calculus one and several variables 10e salas solutions manual ch18
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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-18 JWDD027-Salas-v1 January 4, 2007 18:12
918 SECTION 18.1
CHAPTER 18
SECTION 18.1
1. (a) h(x, y) = y i + x j; r(u) = u i + u2 j, u ∈ [ 0, 1 ]
x(u) = u, y(u) = u2; x′(u) = 1, y′(u) = 2u
h(r(u)) · r′(u) = y(u)x′(u) + x(u) y′(u) = u2(1) + u(2u) = 3u2∫C
h(r) ·dr =∫ 1
0
3u2 du = 1
(b) h(x, y) = y i + x j; r(u) = u3 i − 2u j, u ∈ [ 0, 1 ]
x(u) = u3, y(u) = −2u; x′(u) = 3u2, y′(u) = −2
h(r(u)) · r′(u) = y(u)x′(u) + x(u) y′(u) = (−2u)(3u2) + u3(−2) = −8u3∫C
h(r) ·dr =∫ 1
0
−8u3 du = −2
2. (a)∫C
h ·dr =∫ 1
0
(u i + u2 j) · (i + 2u j) du =∫ 1
0
(u + 2u3) du = 1
(b)∫C
h ·dr =∫ 1
0
(u3 i − 2u j) · (3u2 i − 2 j) du =∫ 1
0
(3u5 + 4u) du =52
3. h(x, y) = y i + x j; r(u) = cosu i − sinu j, u ∈ [ 0, 2π ]
x(u) = cosu, y(u) = − sinu; x′(u) = − sinu, y′(u) = − cosu
h(r(u)) · r′(u) = y(u)x′(u) + x(u) y′(u) = sin2 u− cos2 u∫C
h(r) ·dr =∫ 2π
0
(sin2 u− cos2 u) du = 0
4. (a)∫C
h ·dr =∫ 1
0
(e−u i + 2 j)·(eu i − e−u j) du =∫ 1
0
(1 − 2e−u) du = 2e−1 − 1
(b)∫C
h ·dr =∫ 2
0
2 j·(1 − u) i du =∫ 2
0
0 du = 0
5. (a) r(u) = (2 − u) i + (3 − u) j, u ∈ [ 0, 1 ]∫C
h(r) · dr =∫ 1
0
(−5 + 5u− u2) du = −176
(b) r(u) = (1 + u) i + (2 + u) j, u ∈ [ 0, 1 ]∫C
h(r) · dr =∫ 1
0
(1 + 3u + u2) du =176
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SECTION 18.1 919
6. (a)∫C
h · dr =∫ 4
1
(1√
u(1 + u)i +
1u√
1 + uj)·(
12√ui +
12√
1 + uj)
du =∫ 4
1
1u(1 + u)
du = ln85
(b)∫C
h · dr =∫ 1
0
1(1 + u)3
(i + j) · (i + j) du =∫ 1
0
2(1 + u)3
du =34
7. C = C1 ∪ C2 ∪ C3 where,
C1 : r(u) = (1 − u)(−2 i) + u(2 i) = (4u− 2) i, u ∈ [ 0, 1 ]
C2 : r(u) = (1 − u)(2 i) + u(2 j) = (2 − 2u) i + 2u j, u ∈ [ 0, 1 ]
C3 : r(u) = (1 − u)(2 j) + u(−2 i) = −2u i + (2 − 2u) j, u ∈ [ 0, 1 ]∫C
=∫C1
+∫C2
+∫C3
= 0 + (−4) + (−4) = −8
8. r(u) = (−1 + 2u) i + (1 + u) j, u ∈ [0, 1]∫C
h · dr =∫ 1
0
(e−2+u i + e3u j) · (2i + j) du =∫ 1
0
(2e−2+u + e3u) du =e5 − e2 + 6e− 6
3e2
9. C1 : r(u) = (−1 + 2u) i, u ∈ [ 0, 1 ]
C2 : r(u) = cosu i + sinu j, u ∈ [ 0, π ]
∫C
=∫C1
+∫C2
= 0 + (−π) = −π
10. Bottom: r(u) = u i;∫ 1
0
u3j · i du =∫ 1
0
0 du = 0
Right side: r(u) = i + uj;∫ 1
0
[3ui + (1 + 2u)j]·j du =∫ 1
0
(1 + 2u) du = 2
Top: r(u) = (1 − u)i + j;∫ 1
0
3(1 − u)2i · (−i) du =∫ 1
0
−3(1 − u)2 du = −1
Left: r(u) = (1 − u)j;∫ 1
0
2(1 − u)j·(−j) du =∫ 1
0
−2(1 − u) du = −1∫C
h · dr = sum of the above = 0
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920 SECTION 18.1
11. (a) r(u) = u i + u j + uk, u ∈ [ 0, 1 ]∫C
h(r) · dr =∫ 1
0
3u2 du = 1
(b)∫C
h(r) · dr =∫ 1
0
(2u3 + u5 + 3u6) du =2321
12. (a)∫C
h · dr =∫ 1
0
eu(i + j + k) · (i + j + k) du =∫ 1
0
3eu du = 3(e− 1)
(b)∫C
h · dr =∫ 1
0
(eu i + eu2j + eu
3k)·( i + 2u j + 3u2 k) du=
∫ 1
0
(eu + 2ueu2+ 3u2eu
3) du = 3(e− 1).
13. (a) r(u) = 2u i + 3u j − uk, u ∈ [ 0, 1 ]∫C
h(r) · dr =∫ 1
0
(2 cos 2u + 3 sin 3u + 3u2) du =[sin 2u− cos 3u + u3
]10
= 2 + sin 2 − cos 3
(b)∫C
h(r) · dr =∫ 1
0
(2u cosu2 + 3u2 sinu3 − u4
)du =
[sinu2 − cosu3 − 1
5u5
]10
=45
+ sin 1 − cos 1
14. (a)∫C
h · dr =∫ 1
0
(−2u2 i + 4u3 j − 2u3 k)·(2 i − j + k) du =∫ 1
0
(−4u2 − 6u3) du = −176
(b)∫C
h · dr =∫ 1
0
(i + ue2uj + uk) · (eu i − e−u j + k) du =∫ 1
0
(eu − ueu + u) du = e− 32
15. r(u) = u i + u2 j, u ∈ [ 0, 2 ]∫C
F(r) · dr =∫ 2
0
[(u + 2u2) + (2u + u2)2u
]du =
∫ 2
0
(2u3 + 6u2 + u
)du = 26
16. C1 : r(u) = u i;∫ 1
0
u i · i du =∫ 1
0
u du =12
C2 : r(u) = i + uj;∫ 1
0
(cosui − u sin 1j) · j du =∫ 1
0
−u sin 1 du = −12
sin 1
C3 : r(u) = (1 − u)i + j;∫ 1
0
[(1 − u) cos 1i − sin(1 − u)j] ·(−i) du =∫ 1
0
(u− 1) cos 1 du = −12
cos 1
W =∫C
F · dr =12− 1
2sin 1 − 1
2cos 1 =
12(1 − sin 1 − cos 1)
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SECTION 18.1 921
17. r(u) = (1 − u)(j + 4k) + u(i − 4k)
= ui + (1 − u)j + (4 − 8u)k, u ∈ [ 0, 1 ]∫C
F(r) · dr =∫ 1
0
(−32u + 97u2 − 64u3) du =13
18. C1 : r(u) = u i; F(r(u)) = 0,∫C1
F · dr = 0
C2 : r(u) = i + uj;∫ 1
0
uk · j du =∫ 1
0
0 du = 0
C3 : r(u) = i + j + uk;∫ 1
0
(ui + uj + k) · k du =∫ 1
0
du = 1
W = 1
19. r(u) = cosu i + sinu j + uk, u ∈ [ 0, 2π ]∫C
F(r) ·dr =∫ 2π
0
[− cos2 u sinu + cos2 u sinu + u2
]du =
∫ 2π
0
u2 du =8π3
3
20. Place the origin at the center of the circular path C and use the time parameter t. Motion along C
at constant speed is uniform circular motion
r(t) = r(cosωt i + sinωt j).
Differentiation gives
r′(t) = rω(− sinωti + cosωt j), r′′(t) = −rω2(cosωti + sinωt j).
The force on the object is
F(r(t)) = mr′′(t).
Note that F(r(t)) · r′(t) = 0 for all t, and therefore W is 0 on every time integral.
Physical explanation : At each instant the force on the object is perpendicular to the path of the
object. Thus the component of force in the direction of the motion is always zero.
21.∫C
q ·dr =∫ b
a
[q · r′(u)] du =∫ b
a
d
du[q · r(u)] du
= [q · r (b)] − [q · r (a)]
= q · [r (b) − r (a)]∫C
r ·dr =∫ b
a
[r (u) · r′ (u)] du
=12
∫ b
a
‖ r ‖ d‖ r ‖ (see Exercise 57, Section 14.1)
=12(‖r(b)‖2 − ‖r(a)‖2
)
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922 SECTION 18.1
22. (a) r(u) = (1 − 2u) i;∫C1
h ·dr =∫ 1
0
(1 − 2u)2 i · (−2 i) du =∫ 1
0
−2(1 − 2u)2 du = −23
(b)∫C2
h ·dr =∫ 1
0
(i + u j) · j du +∫ 1
0
[(1 − 2u)2i + j] · (−2i) du +∫ 1
0
[i + (1 − u)j] · (−j) du
=∫ 1
0
u du +∫ 1
0
−2(1 − 2u)2 du +∫ 1
0
−(1 − u) du = −23
(c) r(u) = cosu i + sinu j, u ∈ [0, π]∫C3
h ·dr =∫ π
0
(cos2 u i + sinuj) · (− sinui + cosu j) du =∫ π
0
(− sinu cos2 u + sinu cosu) du = −23
23.∫C
f(r) · dr =∫ b
a
[f(r (u)) · r′(u)] du =∫ b
a
[f(u) i · i] du =∫ b
a
f(u) du
24. Follows from the linearity of the dot product and of ordinary integrals.
25. E : r(u) = a cosu i + b sinu j, u ∈ [ 0, 2π ]
W =∫ 2π
0
[(−1
2b sinu
)(−a sinu) +
(12a cosu
)(b cosu)
]du =
12
∫ 2π
0
ab du = πab
If the ellipse is traversed in the opposite direction, then W = −πab. In both cases |W | = πab = area
of the ellipse.
26. force at time t: mr′′(t) = 2mβ j
work during time interval: W =∫ 1
0
4mβ2t dt = 2mβ2
27. r(t) = αt i + βt2j + γt3 k
r′(t) = α i + 2βt j + 3γt2 k
force at time t = mr′′(t) = m(2βj + 6γtk)
W =∫ 1
0
[m(2β j + 6γtk) · (α i + 2βtj + 3γt2 k)] dt
= m
∫ 1
0
(4β2t + 18γ2t3) dt =(
2β2 +92γ2
)m
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SECTION 18.1 923
28. (a) v ⊥ k, v ⊥ r, ‖ v ‖= ω and ωk, r, v, form a right-handed triple
(b) We can parametrize C counterclockwise by
r(t) = a cos t i + a sin t j, 0 ≤ t ≤ 2π.
Thenr′(t) = −a sin t i + a cos t j
and ∫C
(ω k× r) ·dr =∫ 2π
0
(ω k× r(t)) · r′(t) dt.
Now
k× r(t) = a cos t j − a sin t i.
So
(k× r(t)) · r′(t) = a2(cos2 t + sin2 t) = a2.
Thus ∫C
(ωk× r) · dr +∫ 2π
0
ωa2 dt = ωa2(2π) = 2ω(πa2) = 2ωA.
If C is parametrized clockwise, the circulation is −2ωA.
29. Take C : r(t) = r cos t i + r sin t j, t ∈ [ 0, 2π ]∫C
v(r) ·dr =∫ 2π
0
[v(r(t)) · r′(t)] dt
=∫ 2π
0
[f(x(t), y(t)) r(t) · r′(t)] dt
=∫ 2π
0
f(x(t), y(t)) [r (t) · r′(t)] dt = 0
since for the circle r(t) · r′(t) = 0 identically. The circulation is zero.
30. (a) r(u) = i + u j; W =∫ 2
0
k
1 + u2(i + u j) · j du =
∫ 2
0
ku
1 + u2du =
k
2ln 5
(b) r(u) = ui + j; W =∫ 1
0
k
u2 + 1(ui + j) · i du =
∫ 1
0
ku
u2 + 1du =
k
2ln 2.
31. (a) r(u) = (1 − u)(i + 2k) + u(i + 3 j + 2k) = i + 3u j + 2k, u ∈ [ 0, 1 ].∫C
F(r) ·dr =∫ 1
0
9uk(5 + 9u2)3/2
du =[ −k√
5 + 9u2
]10
=k√5− k√
14
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924 SECTION 18.2
(b) Let C be an arc on the sphere ‖r‖ = r = 5.
∫C
F(r) ·dr =∫C2
kr
‖ r ‖3 · dr
=k
53
∫C2
r ·dr =k
53
∫C2
‖ r ‖ d‖ r ‖ (see Exercise 57, Section 14.1)
=k
53
[12‖ r ‖2
](0,4,3)(3,4,0)
= 0
32. Let (x0, y0, z0) and (x1, y1, z1) be the coordinates of a and b, respectively. Then
W =k√
x21 + y2
1 + z21
− k√x2
0 + y20 + z2
0
33. r(u) = u i + αu(1 − u) j, r′(u) = i + α(1 − 2u) j, u ∈ [ 0, 1 ]
W (α) =∫C
F(r) ·dr =∫ 1
0
[(α2u2(1 − u)2 + 1
]+ [u + αu(1 − u)]α(1 − 2u)] dx
=∫ 1
0
[1 + (α + α2)u− (2α + 2α2)u2 + α2u4
]du = 1 − 1
6α +
130
α2
W ′(α) = − 16
+115
α =⇒ α =156
=52
The work done by F is a minimum when α = 5/2.
34. Suppose that C is the curve r(u), a ≤ u ≤ b.
∫C
∇f · dr =∫ b
a
∇f(r(u)) · r′(u) du =∫ b
a
df
dudu =
[f(r(u))
]ba
= f(r(b)) − f(r(a)).
SECTION 18.2
1. h(x, y) = ∇f(x, y) where f(x, y) = 12 (x2 + y2)
C is closed =⇒∫C
h(r) ·dr = 0
2. x i + y j is a gradient (Exercise 1); we need integrate only y i.
∫C
h(r) ·dr =∫ 2π
0
y(t)x′(t) dt =∫ 2π
0
(b sin t)(−a sin t) dt = −πab
3. h(x, y) = ∇f(x, y) where f(x, y) = x cosπy; r(0) = 0, r(1) = i − j∫C
h(r) ·dr =∫C
∇f(r) ·dr = f(r(1)) − f(r(0)) = f(1,−1) − f(0, 0) = −1
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SECTION 18.2 925
4. h = ∇f with f(x, y) =x3
3+
y3
3− xy, and C is closed, so
∫C
h ·dr = 0
5. h(x, y) = ∇f(x, y) where f(x, y) = 12x
2y2; r(0) = j, r(1) = −j∫C
h(r) ·dr =∫C
∇f(r) ·dr = f(r(1)) − f(r(0)) = f(0,−1) − f(0, 1) = 0 − 0 = 0
6. ey i + xey j is a gradient; we need integrate only i − x j
C = C1 ∪ C2 ∪ C3 ∪ C4 where
C1 : r(u) = (2u− 1)i − j, u ∈ [0, 1]
C2 : r(u) = i + (2u− 1)j, u ∈ [0, 1]
C3 : r(u) = (1 − 2u)i + j, u ∈ [0, 1]
C4 : r(u) = −i + (1 − 2u)j, u ∈ [0, 1]∫C
=∫C1
+∫C2
+∫C3
+∫C4
= 2 + (−2) + (−2) + (−2) = −4
7. h(x, y) = ∇f(x, y) where f(x, y) = x2y − xy2; r(0) = i, r(π) = −i∫C
h(r) ·dr =∫C
∇f(r) ·dr = f(r(π)) − f(r(0)) = f(−1, 0) − f(1, 0) = 0 − 0 = 0
8. h(x, y) = ∇f(x, y) where f(x, y) = (x2 + y4)3/2∫C
h(r) · dr =∫C
∇f(r) ·dr = f(−1, 0) − f(1, 0) = 1 − 1 = 0
9. h(x, y) = ∇f(x, y) where f(x, y) = (x2 + y4)3/2∫C
h(r) ·dr =∫C
∇f(r) ·dr = f(1, 0) − f(−1, 0) = 1 − 1 = 0
10. h = ∇f with f(x, y) = coshx2y; and C is closed, so∫C
h ·dr = 0
11. h(x, y) is not a gradient, but part of it,
2x cosh y i + (x2 sinh y − y)j,
is a gradient. Since we are integrating over a closed curve, the contribution of the gradient part is 0.
Thus
∫C
h(r) ·dr =∫C
(−yi) ·dr.
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926 SECTION 18.2
C1 : r(u) = i + (−1 + 2u)j, u ∈ [ 0, 1 ]
C2 : r(u) = (1 − 2u)i + j, u ∈ [ 0, 1 ]
C3 : r(u) = −i + (1 − 2u)j, u ∈ [ 0, 1 ]
C4 : r(u) = (−1 + 2u)i − j, u ∈ [ 0, 1 ]
∫C
h(r) · dr =∫C1
(−y i) · dr +∫C2
(−y i) · dr +∫C3
(−y i) · dr +∫C4
(−y i) · dr
= 0 +∫ 1
0
−i · (−2 i) du + 0 +∫ 1
0
i · (2 i) du
= 0 +∫ 1
0
2 du + 0 +∫ 1
0
2 du
= 4
12. h(x, y) = ∇(x2y2
2
)
(a)∫ 2
0
(u5 i + u4 j) · (i + 2u j) du =∫ 2
0
3u5 du = 32
(b) f(2, 4) − f(0, 0) = 32 − 0 = 32
13. h(x, y) = (3x2y3 + 2x) i + (3x3y2 − 4y) j;∂P
∂y= 9x2y2 =
∂Q
∂x. Thus h is a gradient.
