]20
= −23
8. ∫ π2
0
∫ π2
z
∫ sin z
0
3x2 sin y dx dy dz =∫ π
2
0
∫ π2
z
[x3 sin y
]sin z
0dy dz =
∫ π2
0
∫ π2
z
sin3 z sin y dy dz
=∫ π
2
0
sin3 z cos z dz =14
sin4 z]π
2
0=
14
9.∫ 0
−π2
∫ 2 sin θ
0
∫ r2
0
r2 cos θ dz dr dθ =∫ 0
−π2
∫ 2 sin θ
0
r4 cos θ dr dθ =∫ 0
−π2
325
sin5 θ cos θ dθ = −1615
10.∫ π
2
−π6
∫ π2
0
∫ 1
0
ρ3 sinϕ cosϕdρ dθ dϕ =∫ π
2
−π6
∫ π2
0
14
sinϕ cosϕdθ dϕ
=∫ π
2
−π6
π
16sin 2ϕdϕ = −
[ π32
cos 2ϕ]π
2
−π6
=3π64
11.∫ 1
0
∫ 1
y
ex2dx dy =
∫ 1
0
∫ x
0
ex2dy dx =
∫ 1
0
ex2y∣∣∣x0dx =
∫ 1
0
xex2dx =
12ex
2∣∣∣10
=e− 1
2
12.∫ 2
0
∫ 1
x2
cos y2 dy dx =∫ 1
0
∫ 2y
0
cos y2 dx dy =∫ 1
0
x cos y2∣∣∣2y0
dy =∫ 1
0
2y cos y2dy = sin y2∣∣∣10
= sin 1
13.∫ 1
0
∫ √1−y2
0
1√1 − y2
dxdy =∫ 1
0
dy = 1
14.∫ 1
0
∫ 1−x
0
y cos(x + y) dy dx =∫ 1
0
∫ 1−y
0
y cos(x + y) dx dy
=∫ 1
0
y sin(x + y)]1−y
0dy =
∫ 1
0
y(sin 1 − sin y) dy
=[12y2 sin 1 + y cos y − sin y
]10
= cos 1 − 12
sin 1
15.∫ 1
0
∫ √1−x2
0
xy dy dx =∫ 1
0
[12xy2]√1−x2
0dx =
∫ 1
0
(12x− 1
2x3
)dx =
18
16.∫ √
3
−√
3
∫ 4−y2
y2/3
(x− y) dx dy =∫ √
3
−√
3
[x2
2− xy)
]4−y2
y2/3dy =
∫ √3
−√
3
(49y4 +
43y3 − 4y2 − 4y + 8
)dy =
48√
35
17.∫ 2
0
∫ 3x−x2
x
(x2 − xy) dy dx =∫ 2
0
[x2y − 1
2xy2]3x−x2
xdx =
∫ 2
0
(2x4 − 2x3 − 1
2x5
)dx = − 8
15
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912 REVIEW EXERCISES
18.∫ 2
0
∫ 2−x
0
x(x− 1)exy dy dx =∫ 2
0
[(x− 1)exy
]2−x
0dx =
∫ 2
0
(x− 1)e2x−x2dx−
∫ 2
0
(x− 1) dx
= −[12e2x−x2
]20−[12x2 − x
]20
= 0
19.∫ 2
0
∫ y
0
∫ √4−y2
0
2xyz dz dx dy =∫ 2
0
∫ 2
x
xyz2∣∣∣√4−y2
0dx dy =
∫ 2
0
∫ 2
0
xy(4 − y2) dx dy
=∫ 2
0
[12y3(4 − y2)
]dy =
83
20.∫∫∫T
z dxdydz = 2∫ 1
0
∫ x
0
∫ 1−x
0
z dz dy dx =∫ 1
0
∫ x
0
(1 − x)2 dy dx =∫ 1
0
(1 − x)2x dx =112
21.∫ 2
0
∫ √4−x2
0
∫ √4−x2−y2
0
xy dz dy dx =∫ 2
0
∫ √4−x2
0
[xyz]√4−x2−y2
0dy dx
=∫ 2
