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Solution manual calculus 8th edition varberg, purcell, rigdon ch14

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CHAPTER

14

14.1 Concepts Review

Vector Calculus
6.

1. vector-valued function of three real variables or a
vector field
2. gradient field
3. gravitational fields; electric fields
4. ∇ ⋅ F, ∇ × F

Problem Set 14.1
1.

7.

2 x − 3 y , −3 x , 2

8. (cos xyz ) yz , xz , xy
9.

f ( x, y, z ) = ln x + ln y + ln z ;

∇f ( x, y, z ) = x –1 , y –1 , z –1

2.


10.

x, y , z

11. e y cos z , x cos z , – x sin z
12. ∇f ( x, y, z ) = 0, 2 ye –2 z , – 2 y 2 e –2 z
= 2 ye –2 z 0, 1, – y

3.
13. div F = 2x – 2x + 2yz = 2yz
curl F = z 2 , 0, – 2 y

14. div F = 2x + 2y + 2z
curl F = 0, 0, 0 = 0
4.

15. div F =∇ ⋅ F = 0 + 0 + 0 = 0
curl F =∇ × F = x − x, y − y, z − z = 0
16. div F = –sin x + cos y + 0
curl F = 0, 0, 0 = 0
17. div F =e x cos y + e x cos y + 1 = 2e x cos y + 1

5.

curl F = 0, 0, 2e x sin y

18. div F = ∇ ⋅ F = 0 + 0 + 0 = 0
curl F = ∇ × F = 1-1,1-1,1-1 = 0

858

Section 14.1
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


19. a.

g.

meaningless

vector field

b. vector field

h. meaningless

c.

vector field

i.

meaningless

d. scalar field

j.


scalar field

e.

vector field

k. meaningless

f.

vector field

20. a.

div(curl F) = div ⋅ Py – N z , M z – Px , N x – M y

= ( Pyx – N zx ) + ( M zy – Pxy ) + ( N xz – M yz ) = 0

b.

curl(grad f ) = curl f x , f y , f z

= f zy – f yz , f xz – f zx , f yx – f xy = 0

c.

div(fF)=div fM , fN , fP = ( fM x + f x M ) + ( fN y + f y N ) + ( fPz + f z P)
= ( f )( M x + N y + Pz ) + ( f x M + f y N + f z P ) = ( f )(div F ) + (grad f ) ⋅ F

d.


curl(fF )=curl fM , fN , fP
= ( fPy + f y P ) – ( fN z + f z N ), ( fM z + f z M ) – ( fPx + f x P), ( fN x + f x N ) – ( fM y + f y M )
= ( f ) Py – N z , M z – Px , N x – M y + f x , f y , f z × M , N , P = (f)(curl F) + (grad f) × F

21. Let f ( x, y, z ) = – c r
grad( f ) = 3c r

–4

–3

23. grad f = f ′(r ) xr –1/ 2 , f ′(r ) yr –1/ 2 , f ′(r ) zr –1/ 2

, so

r
–5
= 3c r r.
r

(

Then curl F = curl ⎡ –c r
⎢⎣

–3

) r ⎤⎥⎦
r ) × r (by 20d)


( )
(
= ( – c r ) (0) + ( 3c r ) (r × r ) = 0 + 0 = 0
div F =div ⎡( –c r ) r ⎤
⎢⎣
⎥⎦
= ( – c r ) (div r ) + ( 3c r r ) ⋅ r (by 20c)
= ( – c r ) (1 + 1 + 1) + ( 3c r ) r
= ( –3c r ) + 3c r = 0
= –c r

–3

(curl r ) + 3c r

–3

–5

–5

–3

–3

–5

–3


–5

–3

2

3

(

)

(

)

−m ⎤
−m
−m−2
22. curl ⎡⎢ −c r
(0) + mc r
( 0)
r ⎥ = −c r


=0
–m ⎤
–m
– m –2 2
div ⎡⎢ – c r

r ⎥ = –c r
(3) + mc r
r



= (m – 3)c r

(if r ≠ 0)
= f ′(r )r –1/ 2 x, y, z = f ′(r )r –1/ 2 r
curl F = [ f (r )][curl r ] + [ f ′(r )r –1/ 2r ] × r
= [ f (r )][curl r ] + [ f ′(r )r –1/ 2 r ] × r
=0+0=0

24. div F = div[f(r)r] = [f(r)](div r) + grad[f(r)] ⋅ r
= [ f (r )](div r ) + [ f ′(r )r –1r ] ⋅ r
= [ f (r )](3) + [ f ′(r )r –1 ](r ⋅ r )
= 3 f (r ) + [ f ′(r )r –1 ](r 2 ) = 3 f (r ) + rf ′(r )
Now if div F = 0, and we let y = f(r), we have the
dy
= 0, which can be
differential equation 3 y + r
dr
solved as follows:
dy
dr
= –3 ; ln y = –3ln r + ln C = ln Cr –3 ,
y
r


for each C ≠ 0. Then y = Cr –3 , or
f (r ) = Cr –3 , is a solution (even for C = 0).