(a) r(u) = u i + eu j, r′(u) = i + eu j, u ∈ [ 0, 1 ]∫C
h(r) ·dr =∫ 1
0
[(3u2e3u + 2u) + 3u3e3u − 4e2u)
]du =
[u3e3u + u2 − 2e2u
]10
= e3 − 2e2 + 3
(b)∂f
∂x= 3x2y3 + 2x =⇒ f(x, y) = x3y3 + x2 + g(y);
∂f
∂y= 3x3y2 + g′(y) = 3x3 − 4y =⇒ g′(y) = −4y =⇒ g(y) = −2y2
Therefore, f(x, y) = x3y3 + x2 − 2y2.
Now, at u = 0, r(0) = 0 i + j = (0, 1); at u = 1, r(1) = i + e j = (1, e) and∫C
h(r) ·dr =[x3y3 + x2 − 2y2
](1,e)(0,1)
= e3 − 2e2 + 3
14. h(x, y) = ∇(x2 sin y − ex)
(a)∫ π
0
[(2 cosu sinu− ecosu) i + (cos2 u cosu)j
]· (− sinu i + j) du = e− e−1
(b) f(−1, π) − f(1, 0) = e− e−1
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SECTION 18.2 927
15. h(x, y) = (e2y − 2xy) i + (2xe2y − x2 + 1) j;∂P
∂y= 2e2y − 2x =
∂Q
∂x. Thus h is a gradient.
(a) r(u) = ueu i + (1 + u) j, r′(u) = (1 + u)eu i + j, u ∈ [ 0, 1 ]∫C
h(r) ·dr =∫ 1
0
[e2(3ue3u + e3u − 2u3e2u − 5u2e2u − 2ue2u + 1
]du
=[e2ue3u − u3e2u − u2e2u + u
]10
= e5 − 2e2 + 1
(b)∂f
∂x= e2y − 2xy =⇒ f(x, y) = xe2y − x2y + g(y).
∂f
∂y= 2xe2y − x2 + g′(y) = 3x3 − 4y =⇒ g′(y) = 1 =⇒ g(y) = y
Therefore, f(x, y) = xe2y − x2y + y.
Now, at u = 0, r(0) = 0 i + j = (0, 1); at u = 1, r(1) = e i + 2 j = (e, 2) and∫C
h(r) ·dr =[xe2y − x2y + y
](e,2)(0,1)
= e5 − 2e2 + 1
16. h(x, y, z) = ∇f with f(x, y, z) = xy2z3
∫C
h ·dr = f(1, 1, 1) − f(0, 0, 0) = 1
17. h(x, y, z) = (2xz + sin y) i + x cos y j + x2 k;
∂P
∂y= cos y =
∂Q
∂x,
∂P
∂z= 2x =
∂R
∂x,
∂Q
∂z= 0 =
∂R
∂y. Thus h is a gradient.
∂f
∂x= 2xz + sin y, =⇒ f(x, y, z) = x2z + x sin y + g(y, z)
∂f
∂y= x cos y +
∂g
∂y= x cos y, =⇒ g(y, z) = h(z) =⇒ f(x, y, z) = x2z + x sin y + h(z)
∂f
∂z= x2 + h′(z) = x2 =⇒ h′(z) = 0 =⇒ h(z) = C
Therefore, f(x, y, z) = x2z + x sin y (take C = 0)
∫C
h(r) · dr =∫C
∇f · dr =[x2z + x sin y
]r(2π)
r(0)=[x2z + x sin y
](1,0,2π)
(1,0,0)= 2π
18. h(x, y, z) = ∇f with f(x, y, z) = yz sinπx∫C
h · dr = f(
12 ,
√3
2 , π3
)− f(1, 0, 0) =
16π√
3
19. h(x, y, z) = (2xy + z2) i + x2 j + 2xz k;
∂P
∂y= 2x =
∂Q
∂x,
∂P
∂z= 2z =
∂R
∂x,
∂Q
∂z= 0 =
∂R
∂y.Thus h is a gradient.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-18 JWDD027-Salas-v1 January 4, 2007 18:12
928 SECTION 18.2
∂f
∂x= 2xy + z2 =⇒ f(x, y, z) = x2y + xz2 + g(y, z)
∂f
∂y= x2 +
∂g
∂y= x2 =⇒ g(y, z) = h(z) =⇒ f(x, y, z) = x2y + xz2 + h(z)
∂f
∂z= 2xz + h′(z) = 2xz =⇒ h′(z) = 0 =⇒ h(z) = C
Therefore, f(x, y, z) = x2y + xz2 (take C = 0)
∫C
h(r) · dr =∫C
∇f · dr =[x2y + xz2
]r(1)r(0)
=[x2y + xz2
](2,3,−1)
(0,2,0)= 14
20. h(x, y, z) = ∇f with f(x, y, z) = z3 − e−x ln y∫C
h ·dr = f(2, e2, 2) − f(1, 1, 1) = 7 − 2e−2
21. F(x, y) = (x + e2y) i + (2y + 2xe2y) j;∂P
∂y= 2e2y =
∂Q
∂x. Thus F is a gradient.
∂f
∂x= x + e2y =⇒ f(x, y) =
12x2 + xe2y + g(y);
∂f
∂y= 2xe2y + g′(y) = 2y + 2xe2y =⇒ g′(y) = 2y =⇒ g(y) = y2 (take C = 0)
Therefore, f(x, y) =12x2 + xe2y + y2.
∫C
F(r) ·dr =∫C
∇f ·dr =[12x2 + xe2y + y2
]r(2π)
r(0)
=[12x2 + xe2y + y2
](3,0)(3,0)
= 0
22. F = ∇f with f(x, y, z) = x2 ln y − xyz W = f(3, 2, 2) − f(1, 2, 1) = 8 ln 2 − 10
23. Set f(x, y, z) = g(x) and C : r(u) = u i, u ∈ [ a, b ].
In this case
∇f(r(u)) = g′(x(u))i = g′(u)i and r′(u) = i,
so that ∫C
∇f(r) ·dr =∫ b
a
[∇f(r(u)) · r′(u)] du =∫ b
a
g′(u) du.
Since f(r(b)) − f(r(a)) = g(b) − g(a),∫C
∇f(r) ·dr = f(r(b)) − f(r(a)) gives∫ b
a
g′(u) du = g(b) − g(a).
24. F(x, y, z) =k
(x2 + y2 + z2)n/2(x i + y j + z j) = ∇f
(a) n = 2 : f(r) = k ln r + C (b) n �= 2 : f(r) = −(
k
n− 2
)1
rn−2+ C
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-18 JWDD027-Salas-v1 January 4, 2007 18:12
SECTION 18.2 929
25. F(r) = kr r = k√x2 + y2 + z2(x i + y j + z k), k > 0 constant.
∂P
∂y=
kxy√x2 + y2 + z2
=∂Q
∂x,
∂P
∂z=
kxz√x2 + y2 + z2
=∂R
∂x
∂Q
∂z=
kyz√x2 + y2 + z2
=∂R
∂y
Therefore, F is a gradient field.∂f
∂x= kx
√x2 + y2 + z2 =⇒ f(x, y, z) =
k
3(x2 + y2 + z2
)3/2 + g(y, z).
∂f
∂y= ky
√x2 + y2 + z2 +
∂g
∂y= ky
√x2 + y2 + z2 =⇒ f(x, y, z) =
k
3(x2 + y2 + z2
)3/2 + h(z)
∂f
∂z= kz
√x2 + y2 + z2 + h′(z) = kz
√x2 + y2 + z2 =⇒ h(z) = C, constant
Therefore, f(x, y, z) = k3
(x2 + y2 + z2
)3/2 + C.
26. Set f(x, y, z) = 12
∫ x2+y2+z2
0
g(u) du. Then
∇f = g(x2 + y2 + z2) [x i + y j + xk] = F(r)
27. F(r) = ∇(mG
r
); W =
∫C
F(r) ·dr = mG
(1r2
− 1r1
)
28. (a) Since the denominator is never 0 in Ω, P and Q are continuously differentiable on Ω.
∂P
∂y=
x2 − y2
(x2 + y2)2=
∂Q
∂x.
(b) Take r(u) = 12 cosu i + 1
2 sinu j.∫C
h ·dr =∫ 2π
0
( 12 sinu
1/4i −
12 cosu1/4
j)
·(−1
2sinu i +
12
cosu j)
du =∫ 2π
0
− du = −2π
Therefore h is not a gradient since the integral over C (a closed curve) is not zero.
(c) Ω : 0 < x2 + y2 < 1 is an open plane region but is not simply connected.
29. F(x, y, z) = 0 i + 0 j +−mGr2
0
(r0 + z)2k;
∂P
∂y= 0 =
∂Q
∂x,
∂P
∂z= 0 =
∂R
∂x,
∂Q
∂z= 0 =
∂R
∂y.
Therefore, F(x, y, z) is a gradient.
∂f
∂x= 0 =⇒ f(x, y, z) = g(y, z);
∂f
∂y=
∂g
∂y= 0 =⇒ g(y, z) = h(z).
Therefore f(x, y, z) = h(z).
Now∂f
∂z= h′(z) =
−mGr20
(r0 + z)2=⇒ f(x, y, z) = h(z) =
mGr20
r0 + z
30. W = f(x, y, 0) − f(x, y, 300) = mGr0 −mGr0
2
r0 + 300= 279.07mG.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-18 JWDD027-Salas-v1 January 4, 2007 18:12
930 SECTION 18.3
SECTION 18.3
1. If f is continuous, then −f is continuous and has antiderivatives u. The scalar fields U(x, y, z) = u(x)
are potential functions for F:
∇U =∂U
∂xi +
∂U
∂yj +
∂U
∂zk =
du
dxi = −f i = −F.
2. d
dt
(12mv2
)=
d
dt
[12m(v ·v)
]= m(v ·a) = v ·ma
= v ·F = v · ec
[v×B] = 0
3. The scalar field U(x, y, z) = αz + d is a potential energy function for F. We know that the total
mechanical energy remains constant. Thus, for any times t1 and t2,
12m[v (t1)]2 + U(r(t1)) = 1
2m[v (t2)]2 + U(r(t2)).
This gives
12m[v (t1)]2 + αz(t1) + d = 1
2m[v (t2)]2 + αz(t2) + d.
Solve this equation for v (t2) and you have the desired formula.
4. Throughout the motion, the total mechanical energy of the object remains constant:
12mv2 − GmM
r= E.
At firing v = v0, r = Re = the radius of the earth and we have
12mv0
2 − GmM
Re= E.
As r → ∞, v → 0 (by assumption) and also −GmM/r → 0.
Thus E = 0 and we have
12mv0
2 =GmM
Reand v0 =
√2GM
Re.
(Note that v0 is independent of the mass of the projectile.)
5. (a) We know that −∇U points in the direction of maximum decrease of U . Thus F = −∇U attempts
to drive objects toward a region where U has lower values.
(b) At a point where u has a minimum, ∇U = 0 and therefore F = 0.
6. We have x(0) = 2, x′(0) = v(0) = 1. Inserting these values in the formula for E we have
E =12m + 2λ.
Since E = 12mv2 + 1
2λx2 is constant, the maximum value of v comes when x = 0. Then
E =12mv2 =
12m + 2λ and v =
√1 + 4λ/m.
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-18 JWDD027-Salas-v1 January 4, 2007 18:12
SECTION 18.4 931
The maximum value of x comes when v = 0 (at the endpoints of the oscillation). Then
E =12λx2 =
12m + 2λ and x =
√m/λ + 4.
7. (a) By conservation of energy 12mv2 + U = E. Since E is constant and U is constant, v is constant.
(b) ∇U is perpendicular to any surface where U is constant. Obviously so is F = −∇U .
8. F(r) =k
r2r = ∇f where f(r) = k ln r
9. f(x, y, z) = − k√x2 + y2 + z2
is a potential function for F. The work done by F moving an object
along C is:
W =∫C
F(r) ·dr =∫ b
a
∇f ·dr = f [r(b)] − f [r(a)].
Since r(a) = (x0, y0, z0) and r(b) = (x1, y1, z1) are points on the unit sphere,
f [r(b)] = f [r(a)] = −k and so W = 0
SECTION 18.4
1. r(u) = u i + 2u j, u ∈ [ 0, 1 ]∫C
(x− 2y) dx + 2x dy =∫ 1
0
{[x(u) − 2y(u)]x′(u) + 2x(u) y′(u)} du =∫ 1
0
u du =12
2. r(u) = ui + 2u2 j, u ∈ [0, 1]∫C
(x− 2y) dx + 2x dy =∫ 1
0
{[x(u) − 2y(u)]x′(u) + 2x(u)y′(u)} du
=∫ 1
0
(u + 4u2) du =116
3. C = C1 ∪ C2
C1 : r(u) = u i, u ∈ [ 0, 1 ]; C2 : r(u) = i + 2u j, u ∈ [ 0, 1 ]∫C1
(x− 2y) dx + 2x dy =∫C1
x dx =∫ 1
0
x(u)x′(u) du =∫ 1
0
u du =12∫
C2
(x− 2y) dx + 2x dy =∫C2
2x dy =∫ 1
0
4 du = 4∫C
=∫C1
+∫C2
=92
4. C = C1 ∪ C2
C1 : r(u) = 2u j, u ∈ [0, 1]; C2 : r(u) = u i + 2 j, u ∈ [0, 1]∫C1
(x− 2y) dx + 2x dy =∫C1
0 dy = 0
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-18 JWDD027-Salas-v1 January 4, 2007 18:12
932 SECTION 18.4∫C2
(x− 2y) dx + 2x dy =∫C2
(x− 4) dx =∫ 1
0
(x(u) − 4)x′(u) du =∫ 1
0
(u− 4) du = −72∫
C
=∫C1
+∫C2
= −72
5. r(u) = 2u2 i + u j, u ∈ [ 0, 1 ]
∫C
y dx + xy dy =∫ 1
0
[y(u)x′(u) + x(u) y(u) y′(u)] du =∫ 1
0
(4u2 + 2u3) du =116
6. r(u) = 2u i + u j, u ∈ [0, 1]∫C
y dx + xy dy =∫ 1
0
[y(u)x′(u) + x(u)y(u)y′(u)] du
=∫ 1
0
(2u + 2u2) du =53
7. C = C1 ∪ C2 C1 : r(u) = u j, u ∈ [ 0, 1 ]; C2 : r(u) = 2u i + j, u ∈ [ 0, 1 ]∫C1
y dx + xy dy = 0
∫C2
y dx + xy dy =∫C2
y dx =∫ 1
0
y(u)x′(u) du =∫ 1
0
2 du = 2
∫C
=∫C1
+∫C2
= 2
8. r(u) = 2u3 i + u j, u ∈ [0, 1]∫C
y dx + xy dy =∫ 1
0
[y(u)x′(u) + x(u)y(u)y′(u)] du
=∫ 1
0
(6u3 + 2u4) du =1910
9. r(u) = 2u i + 4u j, u ∈ [ 0, 1 ]∫C
y2 dx + (xy − x2) dy =∫ 1
0
{y2(u)x′(u) +
[x(u)y(u) − x2(u)
]y′(u)
}du
=∫ 1
0
[(4u)2(2) + (8u2 − 4u2)(4)
]du =
∫ 1
0
48u2 du = 16
10. r(u) = u i + u2 j, u ∈ [0, 2]∫C
y2 dx + (xy − x2) dy =∫ 2
0
[y2(u)x′(u) + (x(u)y(u) − x2(u))y′(u)] du
=∫ 2
0
(3u4 − 2u3) du =565
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-18 JWDD027-Salas-v1 January 4, 2007 18:12
SECTION 18.4 933
11. r(u) =18u2 i + u j, u ∈ [ 0, 4 ]
∫C
y2 dx + (xy − x2) dy =∫ 4
0
{y2(u)x′(u) +
[x(u)y(u) − x2(u)
]y′(u)
}du
=∫ 4
0
[u2(u
4
)+
(u2
8(u) −
(u2
8
)2
(1)
)]du
=∫ 4
0
[38u3 − 1
64u4
]du =
1045
12. C = C1 ∪ C2 C1 : r(u) = 2u i, u ∈ [0, 1]; C2 : r(u) = 2 i + 4u j, u ∈ [0, 1]∫C1
y2 dx + (xy − x2) dy =∫C1
0 dx = 0
∫C2
y2 dx + (xy − x2) dy =∫C2
(2y − 4) dy =∫ 1
0
[2y(u) − 4]y′(u) du =∫ 1
0
16(2u− 1) du = 0∫C
=∫C1
+∫C2
= 0
13. r(u) = u i + u j, u ∈ [ 0, 1 ]
∫C
(y2 + 2x + 1) dx + (2xy + 4y − 1) dy
=∫ 1
0
{[y2(u) + 2x(u) + 1]x′(u) + [2x(u)y(u) + 4y(u) − 1]y′(u)
}du∫ 1
0
[(u2 + 2u + 1) + (2u2 + 4u− 1)
]du =
∫ 1
0
(3u2 + 6u
)du = 4
14. r(u) = ui + u2j, u ∈ [0, 1].∫C
(y2 + 2x + 1) dx + (2xy + 4y − 1) dy
=∫ 1
0
[(y2(u) + 2x(u) + 1)x′(u) + (2x(u)y(u) + 4y(u) − 1)y′(u)
]du
=∫ 1
0
(5u4 + 8u3 + 1) du = 4
15. r(u) = u i + u3 j, u ∈ [ 0, 1 ]∫C
(y2 + 2x + 1) dx + (2xy + 4y − 1) dy
=∫ 1
0
{[y2(u) + 2x(u) + 1]x′(u) + [2x(u)y(u) + 4y(u) − 1]y′(u)
}du
=∫ 1
0
[(u6 + 2u + 1) + (2u4 + 4u3 − 1)3u2
]du =
∫ 1
0
(7u6 + 12u5 − 3u2 + 2u + 1
)du = 4
16. C = C1 ∪ C2 ∪ C3
C1 : r(u) = 4u i, u ∈ [0, 1]; C2 : r(u) = 4 i + 2u j, u ∈ [0, 1];
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-18 JWDD027-Salas-v1 January 4, 2007 18:12
934 SECTION 18.4
C3 : r(u) = (4 − 3u)i + (2 − u)j, u ∈ [0, 1]∫C1
=∫C1
(y2 + 2x + 1) dx =∫ 1
0
4(8u + 1) du = 20
∫C2
=∫C2
(8y + 4y − 1) dy =∫ 1
0
2(24u− 1) du = 22
∫C3
=∫ 1
0
{−3[(2 − u)2 + 2(4 − 3u) + 1
]− [2(4 − 3u)(2 − u) + 4(2 − u) − 1]
}du
=∫ 1
0
(−9u2 + 54u− 62) du = −38.∫C
=∫C1
+∫C2
+∫C3
= 20 + 22 − 38 = 4.