0
[− 1
3x√
4 − x2 − y2]√4−x2
0dx
=∫ 2
0
13x(4 − x2)
32 dx =
3215
22.∫∫∫T
(x2 + 2z) dx dy dz =∫ 2
−2
∫ 4
x2
∫ 4−y
0
(x2 + 2z) dz dy dx
=∫ 2
−2
∫ 4
x2(4x2 − x2y + 16 − 8y + y2) dy dx
=∫ 2
−2
(16x6 − 8x2 +
643
)dx =
210
21
23.∫ 2
0
∫ √4−y2
0
e√
x2+y2dx dy =
∫ π/2
0
∫ 2
0
err dr dθ =π
2
∫ 2
0
rerdr =π
2
[rer − er
]20
=π
2(e2 + 1)
24.∫ 1
−1
∫ √1−x2
0
arctan (y/x) dy dx =∫ π/2
0
∫ 1
0
rθ dr dθ +∫ π
π/2
∫ 1
0
(θ − π)r dr dθ
=∫ π/2
0
θ
2dθ +
∫ π
π/2
12(θ − π) dθ = 0
25. V =∫ 3
0
∫ 2π
0
(9 − r2)r dr dθ = 2π∫ 3
0
(9 − r2)r dr =81π2
26.∫ 1
0
∫ √x
x2(2 − x2 − y2) dy dx = −
∫ 1
0
(2x1/2 − x5/2 − 1
3x3/2 − 2x2 + x4 + 1
3x6)dx =
52105
27. V =∫ 1
0
∫ 1−x
0
(x2 + y2) dy dx =∫ 1
0
[x2 − x3 +
13(1 − x)3
]dx =
[13x3 − 1
4x4 − 1
12(1 − x)4
]10
=16
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REVIEW EXERCISES 913
28. V =∫ 3
0
∫ π/2
0
r2 sin θ dr dθ =∫ π/2
0
9 sin θ dθ = 9
29. M =∫ π/2
−π/2
∫ cosx
0
y dy dx =∫ π/2
−π/2
12
cos2 x dx =π
4
xMM =∫ π/2
−π/2
∫ cosx
0
xy dy dx = 0 by symmetry
yMM =∫ π/2
−π/2
∫ cosx
0
y2 dy dx =∫ π/2
−π/2
13
cos3 x dx =49
The center of mass is: (0, 169π )
30. M =∫ 1
0
∫ y
y22x dx dy =
∫ 1
0
(y2 − y4) dy =215
xMM =∫ 1
0
∫ y
y22x2 dx dy =
∫ 1
0
(23y3 − 2
3y6
)dy =
114
yMM =∫ 1
0
∫ y
y22xy dx dy =
∫ 1
0
(y3 − y5) dy =112
The center of mass is: (15/28, 5/8)
31. M =∫ π/2
0
∫ R
r
u3 du dθ =π
8(R4 − r4); (polar coordinates [u, θ])
By symmetry, x = y.
xMM =∫ π/2
0
∫ R
r
u4 cos θ du dθ =15(R5 − r5); xM =
8(R5 − r5)5π(R4 − r4)
32. M =∫ π
0
∫ 2(1+cos θ)
0
r2 dr dθ =∫ π
0
83(1 + cos θ)3 dθ =
203π
xMM =∫ π
0
∫ 2(1+cos θ)
0
r3 cos θ dr dθ =∫ π
0
4(1 + cos θ)4 cos θ dθ = 14π
yMM =∫ π
0
∫ 2(1+cos θ)
0
r3 sin θ dr dθ =∫ π
0
4(1 + cos θ)4 sin θ dθ = −45(1 + cos θ)5|π0 =
1285
The center of mass is: ( 2110 ,
9625π )
