Calculus one and several variables 10E Salas solutions manual ch18

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



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



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)



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

)



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



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



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



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.



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



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.



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



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.



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.



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



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



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];



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 =∫ 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


x dz =∫ 1

0

du = 1∫C

y dx + 2z dy + x dz = 0 + 0 + 1 = 1



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



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)



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



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



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


x2 dx =∫ 1

0

−(1 − u)2 du = −13∫

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



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



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)



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)



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



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].



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



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



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.



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



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.



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)]



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



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



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



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.



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



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



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



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.



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



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



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



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

)



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


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



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


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


(y + z + x) dxdydz = (y + z + x)V =(

12

+12

+12

)(1) =

32.


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


2 (x + z) dxdydz = 2 (x + z)V = 2 ( 12 + 1

2 )1 = 2



SECTION 18.9 965


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


(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


4y dx dy dz = 4yV = 4(1)(

323

)=

1283



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


(2 + 2x) dx dy dz = V =323π

15. flux =∫∫∫T

(2y + 2y + 3y) dxdydz = 7yV = 0

16. flux =∫∫∫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



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



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.



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



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.



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.



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.)



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·



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



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




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




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




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




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)




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




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




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




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.

Calculus one and several variables 10E Salas solutions manual ch18

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Transcript of Calculus one and several variables 10E Salas solutions manual ch18