17. r(u) = u i + u j + uk, u ∈ [ 0, 1 ]∫C
y dx + 2z dy + x dz =∫ 1
0
[y(u)x′(u) + 2z(u) y′(u) + x(u) z′(u)] du =∫ 1
0
4u du = 2
18.∫C
y dx + 2z dy + x dz =∫ 1
0
[y(u)x′(u) + 2z(u)y′(u) + x(u)z′(u)] du
=∫ 1
0
(u2 + 3u3 + 4u4) du =11360
19. C = C1 ∪ C2 ∪ C3
C1 : r(u) = uk, u ∈ [ 0, 1 ]; C2 : r(u) = u j + k, u ∈ [ 0, 1 ]; C3 : r (u) = u i + j + k, u ∈ [ 0, 1 ]∫C1
y dx + 2z dy + x dz = 0
∫C2
y dx + 2z dy + x dz =∫C2
2z dy =∫ 1
0
2z(u) y′(u) du =∫ 1
0
2 du = 2
∫C3
y dx + 2z dy + x dz =∫C3
y dx =∫ 1
0
y(u)x′(u) du =∫ 1
0
du = 1
∫C
=∫C1
+∫C2
+∫C3
= 3
20. C = C1 ∪ C2 ∪ C3
C1 : r(u) = u i, u ∈ [0, 1]; C2 : r(u) = i + u j, u ∈ [0, 1]; C3 : r(u) = i + j + uk, u ∈ [0, 1]∫C1
y dx + 2z dy + x dz = 0∫C2
y dx + 2z dy + x dz = 0
∫C3
y dx + 2z dy + x dz =∫C3
x dz =∫ 1
0
du = 1∫C
y dx + 2z dy + x dz = 0 + 0 + 1 = 1
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-18 JWDD027-Salas-v1 January 4, 2007 18:12
SECTION 18.4 935
21. r(u) = 2u i + 2u j + 8uk, u ∈ [ 0, 1 ]∫C
xy dx + 2z dy + (y + z) dz
=∫ 1
0
{x(u)y(u)x′(u) + 2z(u)y′(u) + [y(u) + z(u)]z′(u)} du
=∫ 1
0
[(2u)(2u)(2) + 2(8u)(2) + (2u + 8u)(8)] du
=∫ 1
0
(8u2 + 112u
)du =
1763
22. C = C1 ∪ C2 ∪ C3
C1 : r(u) = 2u i, u ∈ [0, 1]; C2 : r(u) = 2 i + 2u j, u ∈ [0, 1];
C3 : r(u) = 2 i + 2 j + 8uk, u ∈ [0, 1]∫C1
xy dx + 2z dy + (y + z) dz = 0∫C2
xy dx + 2z dy + (y + z) dz = 0
∫C3
xy dx + 2z dy + (y + z) dz =∫C2
(y + z) dz =∫ 1
0
8(2 + 8u) du = 48
23. r(u) = u i + u j + 2u2 k, u ∈ [ 0, 2 ]∫C
xy dx + 2z dy + (y + z) dz
=∫ 2
0
{x(u)y(u)x′(u) + 2z(u)y′(u) + [y(u) + z(u)]z′(u)} du
=∫ 2
0
[(u)(u)(1) + 2(2u2)(1) + (u + 2u2)(4u)
]du
=∫ 2
0
(8u3 + 9u2
)du = 56
24. C = C1 ∪ C2
C1 : r(u) = 2u i + 2u j + 2uk, u ∈ [0, 1]; C2 : r(u) = 2 i + 2 j + (2 + 6u)k, u ∈ [0, 1].∫C1
xy dx + 2z dy + (y + z) dz =∫ 1
0
8(u2 + 2u) du =323∫
C2
xy dx + 2z dy + (y + z) dz =∫C2
(y + z) dz =∫ 1
0
6(4 + 6u) du = 42∫C
=∫C1
+∫C2
=1583
25. r(u) = (u− 1) i + (1 + 2u2) j + uk, u ∈ [ 1, 2 ]∫C
x2y dx + y dy + xz dz
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-18 JWDD027-Salas-v1 January 4, 2007 18:12
936 SECTION 18.4
=∫ 2
1
[x2(u)y(u)x′(u) + y(u)y′(u) + x(u)z(u)z′(u)
]du
=∫ 2
1
[(u− 1)2(1 + 2u2)(1) + (1 + 2u2)(4u) + (u− 1)u
]du
=∫ 2
1
(2u4 + 4u3 + 4u2 + u + 1
)du =
117730
26. r(u) =(
2 − u2
2
)i + u
√1 − u2
4j + uk, u ∈ [0, 2]
∫C
y dx + yz dy + z(x− 1) dz =∫ 2
0
[−u2
√1 − u2
4+
u2
2(2 − u2) + u(1 − u2
2)
]du = −π
2− 8
15
27. (a)∂P
∂y= 6x− 4y =
∂Q
∂x
∂f
∂x= x2 + 6xy − 2y2 =⇒ f(x, y) =
13x3 + 3x2y − 2xy2 + g(y)
∂f
∂y= 3x2 − 4xy + g′(y) = 3x2 − 4xy + 2y =⇒ g′(y) = 2y =⇒ g(y) = y2 + C
Therefore, f(x, y) =13x3 + 3x2y − 2xy2 + y2 (take C = 0)
(b) (i)∫C
(x2 + 6xy − 2y2) dx + (3x2 − 4xy + 2y) dy = [f(x, y)](0,4)(3,0) = 7
(ii)∫ ′
C
(x2 + 6xy − 2y2) dx + (3x2 − 4xy + 2y) dy = [f(x, y)](0,3)(4,0) = − 373
28. (a) F = ∇f where f(x, y, z) = x2y + xz2 − y2z
(b) (i)∫C
(2xy + z2) dx + (x2 − 2yz) dy + (2xz − y2) dz = f(3, 2,−1) − f(1, 0, 1) = 25 − 1 = 24
(ii)∫C ′
(2xy + z2) dx + (x2 − 2yz) dy + (2xz − y2) dz = f(1, 0, 1) − f(3, 2 − 1) = −24
29. s′(u) =√
[x′(u)]2 + [y′(u)]2 = a
(a) M =∫C
k(x + y) ds = k
∫ π/2
0
[x(u) + y(u)] s′(u) du = ka2
∫ π/2
0
(cosu + sinu) du = 2ka2
xMM =∫C
kx(x + y) ds = k
∫ π/2
0
x(u) [x(u) + y(u)] s′(u) du
= ka3
∫ π/2
0
(cos2 u + cosu sinu) du =14ka3(π + 2)
yMM =∫C
ky(x + y) ds = k
∫ π/2
0
y(u) [x(u) + y(u)] s′(u) du
= ka3
∫ π/2
0
(sinu cosu + sin2 u) du =14ka3(π + 2)
xM = yM = 18a(π + 2)
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SECTION 18.4 937
(b)I =∫C
k(x + y)y2 ds = k
∫ π/2
0
[x(u)y2(u) + y3(u)
]s′(u) du
= ka4
∫ π/2
0
[sin2 u cos u + sin3 u
]du
= ka4
∫ π/2
0
[sin2 u cos u + (1 − cos2 u) sin u
]du
= ka4[
13 sin3 u− cos u + 1
3 cos3 u]π/20
= ka4
I = 12a
2M .
30. (a) I =∫C
M
Lx2 ds =
Ma2
2π
∫ 2π
0
cos2 u du =12Ma2
(b) I =∫C
M
La2 ds =
Ma
2π
∫C
ds = Ma2
31. (a) Iz =∫C
k(x + y)a2 ds = a2
∫C
k(x + y) ds = a2M = Ma2
(b) The distance from a point (x∗, y∗) to the line y = x is |y∗ − x∗|/√
2. Therefore
I =∫C
k(x + y)[12(y − x)2
]ds =
12k
∫ π/2
0
(a cosu + a sinu)(a sinu− a cosu)2a du
=12ka4
∫ π/2
0
(sinu− cosu)2d
du(sinu− cosu) du
=12ka4
[13(sinu− cosu)3
]π/20
=13ka4.
From Exercise 29, M = 2ka2. Therefore
I = 16 (2ka2)a2 = 1
6Ma2.
32. (a) M =∫C
k ds =∫ 2π
0
k
√sin2 u + (1 − cosu)2 du =
∫ 2π
0
2k sin12u du = 8k
(b) xMM =∫C
kx ds =∫ 2π
0
[(1 − cosu)(2k sin
12u)]du
= 4k∫ 2π
0
sin3 12u du =
323k; xM =
43
yMM =∫C
ky ds =∫ 2π
0
[(u− sinu)(2k sin
12u)]du
= 2k∫ 2π
0
(u sin12u− 2 sin2 1
2u cos
12u) du
= 8πk; yM = π
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JWDD027-18 JWDD027-Salas-v1 January 4, 2007 18:12
938 SECTION 18.4
33. (a) s′(u) =√a2 + b2
L =∫C
ds =∫ 2π
0
√a2 + b2 du = 2π
√a2 + b2
(b) xM = 0, yM = 0 (by symmetry)
zM =1L
∫C
z ds =1
2π√a2 + b2
∫ 2π
0
bu√a2 + b2 du = bπ
(c) Ix =∫C
M
L(y2 + z2) ds =
M
2π
∫ 2π
0
(a2 sin2 u + b2u2) du =16M(3a2 + 8b2π2)
Iy = 16M(3a2 + 8b2π2) similarly
Iz = Ma2 (all the mass is at distance a from the z-axis)
34. (a) s′(u) = 2u2 + 1
L =∫C
ds =∫ a
0
(2u2 + 1) du =23a3 + a =
a(2a2 + 3)3
(b) xM =1L
∫C
x ds =3
a(2a2 + 3)
∫ a
0
(2u3 + u) du =3a(a2 + 1)2(2a2 + 3)
yM =1L
∫C
y ds =3
a(2a2 + 3)
∫ a
0
(2u4 + u2) du =a2(6a2 + 5)5(2a2 + 3)
zM =1L
∫C
z ds =3
a(2a3 + 3)
∫ a
0
(43u5 +
23u3
)du =
a3(4a2 + 3)6(2a3 + 3)
(c) Iz =M
L
∫C
(x2 + y2) ds =3M
a(2a3 + 3)
∫ a
0
[(u2 + u4)(2u2 + 1)] du
=Ma2(30a4 + 63a2 + 35)
35(2a2 + 3)
35. M =∫C
k(x2 + y2 + z2) ds
= k√a2 + b2
∫ 2π
0
(a2 + b2u2) du =23πk√a2 + b2 (3a2 + 4π2b2)
36. C : r = r(u), u ∈ [a, b]
∫C
h(r) ·dr =∫ b
a
[h(r(u)) · r′(u)] du
=∫ b
a
[h(r(u)) · r′(u)
‖ r′(u) ‖
]‖ r′(u) ‖ du
=∫ b
a
[h(r(u)) ·T(r(u))]s′(u) du
=∫C
[h(r) ·T(r)] ds
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JWDD027-18 JWDD027-Salas-v1 January 4, 2007 18:12
SECTION 18.5 939
SECTION 18.5
1. (a) ©∫C
xy dx + x2 dy =∫C1
xy dx + x2 dy +∫C2
xy dx + x2 dy +∫C3
xy dx + x2 dy, where
C1 : r(u) = u i + u j, u ∈ [ 0, 1 ]; C2 : r(u) = (1 − u) i + j, u ∈ [ 0, 1 ]
C3 : r(u) = (1 − u) j, u ∈ [ 0, 1 ].∫C1
xy dx + x2 dy =∫ 1
0
(u2 + u2) du =23∫
C2
xy dx + x2 dy =∫ 1
0
−(1 − u) du = − 12∫
C3
xy dx + x2 dy =∫ 1
0
02(−1) du = 0
Therefore, ©∫C
xy dx + x2 dy =23− 1
2=
16.
(b) ©∫C
xy dx + x2 dy =∫∫
Ω
x dx dy =∫ 1
0
∫ y
0
x dx dy =∫ 1
0
[12x2
]y0
du =12
∫ 1
0
y2 dy =16
2. (a) C = C1 ∪ C3 ∪ C3 ∪ C4
C1 : r(u) = u i, u ∈ [0, 1]; C2 : r(u) = i + u j, u ∈ [0, 1]
C3 : r(u) = (1 − u)i + j, u ∈ [0, 1]; C4 : r(u) = (1 − u)j, u ∈ [0, 1]∫C1
x2y dx + 2y2 dy = 0
∫C2
x2y dx + 2y2 dy =∫C2
2y2 dy =∫ 1
0
2u2 du =23∫
C3
x2y dx + 2y2 dy =∫C3
x2 dx =∫ 1
0
−(1 − u)2 du = −13∫
C4
x2y dx + 2y2 dy =∫C4
2y2 dy =∫ 1
0
−2(1 − u)2 du = −23
©∫C
=∫C1
+∫C2
+∫C3
+∫C4
= −13
(b) ©∫C
x2y dx + 2y2 dy =∫ ∫Ω
[∂
∂x(2y2) − ∂
∂y(x2y)
]dx dy =
∫ 1
0
∫ 1
0
−x2 dx dy = −13
3. (a) C : r(u) = 2 cosu i + 3 sinu j, u ∈ [ 0, 2π ]
©∫C
(3x2 + y) dx + (2x + y3) dy
=∫ 2π
0
[(12 cos2 u + 3 sinu)(− 2 sinu) + (4 cosu + 27 sin3 u)3 cosu
]du
=∫ 2π
0
[−24 cos2 u sinu− 6 sin2 u + 12 cos2 u + 81 sin3 u cosu
]du
=[8 cos3 u− 3u +
32
sin 2u + 6u + 3 sin 2u +814
sin4 u]2π0
= 6π
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940 SECTION 18.5
(b) ©∫C
(3x2 + y) dx + (2x + y3) dy =∫∫
Ω
1 dx dy = area of ellipse Ω = 6π
4. (a) C = C1 ∪ C2
C1 : r(u) = u i + u2 j, u ∈ [0, 1]; C2 : r(u) = (1 − u)i + (1 − u)j, u ∈ [0, 1]∫C1
y2 dx + x2 dy =∫ 1
0
(u4 + 2u3) du =710∫
C2
y2 dx + x2 dy =∫ 1
0
−2(1 − u)2 du = −23; ©∫C
=∫C1
+∫C2
=130
(b) ©∫C
y2 dx + x2 dy =∫ ∫Ω
[∂
∂x(x2) − ∂
∂y(y2)]dx dy =
∫ 1
0
∫ x
x22(x− y) dy dx =
130
5. ©∫C
3y dx + 5x dy =∫ ∫Ω
(5 − 3) dxdy = 2A = 2π
6. ©∫C
5x dx + 3y dy =∫ ∫Ω
0 dx dy = 0
7. ©∫C
x2 dy =∫ ∫Ω
2x dxdy = 2xA = 2(a
2
)(ab) = a2b
8. ©∫C
y2 dx =∫ ∫
Ω
−2y dx dy = −ab2
9. ©∫C
(3xy + y2) dx + (2xy + 5x2) dy =∫ ∫Ω
[(2y + 10x) − (3x + 2y)] dxdy
=∫ ∫Ω
7x dxdy = 7xA = 7(1)(π) = 7π
10. ©∫C
(xy + 3y2) dx + (5xy + 2x2) dy =∫ ∫Ω
(3x− y) dx dy = (3x− y)A = (3 + 2)π = 5π.