33. Introduce a coordinate system as shown in the figure.
(a) A = 12bh; by symmetry, x = 0
y A =∫ 0
−b/2
∫ 2hb (x+ b
2 )
0
y dy dx +∫ b
2
0
∫ − 2hb (x− b
2 )
0
y dy dx
=bh2
6=⇒ y =
h
3 x
y
−b 2 b 2
h
(b) I =∫ 0
−b/2
∫ 2hb (x+ b
2 )
0
λy2 dy dx +∫ b/2
0
∫ − 2hb (x− b
2 )
0
λy2 dy dx =λbh3
12= 1
6Mh2
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914 REVIEW EXERCISES
(c) I = 2∫ b/2
0
∫ − 2hb (x− b
2 )
0
λx2 dx dy =148
λhb3 = 124Mb2
34. Let λ =k√
x2 + y2
(a) M =∫ π
0
∫ R
r
k dr dθ = kπ(R− r)
By symmetry, xM = 0; yMM =∫ π
0
∫ R
r
kr sin θ dr dθ = k(R2 − r2)
The center of mass is: (0, R+rπ )
(b) Ix =∫ π
0
∫ R
r
kr2 sin2 θ dr dθ =k
3(R3 − r3)
∫ π
0
sin2 θ dθ =kπ
6(R3 − r3)
(c) Iy =∫ π
0
∫ R
r
kr2 cos2 θ dr dθ =k
3(R3 − r3)
∫ π
0
cos2 θ dθ =kπ
6(R3 − r3)
35. V =∫ 2
0
∫ x
0
∫ 2x+2y+1
0
dz dy dx =∫ 2
0
∫ x
0
(2x + 2y + 1) dy dx = 10
36. V =∫ 1
0
∫ x
x2
∫ 4(x2+y2)
−1
dz dy dx =∫ 1
0
∫ x
x2(4x2 + 4y2 + 1) dy dx
=∫ 1
0
(x +
163x3 − x2 − 4x4 − 4
3x6
)dx =
107210
37. The curve of intersection of the two surfaces is the circle: x2 + y2 = 4, x = 3
V =∫ 2
−2
∫ √4−x2
−√
4−x2
∫ 12−x2−2y2
2x2+y2dz dy dx =
∫ 2
−2
∫ √4−x2
−√
4−x23(4 − x2 − y2
)dy dx
= 3∫ 2π
0
∫ 2
0
(4 − r2
)r dr dθ
= 3∫ 2π
0
[2r2 − 1
4 r4]20dθ = 12
∫ 2π
0
= 24π
38. V =∫ 1
0
∫ √1−y2
0
∫ (2−y−z)/2
0
dx dz dy =∫ 1
0
∫ π/2
0
2 − r cos θ − r sin θ
2r dθ dr
=∫ 1
0
12(π − 2r)r dr =
3π − 412
39. V =∫ 2π
0
∫ π/3
0
∫ 2
secφ
ρ2 sinφdρ dφ dθ =∫ 2π
0
∫ π/3
0
[13 ρ
3]2secφ
dφ dθ
=13
∫ 2π
0
∫ π/3
0
(8 − sec3φ
)sinφdφ dθ
=13
∫ 2π
0
[− 8 cosφ− 1
2 sec2 φ]π/30
dθ
=13
(52
)(2π) =
5π3
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REVIEW EXERCISES 915
40. V =∫ 4
0
∫ (12−3x)/4
0
∫ 16−x2
0
dz dy dx=∫ 4
0
∫ 12−3x4
0
(16 − x2) dy dx=34
∫ 4
0
(64 − 16x− 4x2 + x3) dx = 80
41. V =∫ 2π
0
∫ π/2
π/4
∫ 1
0
ρ2 sinφdρ dφ dθ =∫ 2π
0
∫ π/2
π/4
13
sinφdφ dθ =√
2π3
42. V =∫ 2π
0
∫ π/4
0
∫ 1
0
ρ2 sinφdρ dφ dθ =2 −
√2
3π
43. (a) V =∫ 1
0
∫ x
0
∫ √1−x2
0
dz dy dx +∫ 1
0
∫ y
0
∫ √1−y2
0
dz dx dy
= 2∫ 1
0
∫ x
0
√1 − x2 dy dx = 2
∫ 1
0
x√
1 − x2dx =23
By symmetry, x = y.
xV =∫ 1
0
∫ x
0
∫ √1−x2
0
x dz dy dx +∫ 1
0
∫ y
0
∫ √1−y2
0
x dz dx dy
For the first integral:
∫ 1
0
∫ x
0
∫ √1−x2
0
x dz dy dx =∫ 1
0
∫ x
0
x√
1 − x2 dy dx
=∫ 1
0
x2√
1 − x2 dx =∫ π/2
0
sin2 u cos2 u du =π
16
x = sinu∧
For the second integral:∫ 1
0
∫ y
0
∫ √1−y2
0
x dz dx dy =∫ 1
0
∫ y
0
x√
1 − y2dx dy =∫ 1
0
12y2√
1 − y2 dy =π
32
Thus, xV =3π32
=⇒ x = y =9π64
Now calculate z:
z V =∫ 1
0
∫ x
0
∫ √1−x2
0
z dz dy dx +∫ 1
0
∫ y
0
∫ √1−y2
0
z dz dx dy;
∫ 1
0
∫ x
0
∫ √1−x2
0
z dz dy dx =∫ 1
0
∫ x
0
12(1 − x2)dy dx =
12
∫ 1
0
(x− x3
)dx =
18
and similarly,∫ 1
0
∫ y
0
∫ √1−y2
0
z dz dy dx =18.
Therefore, zV =14
=⇒ z =38
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916 REVIEW EXERCISES
(b) Iz =∫ 1
0
∫ x
0
∫ √1−x2
0
λ(√
x2 + y2)2
dz dy dx +∫ 1
0
∫ y
0
∫ √1−y2
0
λ(√
x2 + y2)2
dz dx dy;
∫ 1
0
∫ x
0
∫ √1−x2
0
λ(√
x2 + y2)2
dz dy dx =∫ π/4
0
∫ sec θ
0
∫ r sin θ
0
λr3dz dr dθ =320
λ
and∫ 1
0
∫ y
0
∫ √1−y2
0
λ(√x2 + y2)2dz dx dy =
320
λ =⇒ Iz =310
λ
44. (a) V =∫ 2π
0
∫ 1
0
(r − r2)r dr dθ = 2π∫ 1
0
(r2 − r3) dr =π
6By symmetry, x = y = 0
zV =∫ 2π
0
∫ 1
0
∫ r
r2zr dz dr dθ =
π
12and hence z = 1
2
(b) Iz = K
∫ 2π
0
∫ 1
0
∫ r
r2r2 dz dr dθ = 2πK
∫ 1
0
r2(r − r2)dr =πK
10
Here, K is the density of the mass.
45. Denote polar coordinates by [u, θ].
(a) M =∫ 2π
0
∫ r
0
∫ h
0
u3 dz du d θ = 2πh∫ r
0
u3 du =πhr4
2
(b) By symmetry, xM = yM = 0
(c) zMM =∫ 2π
0
∫ r
0
∫ h
0
u3z dz du dθ =πh2r4
4=⇒ zM = h/2
46. λ =√x2 + y2
(a) M =∫ 2π
0
∫ π/2
0
∫ r
0
ρ3 sin2 φdρ dφ dθ =r4π
2
∫ π2
0
sin2 φdφ =r4π2
8
(b) By symmetry, xM = yM = 0
zMM =∫ 2π
0
∫ π/2
0
∫ r
0
ρ4 sin2 φ cosφdρ dφ dθ =2r5π
15=⇒ zM =
16r15π
47. (a) M =∫ 1
0
∫ 2π
0
∫ 1
r
r2 dz dθ dr = 2π∫ 1
0
∫ 1
r
r2 dz dr = 2π∫ 1
0
r2(1 − r) dr =π
6
(b) By symmetry, xM = yM = 0
zMM =∫ 1
0
∫ 2π
0
∫ 1
r
r2z dz dθ dr = π
∫ 1
0
r2(1 − r2) dr =2π15
=⇒ zM =45
(c) Iz =∫ 1
0
∫ 2π
0
∫ 1
r
r4 dz dθ dr =π
15
48. J(u, v) =
∣∣∣∣∣ 2u 2v
−2v 2u
∣∣∣∣∣ = 4u2 + 4v2
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-17 JWDD027-Salas-v1 December 7, 2006 16:38