11. ©∫C
(2x2 + xy − y2) dx + (3x2 − xy + 2y2) dy =∫ ∫Ω
[(6x− y) − (x− 2y)] dxdy
=∫ ∫Ω
(5x + y) dxdy = (5x + y)A = (5a + 0)(πr2) = 5aπr2
12. ©∫C
(x2 − 2xy + 3y2) dx + (5x + 1) dy =∫ ∫Ω
(5 + 2x− 6y) dx dy = (5 + 2x− 6y)A = (5 − 6b)πr2
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-18 JWDD027-Salas-v1 January 4, 2007 18:12
SECTION 18.5 941
13. ©∫C
ex sin y dx + ex cos y dy =∫ ∫Ω
[ex cos y − ex cos y] dxdy = 0
14. ©∫C
ex cos y dx + ex sin y dy =∫ ∫Ω
2ex sin y dx dy =∫ 1
0
∫ π
0
2ex sin y dy dx = 4(e− 1)
15. ©∫C
2xy dx + x2 dy =∫ ∫Ω
[2x− 2x] dxdy = 0
16. ©∫C
y2 dx + 2xy dy =∫ ∫Ω
0 dx dy = 0
17. C : r(u) = a cosu i + a sinu j; u ∈ [ 0, 2π ]
A = ©∫C
−y dx =∫ 2π
0
(−a sinu)(−a sinu) du = a2
∫ 2π
0
sin2 u du = a2
[12u− 1
4sin 2u
]2π0
= π a2
18. C : r(u) = a cos3 u i + a sin3 u j, u ∈ [0, 2π]
A = ©∫C
−y dx =∫ 2π
0
(−a sin3 u)(−3a cos2 u sinu) du = 3a2
∫ 2π
0
sin4 u cos2 u du =38πa2
19. A = ©∫C
x dy, where C = C1 ∪ C2;
C1 : r(u) = u i +4u
j, 1 ≤ u ≤ 4; C2 : r(u) = (4 − 3u) i + (1 + 3u) j, 0 ≤ u ≤ 1.
©∫C1
x dy =∫ 4
1
u
(−4u2
)du = −4
∫ 4
1
1udu = −4 ln 4;
©∫C2
x dy =∫ 1
0
(4 − 3u)3 du =∫ 1
0
(12 − 9u) du =152
.
Therefore, A = 152 − 4 ln 4.
20. A =12©∫C
x dy − y dx, where C = C1 ∪ C2;
C1 : r(u) =√
5 tanu i +√
5 secu j, tan−1(−2/
√5)≤ u ≤ tan−1
(2/√
5)
C2 : (2 − 4u) i + 3 j, 0 ≤ u ≤ 1
12©∫C1
x dy − y dx =52
ln 5,12©∫C2
x dy − y dx = 6 Therefore, A = 6 + 52 ln 5.
21. ©∫C
(ay + b) dx + (cx + d) dy =∫ ∫Ω
(c− a) dxdy = (c− a)A
22. ©∫C
F(r) · d r =∫ ∫Ω
(−5) dx dy = −5A = −158πa2 (by Exercise 18)
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JWDD027-18 JWDD027-Salas-v1 January 4, 2007 18:12
942 SECTION 18.5
23. We take the arch from x = 0 to x = 2πR. (Figure 9.11.1) Let C1 be the line segment from (0, 0) to
(2πR, 0) and let C2 be the cycloidal arch from (2πR, 0) back to (0, 0). Letting C = C1 ∪ C2, we have
A = ©∫C
x dy =∫C1
x dy +∫C2
x dy = 0 +∫C2
x dy
=∫ 0
2π
R(θ − sin θ)(R sin θ) dθ
= R2
∫ 2π
0
(sin2 θ − θ sin θ) dθ
= R2
[θ
2− sin 2θ
4+ θ cos θ − sin θ
]2π0
= 3πR2.
24. ©∫C
y3 dx + (3x− x3) dy =∫ ∫Ω
(3 − 3x2 − 3y2) dx dy = 3∫ ∫
Ω
(1 − x2 − y2) dx dy
The double integral is maximized by
Ω : 0 ≤ x2 + y2 ≤ 1.
(This is the maximal region on which the integral is nonnegative.) The line integral is maximized by
the unit circle traversed counterclockwise.
25. Taking Ω to be of type II (see Figure 18.5.2), we have∫ ∫Ω
∂Q
∂x(x, y) dxdy =
∫ d
c
∫ ψ2(y)
ψ1(y)
∂Q
∂x(x, y) dx dy
=∫ d
c
{Q[ψ2(y), y] −Q[ψ1(y), y]} dy
(∗) =∫ d
c
Q[ψ2(y), y] dy −∫ d
c
Q[ψ1(y), y] dy.
The graph of x = ψ2(y) from x = c to x = d is the curve
C4 : r4(u) = ψ2(u) i + u j, u ∈ [c, d ].
The graph of x = ψ1(y) from x = c to x = d is the curve
C3 : r3(u) = ψ1(u) i + u j, u ∈ [c, d ].
Then
©∫C
Q(x, y) dy =∫C4
Q(x, y) dy −∫C3
Q(x, y) dy
=∫ d
c
Q[ψ2(u), u] du−∫ d
c
Q[ψ1(u), u] du.
Since u is a dummy variable, it can be replaced by y. Comparison with (∗) gives the result.
26. Let h(r) = f(r)∇g(r) + g(r)∇f(r). Then h = ∇(fg)
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SECTION 18.5 943
27. Suppose that f is harmonic. By Green’s theorem,∫C
∂f
∂ydx− ∂f
∂xdy =
∫∫Ω
(−∂2f
∂2x− ∂2f
∂2y
)dxdy =
∫∫Ω
0 dxdy = 0.
28. −λ
3©∫
y3 dx = −λ
3
∫ ∫Ω
(−3y2) dx dy =∫ ∫Ω
λy2 dx dy = Ix
λ
3©∫
x3 dy =λ
3
∫ ∫Ω
3x2 dx dy =∫ ∫Ω
λx2 dx dy = Iy
29. ©∫C1
= ©∫C2
+ ©∫C3
30. Let Ω be the region enclosed by C. Then∫C
f(x) dx + g(y) dy = ±©∫C
f(x) dx + g(y) dy
= ±∫ ∫Ω
0︷ ︸︸ ︷(∂
∂x[g(y)] − ∂
∂y[f(x)]
)dx dy = 0
31.∂P
∂y=
−2xy(x2 + y2)2
=∂Q
∂xexcept at (0, 0)
(a) If C does not enclose the origin, and Ω is the region enclosed by C, then
©∫C
x
x2 + y2dx +
y
x2 + y2dy =
∫ ∫Ω
0 dxdy = 0.
(b) If C does enclose the origin, then
©∫C
= ©∫Ca
where Ca : r(u) = a cosu i + a sinu j, u ∈ [ 0, 2π ] is a small circle in the inner region of C.
In this case
©∫C
=∫ 2π
0
[a cosua2
(−a sinu) +a sinu
a2(a cosu)
]du =
∫ 2π
0
0 du = 0.
The integral is still 0.
32. (a) ©∫C
− y3
(x2 + y2)2dx +
xy2
(x2 + y2)2dy =
∫ ∫Ω
0 dy dx = 0
(b) By Green’s theorem, ©∫C
= ©∫C ′, where C ′ is a circle about the origin. r(u) = a cosu i + a sinu j.
©∫C ′
=∫ 2π
0
(sin4 u + sin2 u cos2 u) du =∫ 2π
0
sin2 u du = π
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JWDD027-18 JWDD027-Salas-v1 January 4, 2007 18:12
944 SECTION 18.6
33. If Ω is the region enclosed by C, then
©∫C
v ·dr = ©∫C
∂φ
∂xdx +
∂φ
∂ydy =
∫ ∫Ω
{∂
∂x
(∂φ
∂y
)− ∂
∂y
(∂φ
∂x
)}dxdy
=∫ ∫Ω
0 dxdy = 0.
equality of mixed partials
34. r(u) = [x1 + (x2 − x1)u]i + [y1 + (y2 − y1)u]j, u ∈ [0, 1]∫C
−y dx + x dy =∫ 1
0
{[−y1 − (y2 − y1)u] (x2 − x1) + [x1 + (x2 − x1)u] (y2 − y1)} du
=∫ 1
0
(x1y2 − x2y1) du = x1y2 − x2y1.
35. A =12©∫C
(−y dx + x dy)
=[∫
C1
+∫C2
+ · · ·∫Cn
]
Now∫Ci
(−y dx + x dy) =∫ 1
0
{[yi + u(yi+1 − yi)] (xi+1 − xi) + [xi + u(xi+1 − xi)] (yi+1 − yi)} du
= xiyi+1 − xi+1yi, i = 1, 2, . . . , n; xn+1 = x1, yn+1 = y1
Thus, A =12
[(x1y2 − x2y1) + (x2y3 − x3y2) + · · · + (xny1 − x1yn)]
36. (a) A =12[0 + (8 − 1) + 0] =
72
(b) A =12[0 + (12 − 2) + (12 − 0) + (0 + 6) + 0] = 14
SECTION 18.6
1. 4[(u2 − v2)i − (u2 + v2)j
+ 2uv k]
2. uk 3. 2(j − i)
4. sinu sin v i + cosu cos v j + (sin2 u sin2 v − cos2 u cos2 v)k
5. r(u, v) = 3 cosu cos v i + 2 sinu cos v j + 6 sin v k, u ∈ [ 0, 2π ], v ∈ [ 0, π/2 ]
6. r(θ, z) = 2 cos θ i + 2 sin θ j + z k, θ ∈ [0, 2π], z ∈ [1, 4].
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SECTION 18.6 945
7. r(u, v) = 2 cosu cos v i + 2 sinu cos v j + 2 sin v k, u ∈ [ 0, 2π ], v ∈ (π/4, π/2 ]
8. r(s, θ) = s cos θ i + s sin θ j + (s cos θ + 2)k, s ∈ [0, 1], θ ∈ [0, 2π].
9. The surface consists of all points of the form (x, g(x, z), z) with (x, z) ∈ Ω. This set of points is
given by
r(u, v) = u i + g(u, v) j + v k, (u, v) ∈ Ω.
10. r(y, z) = h(y, z) i + y j + z k, (y, z) ∈ Γ
11. x2/a2 + y2/b2 + z2/c2 = 1; ellipsoid
12. z =x2
a2+
y2
b2; elliptic paraboloid.
13. x2/a2 − y2/b2 = z; hyperbolic paraboloid
14. (a) See Exercise 53, Section 14.2.
(b) (c)
15. For each v ∈ [a, b ], the points on the surface at level z = f(v) form a circle of radius v.
That circle can be parametrized:
R(u) = v cosu i + v sinu j + f(v)k, u ∈ [ 0, 2π ].
Letting v range over [a, b ], we obtain the entire surface:
r(u, v) = v cosu i + v sinu j + f(v)k; 0 ≤ u ≤ 2π, a ≤ v ≤ b.
16. For the parametrization given in the answer to Exercise 15
N(u, v) = −vf ′(v) cosu i + v j − vf ′(v) sinuk, ‖ N(u, v) ‖= v√
1 + [f ′(v)]2.
ThereforeA =
∫ 2π
0
{∫ b
a
v√
1 + [f ′(v)]2 dv}
du
= 2π∫ b
a
v√
1 + [f ′(v)]2 dv =∫ b
a
2πx√
1 + [f ′(v)]2 dx
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JWDD027-18 JWDD027-Salas-v1 January 4, 2007 18:12
946 SECTION 18.6
17. Since γ is the angle between p and the xy-plane, γ is the angle between the upper normal to p and k.
(Draw a figure.) Therefore, by 18.6.5,
area of Γ =∫ ∫Ω
sec γ dxdy = (sec γ)AΩ = AΩ sec γ.
γ is constant
18. n = i + j + k is an upper normal.
cos γ =n ·k√
3=
1√3, sec γ =
√3 A =
√3πb2
19. The surface is the graph of the function
f(x, y) = c(1 − x
a− y
b
)=
c
ab(ab− bx− ay)
defined over the triangle Ω : 0 ≤ x ≤ a, 0 ≤ y ≤ b(1 − x/a). Note that Ω has area 12ab.
A =∫ ∫Ω
√[f ′
x(x, y)]2 + [f ′y(x, y)]2 + 1 dxdy
=∫ ∫Ω
√c2/a2 + c2/b2 + 1 dxdy
=1ab
√a2b2 + a2c2 + b2c2
∫ ∫Ω
dx dy =12
√a2b2 + a2c2 + b2c2.
20. f(x, y) =√x2 + y2, Ω : 0 ≤ x2 + y2 ≤ 1
A =∫ ∫Ω
√[f ′
x(x, y)]2 + [f ′y(x, y)]2 + 1 dx dy =
∫ ∫Ω
√2 dx dy =
√2π
21. f(x, y) = x2 + y2, Ω : 0 ≤ x2 + y2 ≤ 4
A =∫ ∫Ω
√4x2 + 4y2 + 1 dx dy [ change to polar coordinates ]
=∫ 2π
0
∫ 2
0
√4r2 + 1 r dr dθ
= 2π[
112 (4r2 + 1)3/2
]20
= 16π(17
√17 − 1)
22. f(x, y) =√
2xy, Ω : 0 ≤ x ≤ a, 0 ≤ y ≤ b
A =∫ ∫Ω
x + y√2xy
dx dy =1√2
∫ ∫Ω
(√x/y +
√y/x) dx dy
=1√2
∫ a
0
∫ b
0
(√x/y +
√y/x)dy dx
=23
√2(a + b)
√ab
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JWDD027-18 JWDD027-Salas-v1 January 4, 2007 18:12
SECTION 18.6 947
23. f(x, y) = a2 − (x2 + y2), Ω : 14a
2 ≤ x2 + y2 ≤ a2
A =∫ ∫
Ω
√4x2 + 4y2 + 1 dxdy [ change to polar coordinates ]
=∫ 2π
0
∫ a
a/2
r√
4r2 + 1 dr dθ = 2π[
112
(4r2 + 1)3/2]aa/2
=π
6
[(4a2 + 1)3/2 − (a2 + 1)3/2
]
24. f(x, y) =1√3(x + y)3/2, Ω : 0 ≤ x ≤ 2, 0 ≤ y ≤ 2 − x
A =∫∫
1√2
√3x + 3y + 2 dx dy =
1√2
∫ 2
0
∫ 2−x
0
√3x + 3y + 2 dy dx =
464135
25. f(x, y) = 13 (x3/2 + y3/2), Ω : 0 ≤ x ≤ 1, 0 ≤ y ≤ x
A =∫∫
Ω
12
√x + y + 4 dxdy
=∫ 1
0
∫ x
0
12
√x + y + 4 dy dx =
∫ 1
0
[13(x + y + 4)3/2
]x0
dx
=∫ 1
0
13
[(2x + 4)3/2 − (x + 4)3/2
]dx =
13
[15(2x + 4)5/2 − 2
5(x + 4)5/2
]10
=115
(36√
6 − 50√
5 + 32)
26. f(x, y) = y2, Ω : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1
A =∫ ∫Ω
√4y2 + 1 dx dy =
∫ 1
0
∫ 1
0
√4y2 + 1 dy dx =
14
[2√
5 + ln(2 +√
5)]
27. The surface x2 + y2 + z2 − 4z = 0 is a sphere of radius 2 centered at (0, 0, 2):
x2 + y2 + z2 − 4z = 0 ⇐⇒ x2 + y2 + (z − 2)2 = 4.
The quadric cone z2 = 3(x2 + y2) intersects the sphere at height z = 3:
x2 + y2 + z2 − 4z = 0
z2 = 3(x2 + y2)
}=⇒
3(x2 + y2) + 3z2 − 12z = 0
4z2 − 12z = 0
z = 3. (since z ≥ 2)
The surface of which we are asked to find the area is a spherical segment of width 1 (from z = 3 to
z = 4) in a sphere of radius 2. The area of the segment is 4π. (Exercise 27, Section 9.9.)
A more conventional solution. The spherical segment is the graph of the function
f(x, y) = 2 +√
4 − (x2 + y2), Ω : 0 ≤ x2 + y2 ≤ 3.
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JWDD027-18 JWDD027-Salas-v1 January 4, 2007 18:12
948 SECTION 18.6
Therefore
A =∫ ∫Ω
√√√√( −x√4 − x2 − y2
)2
+
(−y√
4 − x2 − y2
)2
+ 1 dxdy
=∫ ∫Ω
2√4 − (x2 + y2)
dxdy
=∫ 2π
0
∫ √3
0
2r√4 − r2
dr dθ [ changed to polar coordinates ]
= 2π[−2√
4 − r2]√3
0= 4π
28. The spherical segment is the graph of the function
f(x, y) = a +√a2 − (x2 + y2), Ω ≤ x2 + y2 ≤ 2ab− b2.
Therefore
A =∫ ∫Ω
a√a2 − (x2 + y2)
dx dy [change to polar coordinate]
=∫ 2π
0
∫ √2ab−b2
0
ar√a2 − r2
dr dθ
= 2πab.
29. (a)∫ ∫Ω
√[∂g
∂y(y, z)
]2+[∂g
∂z(y, z)
]2+ 1 dydz =
∫ ∫Ω
sec [α(y, z)] dydz
where α is the angle between the unit normal with positive i component and the positive x-axis
(b)∫ ∫Ω
√[∂h
∂x(x, z)
]2+[∂h
∂z(x, z)
]2+ 1 dxdz =
∫ ∫Ω
sec [β(x, z)] dxdz
where β is the angle between the unit normal with positive j component and the positive y-axis
30. (a) ru′ = −a sinu i + a cosu j; r′v = k N(u, v) = r′u × r′v = a cosu i + a sinu j
(b) A =∫ ∫Ω
‖ N(u, v) ‖ du dv =∫ ∫Ω
a du dv =∫ 2π
0
∫ l
0
a dv du = 2πla
31. (a) N(u, v) = v cosu sinα cosα i + v sinu sinα cosα j − v sin2 αk
(b) A =∫ ∫Ω
‖N(u, v)‖ dudv =∫ ∫Ω
v sinαdudv
=∫ 2π
0
∫ s
0
v sinαdv du = πs2 sinα
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SECTION 18.6 949
32. (a) Set x = a cosu sin v, y = a sinu sin v, z = b cos v. Then
x2
a2+
y2
a2+
z2
b2= 1
(b)
(c) N(u, v) = −ab cosu sin2 v i − ab sinu sin2 v j − a2 sin v cos v k,
A =∫ ∫Ω
‖ N(u, v) ‖ du dv =∫ 2π
0
∫ π
0
√a2b2 sin4 v + a4 sin2 v cos2 v dv du
= 2πa∫ π
0
sin v√b2 sin2 v + a2 cos2 v dv
33. (a) Set x = a cosu cosh v, y = b sinu cosh v, z = c sinh v. Then,
x2
a2+
y2
b2− z2
c2= 1.
(b)
(c) A =∫∫
Ω
‖N(u, v)‖ dv du
=∫ 2π
0
∫ ln 2
− ln 2
√64 cos2 u cosh2 v + 144 sin2 u cosh2 v + 36 cosh2 v sinh2 v dv du
34. (a) Set x = a cosu sinh v, y = b sinu sinh v, z = c cosh v. Then,
x2
a2+
y2
b2− z2
c2= −1.
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JWDD027-18 JWDD027-Salas-v1 January 4, 2007 18:12
950 SECTION 18.7
(b)
(c) Assuming c > 0, z = c cosh v > 0 for all v.
35. A =√A1
2 + A22 + A3
2; the unit normal to the plane of Ω is a vector of the formcos γ1 i + cos γ2 j + cos γ3 k.
Note thatA1 = A cos γ1, A2 = A cos γ2, A3 = A cos γ3.
ThereforeA1
2 + A22 + A3
2 = A2[cos2 γ1 + cos2 γ2 + cos2 γ3] = A2.
36. We can parametrize the surface by setting
R(r θ) = r cos θ i + r sin θ j + f(r, θ)k, (r, θ) ∈ Ω.
The integrand is ‖ N(r, θ) ‖ .
37. (a) (We use Exercise 36.) f(r, θ) = r + θ; Ω : 0 ≤ r ≤ 1, 0 ≤ θπ
A =∫ ∫Ω
√r2 [f ′
r(r, θ)]2 + [f ′
θ(r, θ)]2 + r2 drdθ =
∫ ∫Ω
√2r2 + 1 drdθ
=∫ π
0
∫ 1
0
√2r2 + 1 dr dθ =
14
√2π[√
6 + ln (√
2 +√
3)]
(b) f(r, θ) = reθ; Ω : 0 ≤ r ≤ a, 0 ≤ θ ≤ 2π
A =∫ ∫Ω
r√
2e2θ + 1 drdθ =(∫ 2π
0
√2e2θ + 1 dθ
)(∫ a
0
r dr
)
= 12a
2[√
2e4π + 1 −√
3 + ln (1 +√
3) − ln (1 +√
2e4π + 1)]
38. r(u, v) = x(u, v) i + y(u, v) j, (u, v) ∈ ΩStraightforward calculation shows that ‖ N(u, v) ‖= |J(u, v)|.
SECTION 18.7
For Exercises 1–6 we have sec [ γ(x, y)] =√y2 + 1. N(x, y) = −yj + k, so ‖ N(x, y) ‖=
√y2 + 1.
1.∫ ∫S
dσ =∫ 1
0
∫ 1
0
√y2 + 1 dx dy =
∫ 1
0
√y2 + 1 dy =
12[√
2 + ln (1 +√
2)]
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JWDD027-18 JWDD027-Salas-v1 January 4, 2007 18:12
SECTION 18.7 951
2.∫ ∫S
x2 dσ =∫ 1
0
∫ 1
0
x2√y2 + 1 dy dx
=(∫ 1
0
x2 dx
)(∫ 1
0
√y2 + 1 dy
)=
16
[√2 + ln(1 +
√2)]
3.∫ ∫S
3y dσ =∫ 1
0
∫ 1
0
3y√y2 + 1 dy dx =
∫ 1
0
3y√y2 + 1 dy =
[(y2 + 1)3/2
]10
= 2√
2 − 1
4.∫ ∫S
(x− y) dσ =∫ 1
0
∫ 1
0
x√y2 + 1 dy dx−
∫ 1
0
∫ 1
0
y√y2 + 1 dy dx
=(∫ 1
0
x dx
)(∫ 1
0
√y2 + 1 dy
)−∫ 1
0
y√y2 + 1 dy
=14[√
2 + ln(1 +√
2)] − 13(2√
2 − 1)
=13− 5
12
√2 +
14
ln(1 +√
2)
5.∫ ∫S
√2z dσ =
∫ ∫S
y dσ =13(2√
2 − 1) (Exercise 3)
6.∫ ∫S
√1 + y2 dσ =
∫ 1
0
∫ 1
0
(1 + y2) dy dx =∫ 1
0
(1 + y2) dy =43
7.∫ ∫S
xy dσ; S : r(u, v) = (6 − 2u− 3v) i + u j + v k, 0 ≤ u ≤ 3 − 32 v, 0 ≤ v ≤ 2
‖N(u, v) ‖ = ‖ (−2 i + j) × (−3 i + k) ‖ =√
14
∫ ∫S
xy dσ =√
14∫∫
Ω
x(u, v)y(u, v) du dv
=√
14∫∫
Ω
(6 − 2u− 3v)u du dv
=√
14∫ 2
0
∫ 3−3v/2
0
(6u− 2u2 − 3uv) du dv
=√
14[3(3 − 3
2 v)2 − 2
3
(3 − 3
2 v)3 − 3
2 v(3 − 3
2 v)2]
dv =92
√14
8. S is given by z = f(x, y) = 1 − x− y, 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x∫ ∫S
xyz dσ =∫ 1
0
∫ 1−x
0
xy(1 − x− y)√
(−1)2 + (−1)2 + 1 dy dx
=√
3∫ 1
0
∫ 1−x
0
xy(1 − x− y) dy dx =√
3120
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952 SECTION 18.7
9.∫ ∫S
x2z dσ; S : r(u, v) = (cosu i + v j + sinuk, 0 ≤ u ≤ π, 0 ≤ v ≤ 2.
N(u, v) =
∣∣∣∣∣∣∣∣i j k
− sinu 0 cosu
0 1 0
∣∣∣∣∣∣∣∣ = − cosu i − sinuk and ‖N(u, v) ‖ = 1.
∫ ∫S
x2z dσ =∫∫
Ω
cos2 u sinu du dv =∫ 2
0
∫ π
0
cos2 u sinu du dv =43
10.∫ ∫S
(x2 + y2 + z2) dσ =∫ ∫
x2+y2≤1
[x2 + y2 + (x + 2)2]√
2 dx dy
=∫ 2π
0
∫ 1
0
[r2 + (r cos θ + 2)2]√
2 r dr dθ
=∫ 2π
0
∫ 1
0
(r3 + r3 cos2 θ + 4r2 cos θ + 4r) dr dθ =19√
24
π
11.∫ ∫S
(x2 + y2) dσ; S : r(u, v) = cosu cos v i + cosu sin v j + sinuk, 0 ≤ u ≤ π/2, 0 ≤ v ≤ 2π.
N(u, v) =
∣∣∣∣∣∣∣∣i j k
− sinu cos v − sinu sin v cosu
− cosu sin v cosu cos v 0
∣∣∣∣∣∣∣∣ = − cos2 u cos v i + cos2 u sin v j − sinu cosuk;
‖N(u, v) ‖ = cosu.
∫ ∫S
(x2 + y2) dσ =∫∫
Ω
cos2 u cosu du dv =∫ 2π
0
∫ π/2
0
cos3 u du dv =43π
12.∫ ∫S
(x2 + y2) dσ =∫ ∫x2+y2≤1
(x2 + y2)√
4x2 + 4y2 + 1 dx dy =∫ 2π
0
∫ 1
0
r2√
4r2 + 1 r dr dθ
= 2π∫ 1
0
r3√
4r2 + 1 dr =25√
5 + 160
π
For Exercises 13–16 the surface S is given by:
f(x, y) = a− x− y; 0 ≤ x ≤ a, 0 ≤ y ≤ a− x and sec [γ (x, y)] =√
3.
13. M =∫ ∫S
λ(x, y, x) dσ =∫ a
0
∫ a−x
0
k√
3 dy dx =∫ a
0
k√
3 (a− x) dx =12a2k
√3
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SECTION 18.7 953
14.M =
∫ ∫S
k(x + y) dσ =∫ a
0
∫ a−x
0
k(x + y)√
3 dy dx
=12k√
3∫ a
0
(a2 − x2) dx =13
√3a3k
15. M =∫ ∫S
λ(x, y, z) dσ =∫ a
0
∫ a−x
0
kx2√
3 dy dx =∫ a
0
k√
3x2(a− x) dx =112
a4k√
3
16. xA =∫ ∫S
xdσ =∫ a
0
∫ a−x
0
x√
3 dy dx
=√
3∫ a
0
(ax− x2) dx =16
√3a3
A =∫ ∫S
dσ =∫ a
0
∫ a−x
0
√3 dy dx
=√
3∫ a
0
(a− x) dx =12
√3a2
x = xA/A =13a; similarly y = z =
13a
17. S : r(u, v) = a cosu cos v i + a sinu cos v j + a sin v k with 0 ≤ u ≤ 2π, 0 ≤ v ≤ 12π. By a previous
calculation ‖N(u, v)‖ = a2 cos v.
x = 0, y = 0 (by symmetry)
zA =∫ ∫S
z dσ =∫ ∫Ω
z(u, v) ‖N(u, v)‖ dudv =∫ 2π
0
∫ π/2
0
a3 sin v cos v dv du = πa3
z = 12a since A = 2πa2
18. N(u, v) = 2 i + 2 j − 2k
A =∫ ∫S
dσ =∫ ∫Ω
‖ N(u, v) ‖ du dv =∫ 1
0
∫ 1
0
2√
3 du dv = 2√
3
19. N(u, v) = (i + j + 2k) · (i − j) = 2 i + 2 j − 2k
flux in the direction of N =∫ ∫S
(v · N
‖N‖
)dσ =
∫ ∫Ω
[v(x(u), y(u), z(u)) ·N(u, v)] dudv
=∫ ∫Ω
[(u + v)i − (u− v)j] · [2 i + 2 j − 2k] dudv.
=∫∫
Ω
4v dudv = 4∫ 1
0
∫ 1
0
v dv du = 2
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954 SECTION 18.7
20. sec[γ(x, y)] =√x2 + y2 + 1
M =∫ ∫S
kxy dσ = k
∫ 1
0
∫ 1
0
xy√x2 + y2 + 1 dy dx
=13k
∫ 1
0
[x(x2 + 2)3/2 − x(x2 + 1)3/2
]dx
=115
(9√
3 − 8√
2 + 1)k
For Exercises 21–23: n =1a(x i + y j + z k)
S : r(u, v) = a cosu cos v i + a sinu cos v j + a sin v k with 0 ≤ u ≤ 2π, − 12π ≤ v ≤ 1
2π
‖N(u, v)‖ = a2 cos v
21. With v = z k
flux =∫ ∫S
(v ·n) dσ =1a
∫ ∫S
z2 dσ =1a
∫ ∫Ω
(a2 sin2 v)(a2 cos v) dudv
= a3
∫ 2π
0
∫ π/2
−π/2
(sin2 v cos v) du dv =43πa3
22. With v = x i + y j + z k
flux =∫ ∫S
(v ·n) dσ = a
∫ ∫S
dσ = aA = 4πa3
23. With v = y i − x j
flux =∫ ∫S
(v ·n) dσ =1a
∫ ∫S
(yx− xy)︸ ︷︷ ︸0
dσ = 0
24. Ix =∫ ∫S
(y2 + z2) dσ = 2√
3∫ 1
0
∫ 1
0
(5u2 − 2uv + v2) dv du = 3√
3
Iy =∫ ∫S
(x2 + z2) dσ = 2√
3∫ 1
0
∫ 1
0
(5u2 + 2uv + v2) dv du = 5√
3
Iz =∫ ∫S
(x2 + z2) dσ = 4√
3∫ 1
0
∫ 1
0
(u2 + v2) dσ =83
√3
For Exercises 25–27 the triangle S is the graph of the function
f(x, y) = a− x− y on Ω : 0 ≤ x ≤ a, 0 ≤ y ≤ a− x.
The triangle has area A = 12
√3a2.
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SECTION 18.7 955
25. With v = x i + y j + z k
flux =∫ ∫S
(v ·n) dσ =∫ ∫Ω
(−v1f′x − v2f
′y + v3) dxdy
=∫ ∫Ω
[−x(−1) − y(−1) + (a− x− y)] dxdy = a
∫ ∫Ω
dxdy = aA =12
√3a3
26. With v = (x + z)k
flux =∫ ∫S
(v ·n) dσ =∫ ∫Ω
(−v1f′x − v2f
′y + v3) dx dy
=∫ ∫Ω
(a− y) dx dy =∫ a
0
∫ a−x
0
(a− y) dy dx =13a3
27. With v = x2 i − y2 j
flux =∫ ∫S
(v ·n) dσ =∫ ∫Ω
(−v1f′x − v2f
′y + v3) dxdy
=∫∫
Ω
[−x2(−1) − (−y2)(−1) + 0] dxdy =∫ a
0
∫ a−x
0
(x2 − y2) dy dx
=∫ a
0
[ax2 − x3 − 1
3(a− x)3
]dx =
[13ax3 − 1
4x4 +
112
(a− x)4]a0
= 0
28. With v = −xy2i + z j
flux =∫ ∫S
(v ·n) dσ =∫ ∫Ω
(−v1fx′ − v2fy
′ + v3) dx dy
=∫ 1
0
∫ 2
0
(xy3 − x2y) dy dx =43
29. With v = xz j − xy k
flux =∫ ∫S
(v ·n) dσ =∫ ∫Ω
(−v1f′x − v2f
′y + v3) dxdy
=∫ ∫Ω
(−x3y − xy) dxdy =∫ 1
0
∫ 2
0
(−x3y − xy) dy dx
=∫ 1
0
−2(x3 + x) dx = −32
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956 SECTION 18.7
30. With v = x2y i + z2 k
flux =∫ ∫S
(v ·n) dσ =∫ ∫Ω
(−v1f′x − v2f
′y + v3) dx dy
=∫ 1
0
∫ 2
0
(−x2y2 + x2y2) dy dx = 0
31. n =1a(x i + y j)
flux =∫ ∫S
(v ·n) dσ =1a
∫ ∫S
[(x i + y j + z k) · (x i + y j)] dσ
=1a
∫ ∫S
(x2 + y2) dσ = a
∫ ∫S
dσ = a (area of S) = a (2πal) = 2πa2l
32. flux =∫ ∫S
(GmM
rr3
· rr
)dσ = GmM
∫ ∫S
1r2
dσ
= GmM
∫ ∫S
1a2
dσ =GmM
a2
∫ ∫S
dσ =GmM
a2(4πa2) = 4πGmM
In Exercises 33–36, S is the graph of f(x, y) =23(x3/2 + y3/2), 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x
We use
flux =∫ ∫S
(v ·n) dσ =∫ ∫Ω
(−v1f′x − v2f
′y + v3) dx dy.
33. With v = x i − y j + 32 z k
flux =∫ ∫S
(v ·n) dσ =∫ ∫Ω
(−v1f′x − v2f
′y + v3) dxdy =
∫ ∫Ω
2y3/2 dxdy
=∫ 1
0
∫ 1−x
0
2y3/2 dy dx =∫ 1
0
45(1 − x)5/2 dx =
835
34. With v = x2 i,
flux =∫ 1
0
∫ 1−x
0
−x5/2 dy dx = − 463
.
35. With v = y2 j
flux =∫ ∫S
(v ·n) dσ =∫ ∫Ω
(−v1f′x − v2f
′y + v3) dxdy =
∫ ∫Ω
−y5/2 dσ
=∫ 1
0
∫ 1−x
0
−y5/2 dy dx =∫ 1
0
−27(1 − x)7/2 dx = − 4
63
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JWDD027-18 JWDD027-Salas-v1 January 4, 2007 18:12
SECTION 18.7 957
36. With v = y i −√xy j,
flux =∫ 1
0
∫ 1−x
0
(−yx1/2 +√xyy1/2) dy dx = 0.
37. x = 0, y = 0 by symmetry. You can verify that ‖N(u, v)‖ = v sinα.
zA =∫ ∫S
z dσ =∫ ∫Ω
(s cosα)(v sinα) dudv = sinα cosα∫ 2π
0
∫ s
0
v2 dv du = 23π sinα cosα s3
z = 23s cosα since A = πs2 sinα
38. M =∫ ∫Ω
k√x2 + y2 dσ =
∫ ∫Ω
k√x2 + y2
√2 dx dy
= k√
2∫ 2π
0
∫ 1
0
r2 dr dθ =23
√2πk
39. f(x, y) =√x2 + y2 on Ω : 0 ≤ x2 + y2 ≤ 1; λ(x, y, z) = k
√x2 + y2
xM = 0, yM = 0 (by symmetry)
zMM =∫ ∫S
zλ(x, y, z) dσ =∫ ∫Ω
k(x2 + y2) sec [γ(x, y)] dxdy
= k√
2∫ ∫Ω
(x2 + y2) dxdy
= k√
2∫ 2π
0
∫ 1
0
r3 dr dθ =12
√2πk
zM = 34 since M = 2
3
√2πk (Exercise 38)
40. (a) Ix =∫ ∫Ω
k√x2 + y2(y2 + z2) dσ =
∫ ∫Ω
k√x2 + y2(y2 + x2 + y2) dσ
= k√
2∫ ∫Ω
[y2(x2 + y2)1/2 + (x2 + y2)3/2
]dx dy
= k√
2∫ 2π
0
∫ 1
0
(r4 sin2 θ + r4) dr dθ =3√
25
πk
(b) Iy = Ix by symmetry
(c) Iz =∫ ∫Ω
k√x2 + y2(x2 + y2) dσ =
∫ ∫S
k(x2 + y2)3/2 dσ
= k√
2∫ ∫Ω
(x2 + y2)3/2 dx dy = k√
2∫ 2π
0
∫ 1
0
r4 dr dθ =25
√2πk
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JWDD027-18 JWDD027-Salas-v1 January 4, 2007 18:12
958 SECTION 18.7
41. no answer required
42. M =∫ ∫S
k(y2 + z2) dσ = 2√
3k∫ ∫S
[(u− v)2 + 4u2] du dv
= 2√
3k∫ 1
0
∫ 1
0
(5u2 − 2uv + v2) dv du = 3√
3k
43. xMM =∫ ∫S
xλ (x, y, z) dσ =∫ ∫S
kx(y2 + z2) dσ
= 2√
3 k∫ ∫Ω
(u + v)[(u− v)2 + 4u2
]dudv
= 2√
3 k∫ 1
0
∫ 1
0
(5u3 − 2u2v + uv2 + 5u2v − 2uv2 + v3) dv du
= 2√
3 k∫ 1
0
(5u3 − u2 +
13u +
52u2 − 2
3u +
14
)du =
113
√3k
xM =119
since M = 3√
3k (Exercise 42)
44. Iz =∫ ∫S
λ(x, y, z)(x2 + y2) dσ =∫ ∫S
k(y2 + z2)(x2 + y2) dσ
= 2√
3k∫ ∫Ω
[(u− v)2 + 4u2
] [(u + v)2 + (u− v)2
]du dv
= 4√
3k∫ 1
0
∫ 1
0
(5u4 − 2u3v + 6u2v2 − 2uv3 + v4) du dv =82√
315
k.
45. Total flux out of the solid is 0. It is clear from a diagram that the outer unit normal to the cylindrical
side of the solid is given by n = x i + y j in which case v · n = 0. The outer unit normals to the top
and bottom of the solid are k and −k respectively. So, here as well, v · n = 0 and the total flux is 0.
46. The flux through the upper boundary is 0:
(yi − xj) · k = 0.
The flux through the lower boundary is 0:
v ·(∂f
∂xi +
∂f
∂yj − k
)= (y i − x j) · (2x i + 2y j − k) = (2xy − 2xy) = 0.
Thus the total flux out of the solid is 0.
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JWDD027-18 JWDD027-Salas-v1 January 4, 2007 18:12
SECTION 18.7 959
47. The surface z =√
2 − (x2 + y2) is the upper half of the sphere x2 + y2 + z2 = 2. The surface intersects
the surface z = x2 + y2 in a circle of radius 1 at height z = 1. Thus the upper boundary of the solid,
call it S1, is a segment of width√
2 − 1 on a sphere of radius√
2. The area of S1 is therefore 2π√
2(√
2 − 1). (Exercise 27, Section 9.9). The upper unit normal to S1 is the vector
n =1√2(x i + y j + z k).
Therefore
flux through S1 =∫ ∫S1
(v ·n) dσ =1√2
∫ ∫S1
2︷ ︸︸ ︷(x2 + y2 + z2) dσ
=√
2∫ ∫S1
dσ =√
2 (area of S1) = 4π(√
2 − 1).
The lower boundary of the solid, call it S2, is the graph of the function
f(x, y) = x2 + y2 on Ω : 0 ≤ x2 + y2 ≤ 1.
Taking n as the lower unit normal, we have
flux through S2 =∫ ∫S2
(v ·n) dσ =∫ ∫Ω
(v1f
′x + v2f
′y − v3
)dxdy
=∫ ∫Ω
(x2 + y2) dxdy =∫ 2π
0
∫ 1
0
r3 dr dθ =12π.
The total flux out of the solid is 4π(√
2 − 1) +12π = (4
√2 − 7
2)π.
48. face n v · n flux
x = 0 −i −xz = 0 0
x = 1 i xz = 1 12
y = 0 −j −4xyz2 = 0 0 total flux = 12 + 2
3 + 2 = 196
y = 1 j 4xyz2 = 4xz2 23
z = 0 −k −2z = 0 0
z = 1 k 2z = 2 2
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JWDD027-18 JWDD027-Salas-v1 January 4, 2007 18:12
960 SECTION 18.8
SECTION 18.8
1. ∇ · v = 2, ∇×v = 0 2. ∇ · v = 0, ∇×v = 0
3. ∇ · v = 0, ∇×v = 0 4. ∇ · v = − 4xy(x2 + y2)2
, ∇×v =2(y2 − x2)(x2 + y2)2
k
5. ∇ · v = 6, ∇×v = 0 6. ∇ · v = 0, ∇×v = 0
7. ∇ · v = yz + 1, ∇×v = −x i + xy j + (1 − x)z k
8. ∇ · v = 2y(x + z), ∇×v = (2xy − y2) i − y2 j − x2 k
9. ∇ · v = 1/r2, ∇×v = 0
10. ∇ · v = ex(3 + x), ∇×v = −exz j + exyk
11. ∇ · v = 2(x + y + z)er2, ∇×v = 2er
2[(y − z)i − (x− z) j + (x− y)k]
12. ∇ · v = 0, ∇×v = −2[zez
2i + xex
2j + yey
2k]
13. ∇ · v = f ′(x), ∇×v = 0
14. each partial derivative that appears in the curl is 0
15. use components.
16. ∇ · F = −GmM[∇ · (r−3r)
]= −GmM(0) = 0
linearity (Exercise 15) (17.8.8)
∇×F = −GmM [∇× (r−3r)] = −GmM(0) = 0
linearity (Exercise 15) (17.8.8)
17. ∇ · v =∂P
∂x+
∂Q
∂y+
∂R
∂z= 2 + 4 − 6 = 0
18. ∇ · v =∂
∂x(3x2) +
∂
∂y(−y2) +
∂
∂z(2yz − 6xz) = 6x− 2y + (2y − 6x) = 0
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JWDD027-18 JWDD027-Salas-v1 January 4, 2007 18:12
SECTION 18.8 961
19. ∇×F =
∣∣∣∣∣∣∣∣∣∣
i j k
∂
∂x
∂
∂y
∂
∂z
x y −2z
∣∣∣∣∣∣∣∣∣∣= 0
20. F(x, y, z) = (2x + y + 2z)i + (x + 4y − 3z)j + (2x− 3y − 6z)k
∇×F =
∣∣∣∣∣∣∣∣∣∣
i j k
∂
∂x
∂
∂y
∂
∂z
2x + y + 2z x + 4y − 3z 2x− 3y − 6z
∣∣∣∣∣∣∣∣∣∣= (−3 + 3)i − (2 − 2)j + (1 − 1)k = 0
21. ∇2f = 12(x2 + y2 + z2)
22. ∇2f = ∇ ·∇f = ∇ · (yzi + xzj + xyk) = 0
23. ∇2f = 2y3z4 + 6x2yz4 + 12x2y3z2
24. Note that for r =√x2 + y2 + z2,
∂r
∂x=
x
r
Then∂2
∂x2(cos r) =
∂
∂x
(−x sin r
r
)=
−r2 sin r − x2r cos r + x2 sin r
r3,
With similar formulas for y and z. Therefore
∇2f =∂2
∂x2cos r +
∂2
∂y2cos r +
∂2
∂z2cos r
=−3r2 sin r − (x2 + y2 + z2)r cos r + (x2 + y2 + z2) sin r
r3
= − cos r − 2r−1 sin r
25. ∇2f = er(1 + 2r−1)
26.∂2
∂x2ln r =
∂
∂x
( x
r2
)=
r2 − 2x2
r4, with similar formula for y and z.
Then ∇2f =3r2 − 2(x2 + y2 + z2)
r4=
1r2
27. (a) 2r2 (b) −1/r
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JWDD027-18 JWDD027-Salas-v1 January 4, 2007 18:12
962 SECTION 18.8
28. (a) (u ·∇)r = (u ·∇x)i + (u ·∇y)j + (u ·∇z)k
= (u · i)i + (u · j)j + (u ·k)k = u
(b) (r ·∇)u = (r ·∇yz)i + (r ·∇xz)j + (r ·∇xy)k
= [r · (zj + yk)]i + [r · (zi + xk)]j + [r · (yi + xj)]k
= (yz + zy)i + (xz + zx)j + (xy + yx)k
= 2(yzi + xzj + xyk) = 2u
29. ∇2f = ∇2g(r) = ∇ · (∇g(r)) = ∇ ·(g′(r)r−1r
)=[(∇g′(r)) · r−1r
]+ g′(r)
(∇ · r−1r
)={[g′′(r)r−1r
]· r−1r
}+ g′(r)(2r−1)
= g′′(r) + 2r−1g′(r)
30. (a) ∇ · (fv) =∂
∂x(fv1) +
∂
∂y(fv2) +
∂
∂z(fv3)
=(f∂v1
∂x+
∂f
∂xv1
)+(f∂v2
∂y+
∂f
∂yv2
)+(f∂v3
∂z+
∂f
∂zv3
)
=(∂f
∂xi +
∂f
∂yj +
∂f
∂zk)
· v + f
(∂v1
∂x+
∂v2
∂y+
∂v3
∂z
)
= (∇f) ·v + f(∇ ·v)
(b) ∇× (fv)
=[∂
∂y(fv3) −
∂
∂z(fv2)
]i +[∂
∂z(fv1) −
∂
∂x(fv3)
]j +[∂
∂x(fv2) −
∂
∂y(fv1)
]k
=[f∂v3
∂y+
∂f
∂yv3 − f
∂v2
∂z− ∂f
∂zv2
]i + etc.
(c) ∇×v =
∣∣∣∣∣∣∣∣∣i j k∂
∂x
∂
∂y
∂
∂z
v1 v2 v3
∣∣∣∣∣∣∣∣∣=(∂v3
∂y− ∂v2
∂z
)i +(∂v1
∂z− ∂v3
∂x
)j +(∂v2
∂x− ∂v1
∂y
)k
i-component of ∇× (∇×v) =∂
∂y
(∂v2
∂x− ∂v1
∂y
)− ∂
∂z
(∂v1
∂z− ∂v3
∂x
)
=∂2v2
∂y∂x− ∂2v1
∂y2− ∂2v1
∂z2+
∂2v3
∂z∂x
=∂
∂x
(∂v2
∂y+
∂v3
∂z
)−(∂2v1
∂y2+
∂2v1
∂z2
)
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JWDD027-18 JWDD027-Salas-v1 January 4, 2007 18:12
SECTION 18.9 963
Adding and subtracting∂2v1
∂x2, we get
∂
∂x
(∂v1
∂x+
∂v2
∂y+
∂v3
∂z
)−(∂2v1
∂x2+
∂2v1
∂y2+
∂2v1
∂z2
)=
∂
∂x(∇ ·v) − ∇2v1 = the i-component of ∇2v.
Equality of the other components can be obtained in a similar manner.
31.∂f
∂x= 2x + y + 2z,
∂2f
∂x2= 2;
∂f
∂y= 4y + x− 3z,
∂2f
∂y2= 4;
∂f
∂z= −6z + 2x− 3y,
∂2f
∂z2= −6;
∂2f
∂x2+
∂2f
∂y2+
∂2f
∂z2= 2 + 4 − 6 = 0
32. f(r) =1r.
∂2
∂x2
(1r
)=
∂
∂x
(−x
r3
)=
−r2 + 3x2
r5, with similar formulas for y and z
Then ∇2f =−3r2 + 3(x2 + y2 + z2)
r5= 0.
33. n = −1
34. Since ∇ · (∇f) = ∇2f = 0, the gradient field ∇f is solenoidal.
∇f is irrotational by Theorem 18.8.4
SECTION 18.9
1.∫∫S
(v ·n) dσ =∫∫∫T
(∇ ·v) dxdydz =∫∫∫T
3 dxdydz = 3V = 4π
2.∫∫S
(v ·n) dσ =∫∫∫T
(∇ ·v) dx dy dz =∫∫∫T
(−3) dx dy dz = −3V = −4π
3.∫∫S
(v ·n) dσ =∫∫∫T
(∇ ·v) dxdydz =∫∫∫T
2(x + y + z) dxdydz.
The flux is zero since the function f(x, y, z) = 2(x + y + z) satisfies the relation f(−x,−y,−z) =−f(x, y, z) and T is symmetric about the origin.
4.∫∫S
(v ·n) dσ =∫∫∫T
(∇ ·v) dx dy dz =∫∫∫T
(−2x + 2y + 1) dx dy dz =∫∫∫T
dx dy dz = V =43π
by symmetry
P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU
JWDD027-18 JWDD027-Salas-v1 January 4, 2007 18:12
964 SECTION 18.9
5. face n v ·n flux
x = 0 −i 0 0
x = 1 i 1 1
y = 0 −j 0 0 total flux = 3
y = 1 j 1 1
z = 0 −k 0 0
z = 1 k 1 1
∫∫∫T
(∇ ·v) dxdydz =∫∫∫T
3 dxdydz = 3V = 3
6. face n v · n flux
x = 0 −i −xy = 0 0
x = 1 i xy = y 1/2
y = 0 −j −yz = 0 0 total flux =32
y = 1 j yz = z 1/2
z = 0 −k −xz = 0 0
z = 1 k xz = x 1/2
∫∫∫T
(∇ ·v) dxdydz =∫∫∫T
(y + z + x) dxdydz = (y + z + x)V =(
12
+12
+12
)(1) =
32.
7. face n v ·n flux
x = 0 −i 0 0
x = 1 i 1 1
y = 0 −j xzfluxes add up to 0 total flux = 2
y = 1 j −xz
z = 0 −k 0 0
z = 1 k 1 1
∫∫∫T
(∇ ·v) dxdydz =∫∫∫T
2 (x + z) dxdydz = 2 (x + z)V = 2 ( 12 + 1
2 )1 = 2
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JWDD027-18 JWDD027-Salas-v1 January 4, 2007 18:12
SECTION 18.9 965
8. face n v ·n flux
x = 0 −i −x = 0 0
x = 1 i x = 1 1
y = 0 −j −xy = 0 0 total flux =74
y = 1 j xy = x 1/2
z = 0 −k −xyz = 0 0
z = 1 k xyz = xy 1/4
∫∫∫T
(∇ ·v) dxdydz =∫∫∫T
(1 + x + xy) dxdydz =∫ 1
0
∫ 1
0
∫ 1
0
(1 + x + xy) dx dy dz =74.
9. flux =∫∫∫T
(1 + 4y + 6z) dxdydz = (1 + 4y + 6z)V = (1 + 0 + 3) 9π = 36π
10. flux =∫∫∫T
(∇ ·v) dxdydz =∫ 1
0
∫ 1−x
0
∫ 1−x−y
0
(y + z + x) dz dy dx
= (x + y + z)V =(
34
)(16
)=
18
11. flux =∫∫∫T
(2x + x− 2x) dxdydz∫∫∫T
x dxdydz
=∫ 1
0
∫ 1−x
0
∫ 1−x−y
0
x dz dy dx
=∫ 1
0
∫ 1−x
0
(x− x2 − xy
)dy dx
=∫ 1
0
[xy − x2y − 1
2xy2
]1−x
0
dx
=∫ 1
0
(12x− x2 +
12x3
)dx =
124
12. flux =∫∫∫T
(∇ ·v) dxdydz =∫∫∫T
4y dx dy dz = 4yV = 4(1)(
323
)=
1283
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966 SECTION 18.9
13. flux =∫∫∫T
2(x + y + z) dxdydz =∫ 4
0
∫ 2
0
∫ 2π
0
2(r cos θ + r sin θ + z)r dθ dr dz
=∫ 4
0
∫ 2
0
4π rz dr dz
=∫ 4
0
8π z dz = 64π
14. flux = 12
∫∫S
(v ·n) dσ = 12
∫∫∫T
(∇ ·v) dxdydz =∫∫∫T
(2 + 2x) dx dy dz = V =323π
15. flux =∫∫∫T
(2y + 2y + 3y) dxdydz = 7yV = 0
16. flux =∫∫∫T
(∇ ·v) dxdydz =∫∫∫T
7y dx dy dz = 7 y V = 7(a
2
)a3 =
72a4
17. flux =∫∫∫T
(A + B + C) dxdydz = (A + B + C)V
18.∫∫S
(∇f ·n) dσ =∫∫∫T
[∇ · (∇f)] dxdydz
=∫∫∫T
(∇2f) dxdydz =∫∫∫T
0 dxdydz = 0
19. Let T be the solid enclosed by S and set n = n1i + n2j + n3k.
∫∫S
n1 dσ =∫∫S
(i ·n) dσ =∫∫∫T
(∇ · i) dxdydz =∫∫∫T
0 dxdydz = 0.
Similarly
∫ ∫S
n2 dσ = 0 and∫ ∫S
n3 dσ = 0.
20. (a) The identity follows from setting v = ∇f in (17.8.6).∫∫S
(ff′n) dσ =
∫∫S
(f∇f ·n) dσ =∫∫∫T
[∇ · (f∇f)] dxdydz
=∫∫∫T
[‖ ∇f ‖2 +f(∇2f)
]dxdydz
=∫∫∫T
‖ ∇f ‖2 dxdydz since ∇2f = 0
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SECTION 18.9 967
(b)∫∫S
(gf′n) dσ =
∫∫S
(g∇f ·n) dσ =∫∫∫T
[∇ · (g∇f)] dxdydz
=∫∫∫T
{(∇g ·∇f) + g[∇ · (∇f)]} dxdydz
=∫∫∫T
[(∇g ·∇f) + g(∇2f)] dxdydz
21. A routine computation shows that ∇ · (∇f ×∇g) = 0. Therefore∫∫S
[(∇f ×∇g) ·n] dσ =∫∫∫T
[∇ · ( ∇f ×∇g)] dxdydz = 0.
22. Since ∇ · r = 3, we can write
V =∫∫∫T
dxdydz =∫∫∫T
(∇ · r
3
)dxdydz =
∫∫S
(13r ·n)
dσ, by the divergence theorem.
23. Set F = F1i + F2j + F3k.
F1 =∫ ∫S
[ρ(z − c)i ·n] dσ =∫∫∫T
[∇ · ρ(z − c)i] dxdydz
=∫∫∫T
∂
∂x[ρ(z − c)]︸ ︷︷ ︸ dxdydz = 0.
Similarly F2 = 0.
F3 =∫∫S
[ ρ(z − c)k ·n] dσ =∫∫∫T
[∇ · ρ(z − c)k] dxdydz
=∫∫∫T
∂
∂z[ρ(z − c)] dxdydz
=∫∫∫T
ρ dxdydz = W.
24. τTot · i =∫ ∫S
{[r× ρ(c− z)n] · i} dσ
(12.5.6)
= −∫∫S
[(i× r) · ρ(c− z)n] dσ
= ρ
∫∫S
(z − c)[(i× r) ·n] dσ
divergence theorem
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968 SECTION 18.10
= ρ
∫∫∫T
[∇ · (z − c)(i×r)] dxdydz
i× r = i×(xi + yj + zk) = −zj + yk
(z − c)(i×r) = (z − c)(−zj + yk) = (cz − z2)j + (yz − cy)k
∇ · (z − c)(i×r) = y
τTot · i = ρ
∫∫∫T
y dxdydz = ρyV = y(ρV ) = yW
(r×F) · i = [(xi + yj + zk)×Wk] · i
= (−xW j + yW i) · i = yW = τTot · i
Equality of the other components can be shown in a similar manner.
PROJECT 18.9
1. For r �= 0, ∇ ·E = ∇ · qr−3r = q(−3 + 3)r−3 = 0 by (17.8.8)
2. By the divergence theorem, flux of E out of S =∫∫∫T
(∇ ·E) dx dy dz =∫∫∫T
0 dx dy dz = 0
3. On Sa,n =rr, and thus E ·n = q
rr3
· rr
=q
r2=
q
a2
Thus flux of E out of Sa =∫∫Sa
(E ·n) dσ =∫∫Sa
q
a2dσ =
q
a2(area of Sa) =
q
a2(4πa2) = 4πq.
SECTION 18.10
For Exercises 1–4: n = xi + yj + zk and C : r(u) = cosu i + sinu j, u ∈ [ 0, 2π ].
1. (a)∫∫S
[(∇×v) ·n] dσ =∫∫S
(0 ·n) dσ = 0
(b) S is bounded by the unit circle C : r(u) = cosu i + sinu j, u ∈ [ 0, 2π ].∮C
v(r) ·dr = 0 since v is a gradient.
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SECTION 18.10 969
2. (a)∫∫S
[(∇×v) ·n] dσ =∫∫S
(−2k ·n) dσ = −2∫∫S
z dσ = −2zA = −2(
12
)2π = −2π
Exercise 17, Section 17.7
(b)∮C
v(r) ·dr =∮C
y dx− x dy =∫ 2π
0
(− sin2 u− cos2 u) du = −2π
3. (a)∫∫S
[(∇×v) ·n] dσ =∫∫S
[(−3y2i + 2zj + 2k) ·n
]dσ
=∫∫S
(−3xy2 + 2yz + 2z) dσ
=∫∫S
(−3xy2) dσ
︸ ︷︷ ︸0
+∫∫S
2yz dσ
︸ ︷︷ ︸0
+2∫∫S
z dσ = 2zV = 2(12)2π = 2π
Exercise 17, Section 17.7
(b)∮C
v(r) ·dr =∮C
z2 dx + 2x dy =∮C
2x dy =∫ 2π
0
2 cos2 u du = 2π
4. (a)∫∫S
[(∇×v) ·n] dσ =∫∫S
[(−6yi + 6xj − 2xk) ·n] dσ
=∫∫S
(−2xz) dσ = 0 by symmetry
∮C
v(r) ·dr =∫C
6xz dx− x2 dy = −∫C
x2 dy
= −∫ 2π
0
cos3 u du = −∫ 2π
0
(cosu− sin2 u cosu) du = 0
For Exercises 5–7 take S : z = 2 − x− y with 0 ≤ x ≤ 2, 0 ≤ y ≤ 2 − xand C as the triangle (2, 0, 0), (0, 2, 0), (0, 0, 2). Then C = C1 ∪ C2 ∪ C3 with
C1 : r1(u) = 2(1 − u) i + 2uj, u ∈ [ 0, 1 ],
C2 : r2(u) = 2(1 − u) j + 2uk, u ∈ [ 0, 1 ],
C3 : r3(u) = 2(1 − u)k + 2ui, u ∈ [ 0, 1 ].
n = 13
√3(i + j + k) area of S : A = 2
√3 centroid:
(23 ,
23 ,
23
)
5. (a)∫∫S
[(∇×v) ·n] dσ =∫∫S
13
√3 dσ =
13
√3A = 2
(b)∮C
v(r) ·dr =(∫
C1
+∫C2
+∫C3
)v(r) ·dr = −2 + 2 + 2 = 2
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970 SECTION 18.10
6. (a)∫∫S
[(∇×v) ·n] dσ =∫∫S
[(−2xj − 2yk) ·n] dσ
= −23
√3∫∫S
(x + y) dσ = −23
√3(x + y)A = −16
3
(b)∮C
v(r) ·dr =(∫
C1
+∫C2
+∫C3
)v(r) ·dr = −8
3+ 0 − 8
3= −16
3
7. (a)∫∫S
[(∇×v) ·n] dσ =∫∫S
(yk ·n) dσ =13
√3∫∫S
y dσ =13
√3yA =
43
(b)∮C
v(r) ·dr =(∫
C1
+∫C2
+∫C3
)v(r) ·dr =
(43− 32
5
)+
325
+ 0 =43
8. By (18.10.2) v is a gradient: v = ∇φ. Therefore∫C
v(r) ·dr =∫C
[(∇φ) · dr] = 0 by (18.2.2).
9. The bounding curve is the set of all (x, y, z) with
x2 + y2 = 4 and z = 4.
Traversed in the positive sense with respect to n, it is the curve −C where
C : r(u) = 2 cosu i + 2 sinu j + 4k, u ∈ [0, 2π].
By Stokes’s theorem the flux we want is
−∫C
v(r) ·dr = −∫C
y dx + z dy + x2z2 dz
= −∫ 2π
0
(−4 sin2 u + 8 cosu
)du = 4π.
10. The bounding curve is the set of all (x, y, z) with
x2 + z2 = 9, y = −8.
Traversed in the positive direction with respect to n, it is the curve −C where
C : r(u) = 3 cosu i − 8 j + 3 sinuk, u ∈ [0, 2π].
By Stokes’s theorem the flux we want is
−∫C
v(r) · dr = −∫C
12y dx + 2xz dy − 3x dz
= −∫ 2π
0
(12 sinu− 27 cos2 u) du = 27π.
11. The bounding curve C for S is the bounding curve of the elliptical region Ω : 14x
2 + 19y
2 = 1. Since
∇×v = 2x2yz2i − 2xy2z2j
is zero on the xy-plane, the flux of ∇×v through Ω is zero, the circulation of v about C is zero, and
therefore the flux of ∇×v through S is zero.
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SECTION 18.10 971
12. Let T be the solid enclosed by S. By our condition on v, ∇×v is continuously differentiable on T .
Therefore by the divergence theorem∫∫S
[(∇×v) ·n] dσ =∫∫∫T
[∇ · (∇×v)] dxdydz.
This is zero since the divergence of a curl is zero.
13. C bounds the surface
S : z =√
1 − 12 (x2 + y2), (x, y) ∈ Ω
with Ω : x2 + (y − 12 )2 ≤ 1
4 . Routine calculation shows that ∇×v = yk. The circulation of v with
respect to the upper unit normal n is given by∫∫S
(yk ·n) dσ =∫∫Ω
y dxdy = yA =12
(π4
)=
18π.
(18.7.9)
If −n is used, the circulation is − 18π. Answer: ± 1
8π.
14. ∇×v = i − 2j − 2k. Since the plane x + 2y + z = 0 passes through the origin, it intersects the sphere
in a circle of radius a. The surface S bounded by this circle is a disc of radius a with upper unit
normal
n =16
√6(i + 2j + k).
The circulation of v with respect to n is given by∫∫S
[(∇×v) ·n] dσ =∫∫S
(−5
6
√6)dσ = −5
6
√6A = −5
6
√6πa2.
If −n is used, the circulation is 56
√6πa2 . Answer: ± 5
6
√6πa2.
15. ∇×v = i + 2j + k. The paraboloid intersects the plane in a curve C that bounds a flat surface S that
projects onto the disc x2 + (y − 12 )2 = 1
4 in the xy-plane. The upper unit normal to S is the vector
n = 12
√2 (−j + k). The area of the base disc is 1
4π. Letting γ be the angle between n and k, we
have cos γ = n ·k = 12
√2 and sec γ =
√2. Therefore the area of S is 1
4
√2π. The circulation of v with
respect to n is given by∫∫S
[(∇×v) ·n] dσ =∫∫S
−12
√2 dσ =
(−1
2
√2)
(area of S) = −14π.
If −n is used, the circulation is 14π. Answer: ± 1
4π.
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972 SECTION 18.10
16. ∇×v = −yi − zj − xk. The curve C bounds a flat surface S that projects onto the disc x2 + y2 = b2
in the xy-plane. The upper unit normal to S is the vector n = 12
√2(j + k). The area of the base disc
is πb2. Letting γ be the angle between n and k, we have cos γ = n ·k = 12
√2 and sec γ =
√2.
Therefore the area of S is πb2√
2. The circulation of v with respect to n is given by
∫ ∫S
[(∇×v) ·n] dσ = −12
√2∫ ∫S
(x + z) dσ = −12
√2∫ ∫S
z dσ = −12
√2zA.
by symmetry
It’s clear by symmetry that z = a2, the height at which S intersects the xz-plane. Since A = πb2√
2,
the circulation is −πa2b2. If −n is used, the circulation becomes πa2b2, Answer: ±πa2b2.
17. Straightforward calculation shows that
∇× (a× r) = ∇× [(a2z − a3y) i + (a3x− a1z) j + (a1y − a2x)k] = 2a.
18. ∇× (φ∇ψ) = (∇φ×∇ψ) + φ[∇×∇ψ] = ∇φ×∇ψ
(18.8.7)
since the curl of a gradient is zero. Therefore the result follows from Stokes’s theorem.
19. In the plane of C, the curve C bounds some Jordan region that we call Ω. The surface S ∪ Ω is a
piecewise–smooth surface that bounds a solid T. Note that ∇×v is continuously differentiable on T.
Thus, by the divergence theorem,∫∫∫T
[∇ · (∇×v)] dxdydz =∫∫S∪Ω
[(∇×v) ·n] dσ
where n is the outer unit normal. Since the divergence of a curl is identically zero, we have∫ ∫S∪Ω
[(∇×v) · n] dσ = 0.
Now n is n1 on S and n2 on Ω. Thus∫ ∫S
[(∇×v) · n1] dσ +∫ ∫Ω
[(∇×v) · n2] dσ = 0.
This gives
∫ ∫S
[(∇×v) ·n1] dσ =∫ ∫Ω
[(∇×v) · (−n2)] dσ =∮C
v(r) ·dr
where C is traversed in a positive sense with respect to −n2 and therefore in a positive sense with
respect to n1. (−n2 points toward S.)
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SECTION 18.10 973
20. By the chain ruledx
dt=
d
dt[x(u(t), v(t))] =
∂x
∂uu′(t) +
∂x
∂vv′(t).
Thus ∫C1
v1 dx =∫ b
a
(v1
dx
dt
)dt =
∫ b
a
[v1
∂x
∂uu′(t) + v1
∂x
∂vv′(t)]dt
=∫Cr
v1∂x
∂udu + v1
∂x
∂vdv
by Green’s theorem
=∫ ∫Γ
[∂
∂u
(v1
∂x
∂v
)− ∂
∂v
(v1
∂x
∂u
)]du dv.
The integrand can be written
∂v1
∂u
∂x
∂v+ v1
∂2x
∂u∂v− ∂v1
∂v
∂x
∂v− v1
∂2x
∂v∂u=
∂v1
∂u
∂x
∂v− ∂v1
∂v
∂x
∂u.
equality of partials
Thus we have
∫C
v1 dx =∫ ∫Γ
[∂v1
∂u
∂x
∂v− ∂v1
∂v
∂x
∂u
]du dv.
By our previous choice of unit normal, n = N/ ‖ N ‖ . Therefore
∫ ∫S
[(∇× v1i) ·n] dσ =∫ ∫Γ
[(∇× v1i) ·N] du dv.
Note that
(∇× v1i) ·N =
∣∣∣∣∣∣∣∣∣∣
i j k
∂
∂x
∂
∂y
∂
∂z
v1 0 0
∣∣∣∣∣∣∣∣∣∣·
∣∣∣∣∣∣∣∣∣∣∣
i j k
∂x
∂u
∂y
∂u
∂z
∂u∂x
∂v
∂y
∂v
∂z
∂v
∣∣∣∣∣∣∣∣∣∣∣=(∂v1
∂zj − ∂v1
∂yk)
·[(
∂x
∂v
∂z
∂u− ∂x
∂u
∂z
∂v
)j +(∂x
∂u
∂y
∂v− ∂x
∂v
∂y
∂u
)k]
=∂v1
∂z
(∂x
∂v
∂z
∂u− ∂x
∂u
∂z
∂v
)+
∂v1
∂y
(∂x
∂v
∂y
∂u− ∂x
∂u
∂y
∂v
)
=(∂v1
∂z
∂z
∂u+
∂v1
∂y
∂y
∂u
)∂x
∂v−(∂v1
∂z
∂z
∂v+
∂v1
∂y
∂y
∂v
)∂x
∂u·
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974 REVIEW EXERCISES
Now, by the chain rule,
∂v1
∂u=
∂v1
∂x
∂x
∂u+
∂v1
∂y
∂y
∂u+
∂v1
∂z
∂z
∂u,
∂v1
∂v=
∂v1
∂x
∂x
∂v+
∂v1
∂y
∂y
∂v+
∂v1
∂z
∂z
∂v.
Therefore
(∇× v1i) ·N =(∂v1
∂u− ∂v1
∂x
∂x
∂u
)∂x
∂v−(∂v1
∂v− ∂v1
∂x
∂x
∂v
)∂x
∂u
=∂v1
∂u
∂x
∂v− ∂v1
∂v
∂x
∂u
and, as asserted,
∫ ∫S
[(∇× v1i) ·n] dσ =∫ ∫Γ
[∂v1
∂u
∂x
∂v− ∂v1
∂v
∂x
∂u
]du dv.
REVIEW EXERCISES
1. (a) r(u) = ui + uj, 0 ≤ u ≤ 1;∫C
h · dr =∫ 1
0
(u3 − u2) du = − 112
(b)∫C
h · dr =∫ 1
0
(2u8 − 3u7) du = −1172
2. (a)∫C
h · dr =∫ π/2
0
(− cos3 u sinu + sin3 u cosu) du = 0
(b)∫C
h · dr =∫ π/2
0
(−3 cos11 u sinu + 3 sin11 u cosu) du = 0
3. Since h(x, y) = ∇f where f(x, y) = x2y2 + 12x
2 − y,∫C
h(r) ·dr = f(2, 4) − f(−1, 2) =1192
for any curve C beginning at (−1, 2) and ending at (2, 4).
4. h(x, y) is a gradient:∂P
∂y=
y2 − x2
(x2 + y2)2=
∂Q
∂x; h(x, y) = ∇ arctan (y/x).
Therefore the integrals in (a), (b) and (c) all have the same value.
(a) r(u) = 2 cos u i + 2 sin u j, 0 ≤ u ≤ 34π;
∫C
h · dr =∫ 3π/4
0
[−2 sin u
4(−2 sin u) +
2 cos u
4(2 cos u)
]du =
∫ 3π/4
0
1 du =3π4
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REVIEW EXERCISES 975
5. h(x, y, z) = sin y i + xexy j + sin z k; r(u) = u2 i + u j + u3 k, u ∈ [ 0, 3 ]
x(u) = u2 y(u) = u z(u) = u3, x′(u) = 2u, y′(u) = 1, z′(u) = 3u2
h(r(u)) · r′(u) = 2u sinu + u2eu3+ 3u2 sinu3
∫C
h(r) ·dr =∫ 3
0
(2u sinu + u2eu
3+ 3u2 sinu3
)du
=[− 2u cosu + 2 sinu + 1
3 eu3 − cosu3
]30
= 23 − 6 cos 3 + 2 sin 3 + 1
3 e27 − cos 27
6. h(x, y, z) = x2 i + xy j + z2 k; r(u) = cosu i + sinu j + u2 k, u ∈ [0, π/2]
h(r(u)) · r′(u) = 2u5;∫C
h(r) ·dr =∫ π/2
0
2u5 du =13
(π2
)6
7. F(x, y, z) = xy i + yz j + xz k; r(u) = u i + u2 j + u3 k.
F(r(u)) · r′ = u3 + 5u6; W =∫ 2
−1
(u3 + 5u6)du =[14u4 +
57u7]2−1
=268528
8. F(x, y) = x i + (y − 2) j; r(u) = (u− sin u) i + (1 − cos u) j, 0 ≤ u ≤ 2π.
F(r(u)) · r′ = u− u cosu− 2 sinu;
W =∫ 2π
0
(u− u cosu− 2 sinu)du =[12u2 + u sin u + 3 cos u
]2π0
= 2π2
9. A vector equation for the line segment is: r(u) = (1 + 2u) i + 4uk, u ∈ [ 0, 1 ].
F(r(u)) · r′ = C2 + 20u√
1 + 4u + 20u2;∫C
F · dr = C
∫ 1
0
(20u + 2)√1 + 4u + 20u2
du = 4C
10. Suppose that the path C of the object is given by the vector function r = r(u), a ≤ u ≤ b. Then
r′ = v is the velocity of the object and F ·v = 0. The work done by F is
∫C
F(r) ·dr =∫ b
a
F(r(u)) · r′(u) du =∫ b
a
F(r(u)) ·v(u) du = 0.
11.∂(yexy + 2x)
∂y= exy + xyexy =
∂(xexy − 2y)∂x
=⇒ h is a gradient.
(a) h(r(u)) · r′ = 3u2eu3 − 4u3 + 2u;
∫C
h · dr =∫ 2
0
(3u2eu
3 − 4u3 + 2u)du = e8 − 13
(b) Let f(x, y) = exy + x2 − y2. Then ∇f = h and∫C
h · dr = f(2, 4) − f(0, 0) = e8 − 13
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976 REVIEW EXERCISES
12.∂P
∂y= 4xy + 2 =
∂Q
∂x=⇒ h is a gradient.
(a) h(r(u)) =∫C
h · dr =∫ 1
0
(6 + 66u + 216u2 + 576u3
)du =
[6u + 33u2 + 72u3 + 144u4
]10
= 255
(b) Let f(x, y) = (x2y2 + 2xy). Then ∇f = h and∫C
h · dr = f(3, 5) − f(0, 1) = 255
13. h(x, y, z) = ∇f where f(x, y, z) = x4y3z2.
(a) h(r(u)) = 4u15 i + 3u14 j + 2u13 k; r′(u) = i + 2u j + 3u2 k∫C
h(r) ·dr =∫ 1
0
16u15 du = 1.
(b)∫C
h(r) · dr = f(r(1)) − f(r(0)) = f(1, 1, 1) − f(0, 0, 0) = 1.
14. (a) r(u) = ui + 4uj, 0 ≤ u ≤ 2∫C
y2 dx + (x2 − xy) dy =∫ 2
0
[16u2 + 4(u2 − 4u2)] du =∫ 2
0
4u2 du =323
(b) C1 : r(u) = ui, 0 ≤ u ≤ 2; C2 : r(u) = 2i + uj, 0 ≤ u ≤ 8∫C
y2 dx + (x2 − xy) dy =∫C1
y2 dx + (x2 − xy) dy +∫C2
y2 dx + (x2 − xy) dy = 0 +∫ 8
0
(4 − 2u)du = −32
(c) C : r(u) = ui + u3j, 0 ≤ u ≤ 2∫C
y2dx + (x2 − xy)dy =∫ 2
0
(3u4 − 2u6) du = −60835
15. (a) r(u) = (1 − u)i + u j, 0 ≤ u ≤ 1.∫C
2xy1/2 dx + yx1/2 dy =∫ 1
0
[2(1 − u)u1/2(−1) + u(1 − u)1/2
]du
= −2∫ 1
0
(1 − u)u1/2 du +∫ 1
0
u(1 − u)1/2 du
= −∫ 1
0
(1 − u)u1/2 du = − 415
(b) r1 = i + u j, 0 ≤ u ≤ 1; r2 = (1 − u) i + j∫C
2xy1/2dx + yx1/2dy =∫ 1
0
u du +∫ 1
0
−2(1 − u) du = −12
(c) r = cos u i + sin u j, 0 ≤ u ≤ π/2∫C
2xy1/2dx + yx1/2dy =∫ π/2
0
(−2 sin3/2 u cos u + cos3/2 u sin u
)du = −2
5
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REVIEW EXERCISES 977
16.∫
zdx + xdy + ydz =∫ 2π
0
(a2 cos2 u− au sinu + a sinu)du = πa2 + 2πa
17.∫C
yexy dx + cosx dy + (xy
z) dz =
∫ 2
0
(u2eu
3+ 2u cosu + 3u2
)du
=[
13 e
u3+ 2u sinu + 2 cosu + u3
]20
=13e8 +
173
+ 4 sin 2 + 2 cos 2
18. r = cos u i + sin u j; λ(x, y) = k; s′(u) = ‖r′‖ = 1.
(a) M =∫C
λ(x, y) ds =∫ π
0
k du = kπ
By symmetry, xM = 0.
yM M =∫C
yλ(x, y) ds =∫ π
0
k sin u du =[− k cos u
]π0
= 2k; yM =2π
(b)I =∫C
λ(x, y)R2(x, y) ds =∫C
kx2 ds
=∫ π
0
k cos2 u du =k
2
∫ π
0
(1 + sin 2u) du =12kπ
19. (a) Set C1 : r(u) = u i + u2 j, 0 ≤ u ≤ 1; C2 : r(u) = (1 − u) i +√
1 − u j, 0 ≤ u ≤ 1.
Then, C = C1 + C2.∮C
xy2 dx− x2y dy =∫C1
xy2 dx− x2y dy +∫C2
xy2 dx− x2y dy
=∫ 1
0
(u5 − 2u5) du +∫ 1
0
[−(1 − u)2 + 1
2 (1 − u)2]du
=∫ 1
0
(−u5)du− 1
2
∫ 1
0
(1 − u)2 du =[− 1
6u6 + 1
6 (1 − u)3]10
= − 13
(b) P = xy2; Q = −x2y
∮C
xy2 dx− x2y dy =∫ 1
0
∫ √x
x2(−4xy) dy dx =
∫ 1
0
(2x2 − 2x5
)dx = −1
3
20. (a)∮C
(x2 + y2) dx + (x2 − y2) dy =∫∫
Ω
(2x− 2y) dx dy =∫ 2π
0
∫ 1
0
(2r cos θ − 2r sin θ)r dr dθ = 0
(b)∮C
(x2 + y2) dx + (x2 − y2) dy =∫ 2π
0
(− sinu + cos 2u cosu) du = 0
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978 REVIEW EXERCISES
21. P = x− 2y2; Q = 2xy∮C
(x− 2y2) dx + 2xy dy =∫ 2
0
∫ 1
0
6y dy dx = 6
22.∮C
xy dx +(
12x
2 + xy)dy =
∫∫Ω
y dx dy =∫ 1
−1
∫ √1−x22
0
y dy dx =∫ 1
−1
18 (1 − x2)dx = 1
6
23. P = ln(x2 + y2); Q = ln(x2 + y2);∂Q
∂x− ∂P
∂y=
2x− 2yx2 + y2∮
C
ln(x2 + y2) dx + ln(x2 + y2) dy =∫∫
Ω
2x− 2yx2 + y2
dx dy
=∫ π
0
∫ 2
1
2r cos θ − 2r sin θ
r2r dr dθ
= 2∫ π
0
∫ 2
1
(cos θ − sin θ) dr dθ = −4
24. P = 1/y, Q = 1/x,∂Q
∂x− ∂P
∂y= − 1
x2+
1y2∮
C
(1/y) dx + (1/x) dy =∫∫
Ω
(−x−2 + y−2
)dxdy =
∫ 4
1
∫ √x
1
(−x−2 + y−2
)dy dx
=∫ 4
1
(−x−3/2 − x−1/2 + x−2 + 1
)dx =
34
25.∮
y2dx =∫∫
Ω
−2ydxdy =∫ 2π
0
∫ 1+sin θ
0
−2r2 sin θ dr dθ =∫ 2π
0
(− 23 )(1 + sin θ)3 sin θ dθ = −5π
2
26. P = ey cosx, Q = −ey sinx,∂Q
∂x− ∂P
∂y= −2ey cosx
∮C
ey cosx dx− ey sinx dy =∫∫
Ω
(−2ey cosx) dxdy =∫ π/2
0
∫ 1
0
(−2ey cosx) dy dx = 2(1 − e)
27. C1 : r(u) = −u i + (4 − u2) j, −2 ≤ u ≤ 2; C2 : r(u) = u i, −2 ≤ u ≤ 2; C = C1 ∪ C2
A =12
∫C
(−y dx + x dy) =12
∫C1
(−y dx + x dy) +12
∫C2
(−y dx + x dy)
=12
∫ 2
−2
−(4 − u2)(−1) du− u(−2u) du +12
∫ 2
−2
0 du
=∫ 2
−2
(4 + u2) du =323
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REVIEW EXERCISES 979
28. C1 : r(u) = (3 − 2u) i + (1 + 2u) j, 0 ≤ u ≤ 1; C2 : r(u) = u i + (3/u) j, 1 ≤ u ≤ 3;
C = C1 ∪ C2
A =12
∫C
(−y dx + x dy) =12
∫C1
(−y dx + x dy) +12
∫C2
(−y dx + x dy)
=12
∫ 1
0
[−(1 + 2u)(−2) + (3 − 2u)2] du +12
∫ 3
1
[−(3/u) + u(−3/u2)] du
=12
∫ 1
0
8 du +12
∫ 3
1
(−6/u) du = 4 − 3 ln 3
29. By symmetry, it is sufficient to consider the upper part of the sphere: z =√
4 − x2 − y2
∂z
∂x=
−x√4 − x2 − y2
,∂z
∂y=
−y√4 − x2 − y2
Let Ω be the projection of the sphere onto the xy plane, then
S = 2∫∫
Ω
√(zx)2 + (zy)2 + 1 dx dy = 2
∫∫Ω
2√4 − x2 − y2
dx dy
= 4∫ π/2
−π/2
∫ 2 cos θ
0
1√4 − r2
r dr dθ
= 4∫ π/2
−π/2
(2 − 2
√1 − cos2 θ
)dθ = 8(π − 2)
30. From x + y + 2z = 4, we get z =4 − x− y
2and zx = − 1
2 , zy = − 12 .
area of S =∫∫
Ω
√z2x + z2
y + 1 dx dy =
√32
∫ 2π
0
∫ 2
0
r dr dθ = 2√
6π.
31.∂z
∂x=
x√x2 + y2
,∂z
∂y=
y√x2 + y2
.
The projection Ω of the surface onto the xy plane is the disk x2 + y2 ≤ 9.
S =∫∫
Ω
√(zx)2 + (zy)2 + 1 dx dy =
∫∫Ω
√2 dx dy =
∫ 2π
0
∫ 3
0
√2 r dr dθ = 9π
√2
32. A = 2∫∫
Ω
√z2x + z2
y + 1 dxdy = 2∫∫
Ω
√4x2 + 4y2 + 1 dxdy
= 2∫ 2π
0
∫ 3
0
√1 + 4r2 r dr dθ = 4π
∫ 3
0
√1 + 4r2 r dr
=π
3(373/2 − 1)
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980 REVIEW EXERCISES
33.∫∫
S
yz dσ =√
2∫ 2π
0
∫ 1
0
r2 sin θ(r sin θ + 4) dr dθ =√
2π4
34.∫∫
S
xz dσ =∫∫
S
x(1 − x− y) dσ =∫∫
Ω
x(1 − x− y)√
3 dx dy =√
3∫ 1
0
∫ 1−x
0
x(1 − x− y)dy dx =
√3
24
35. The cylindrical surface S1 is parametrized by: x = u, y = 2 cos v, z = 2 sin v, 0 ≤ u ≤ 2, 0 ≤ v ≤ 2π.
N(u, v) = −2 cos v i − 2 sin v j, ‖|N(u, v)‖| = 2
∫∫S1
(x2 + y2 + z2
)dσ =
∫ 2
0
∫ 2π
0
(u2 + 4
)2 dv du =
128π3
The disc S2 : x = 0, y2 + z2 is parametrized by: x = 0, y = u cos v, z = u sin v, 0 ≤ u ≤ 2,
0 ≤ v ≤ 2π.
N = u i, ‖|N(u, v)‖| = u;∫∫
S2
(x2 + y2 + z2
)dσ =
∫ 2
0
∫ 2π
0
(0 + u2
)u dv du = 8π
The disc S3 : x = 2, y2 + z2 is parametrized by: x = 2, y = u cos v, z = u sin v, 0 ≤ u ≤ 2,
0 ≤ v ≤ 2π.
N = u i, ‖|N(u, v)‖| = u;∫∫
S2
(x2 + y2 + z2
)dσ =
∫ 2
0
∫ 2π
0
(4 + u2
)u dv du = 24π
Thus,∫∫
S
(x2 + y2 + z2
)dσ =
128π3
+ 8π + 24π =224π
3
36. The cylindrical surface S is parametrized by:
x = 2 cos u, y = 2 sinu, z = v, 0 ≤ u ≤ 2π, 0 ≤ v ≤ 1.
N(u, v) = 2 cosu i − 2 sinu j, ‖|N(u, v)‖| = 2;∫∫
S
xz dσ =∫ 1
0
∫ 2π
0
2v cos u(2) dv du = 0.
37. ∇ · v = 4x, ∇ × v = 2yk 38 ∇ · v = 0, ∇ × v = 0
39. ∇ · v = 1 + xy, ∇ × v = (xz − x)i − yzj + zk
40. ∇ · v = yz + x sinxy, ∇ × v = x cosxyi + (xy − y cosxy)j + (y sinxy − xz)k
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REVIEW EXERCISES 981
41. (a) ∇ · v = z − x + y∫ 1
0
∫ 1
0
∫ 1
0
(z − x + y)dzdydx =12
(b) at x = 0, n = −i,v · n = 0,∫ 1
0
∫ 1
0
0dydz = 0
at x = 1, n = i,v · n = z,
∫ 1
0
∫ 1
0
zdydz = 1/2
at y = 0, n = −j,v · n = xy = 0,∫ 1
0
∫ 1
0
0dxdz = 0
at y = 1, n = j,v · n = −xy = −x,
∫ 1
0
∫ 1
0
−xdxdz = −1/2
at z = 0, n = −k,v · n = 0,∫ 1
0
∫ 1
0
0dydx = 0
at z = 1, n = k,v · n = yz,
∫ 1
0
∫ 1
0
ydydx = 1/2
The sum is 1/2
42. (a) ∇ · v = 3∫∫∫T
3 dxdydz =∫ 4
0
∫ 2π
0
∫ 1
0
3r dr dθ dx = 4(2π)( 32 ) = 12π
(b) at x = 0, n = −i, v · n = −z,
∫∫S
−z dydz = 0 (by symmetry)
at x = 4, n = i, v · n = 4 + z,
∫∫S
(4 + z) dydz =∫∫
S
4 dydz = 4π
for z =√
1 − y2, 0 ≤ x ≤ 4, n = −y j +√
1 − y2 k and
v · n = 1 − 2y2 − y√
1 − y2 + x√
1 − y2
∫ 1
−1
∫ 4
0
(1 − 2y2 − y√
1 − y2 + x√
1 − y2)dxdy = 8 − 163
+ 4π
for z = −√
1 − y2, 0 ≤ x ≤ 4, n = y j +√
1 − y2 k and
v · n = −1 + 2y2 − y√
1 − y2 + x√
1 − y2
∫ 1
−1
∫ 4
0
(−1 + 2y2 − y√
1 − y2 + x√
1 − y2)dxdy = −8 +163
+ 4π
The sum is 12π
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982 REVIEW EXERCISES
43. The projection of S onto the xy-plane is: Ω : x2 + y2 ≤ 9.∫∫S
v ·n dσ =∫∫
Ω
(4x2 + 2xyz + z2
)dx dy
=∫∫
Ω
(4x2 + 2xy
[9 − x2 − y2
]+[9 − x2 − y2
]2)dx dy
=∫ 2π
0
∫ 3
0
[4r2 cos2 θ + r2(9 − r2) sin 2θ + (9 − r2)2
]rdr dθ = 324π
44. On x = 0, n = −i, v ·n = −x2 = 0, the flux is 0;
on x = a, n = i, v ·n = a2, the flux is a4;
on y = 0, n = −j, v ·n = xz, the flux is∫ a
0
∫ a
0
xz dx dz =14a2;
on y = a, n = j, v ·n = −xz, the flux is∫ a
0
∫ a
0
−xz dx dz = −14a2;
on z = 0, n = −k, v ·n = 0, the flux is 0;
on z = a, n = k, v ·n = a2, the flux is a4.
Hence the total flux is 2a4.
45. (a) (∇ × v) ·n = (i + j + k) ·(−1
2x i − 1
2y j +
√4 − x2 − y2
2k
)= −1
2x− 1
2y +
√4 − x2 − y2
2
∫∫S
(−1
2x− 1
2y +
√4 − x2 − y2
2
)dσ =
∫∫S
(−1
2x− 1
2y +
√4 − x2 − y2
2
)2√
4 − x2 − y2dx dy
=∫∫ ( −x√
4 − x2 − y2− −y√
4 − x2 − y2+ 1
)dx dy
=∫ 2π
0
∫ 2
0
(− r cos θ√
4 − r2− r sin θ√
4 − r2+ 1)r dr dθ = 4π
(b) r(θ) = 2 cos θ i + 2 sin θ j, 0 ≤ θ ≤ 2π
∫∫S
[(∇ × v) · n]dσ =∮C
v(r) · dr =∫ 2π
0
4 cos2 θdθ = 4π
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REVIEW EXERCISES 983
46. (a) v = z3 i + x j + y2 k; n =2x i + 2y j + k√1 + 4x2 + 4y2
;
∫∫S
[(∇× v) ·n] dσ =∫∫
S
1√1 + 4x2 + 4y2
(4xy + 6yz2 + 1) dσ
=∫∫
Ω
(4xy + 6yz2 + 1) dx dy
=∫∫
Ω
[4xy + 6y(9 − x2 − y2)2 + 1] dx dy
=∫ 2π
0
∫ 3
0
[4r2 cos θ sin θ + 6r sin θ(9 − r2)2 + 1
]r dr dθ
=∫ 3
0
2πr dr = 9π
(b) The boundary of the surface is the curve x2 + y2 = 9, z = 0; r(u) = 3 cos u i + 3 sin u j + 0k;
v(r(u)) = 3 cos u j + 9 sin2 uk; r′(u) = −3 sinu i + 3 cos u j∮C
v ·dr =∫ 2π
0
9 cos2 u du = 9π.
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