–m

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Let P = ( x0 , y0 ).
div F = div H = 0 since there is no tendency
toward P except along the line x = x0 , and
along that line the tendencies toward and
away from P are balanced; div G < 0 since
there is no tendency toward P except along
the line x = x0 , and along that line there is
more tendency toward than away from P;
div L > 0 since the tendency away from P is
greater than the tendency toward P.

25. a.

b. No rotation for F, G, L; clockwise rotation
for H since the magnitudes of the forces to
the right of P are less than those to the left.

b. If a paddle wheel is placed at the point (with

its axis perpendicular to the plane), the
velocities over the top half of the wheel will
exceed those over the bottom, resulting in a
net clockwise motion. Using the right-hand
rule, we would expect curl F to point into the
plane (negative z). By calculating
curl F = (0 − 0) i + (0 − 0) j + (0 − 1) k = − k
27.

F ( x, y, z ) = M i + N j + P k , where
M =−

x
(1 + x + y )
2

2 32

, N =−

y
(1 + x + y 2 )3 2
2

,

P=0
1

div F = 0; curl F = 0


c.

2

div G = –2 ye – y < 0 since y > 0 at P; curl
G=0

0.5

div L = ( x 2 + y 2 ) –1/ 2 ; curl L = 0

div H = 0; curl H = 0, 0, – 2 xe – x

2

which
-1

-0.5

0.5

1

points downward at P, so the rotation is
clockwise in a right-hand system.
-0.5

26.


F ( x, y , z ) = M i + N j + P k ,
where
M ( x, y, z ) = y, N ( x, y, z ) = 0, P( x, y, z ) = 0

-1

2

0.5

a. Since all the vectors are directed toward the
origin, we would expect accumulation at that
point; thus div F (0, 0, 0) .should be negative.
Calculating,
3( x 2 + y 2 )
div F ( x, y, z ) =
(1 + x 2 + y 2 )5 2
2

+0
2
(1 + x + y 2 )3 2
so that div F (0, 0, 0) = −2

a. Since the velocity into (1, 1, 0) equals the
velocity out, there is no tendency to diverge
from or accumulate to the point.
Geometrically, it appears that
div F (1,1, 0) = 0 . Calculating,


b. If a paddle wheel is placed at the origin (with
its axis perpendicular to the plane), the force
vectors all act radially along the wheel and so
will have no component acting tangentially
along the wheel. Thus the wheel will not turn
at all, and we would expect curl F = 0 . By
calculating
curl F = (0 − 0) i + (0 − 0) j

1.5

1

1.2

1.4

div F( x, y, z ) =

1.6

1.8

2

∂M ∂N ∂P
+
+
= 0+0+0 = 0

∂x ∂y ∂z



3 yx
3 xy
⎟k
+⎜

⎜ (1 + x 2 + y 2 )3 2 (1 + x 2 + y 2 )3 2 ⎟


=0

860
Section 14.1
Instructor’s Resource Manual
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


30. ∇ 2 f = div(grad f ) = div f x , f y , f z

28. div v = 0 + 0 + 0 = 0;
curl v = 0, 0, w + w = 2ωk
29. ∇f ( x, y, z ) =

= f xx + f yy + f zz

1

mω 2 2 x, 2 y, 2 z = mω 2 x, y, z
2

a.

∇2 f = 4 – 2 – 2 = 0

b.

∇2 f = 0 + 0 + 0 = 0

c.

∇2 f = 6 x – 6 x + 0 = 0

d.

∇ 2 f = div(grad f ) = div grad r

= F(x, y, z)

(

= div – r

–3

)

(


–1

)

r = 0 (by problem 21)

Hence, each is harmonic.
31. a.

F × G = ( f y g z − f z g y )i − ( f x g z − f z g x ) j + ( f x g z − f z g x )k . Therefore,



( f y gz − fz g y ) − ( fx gz − fz gx ) + ( fx g y − f y gx ) .
∂x
∂y
∂z
Using the product rule for partials and some algebra gives
∂f y ⎤
⎡ ∂f
⎡ ∂f y ∂f x ⎤
⎡ ∂f x ∂f z ⎤

+ gz ⎢

div(F × G ) = g x ⎢ z −
⎥ + gy ⎢



⎣ ∂z ∂x ⎦
⎣ ∂y ∂z ⎦
⎣ ∂x ∂y ⎦
∂g y ⎤
⎡ ∂g
⎡ ∂g y ∂g x ⎤
⎡ ∂g x ∂g z ⎤
+ fx ⎢ z −

+ fz ⎢

⎥ + fy ⎢


∂z ⎦
∂x ⎦
∂y ⎦
⎣ ∂z
⎣ ∂y
⎣ ∂x
= G ⋅ curl (F ) − F ⋅ curl (G )
div(F × G ) =

b.

32.

⎛ ∂f ∂g ∂f ∂g ⎞ ⎛ ∂f ∂g ∂f ∂g ⎞ ⎛ ∂f ∂g ∂f ∂g ⎞
∇f × ∇ g = ⎜




⎟i − ⎜
⎟k
⎟ j+ ⎜
⎝ ∂y ∂z ∂z ∂y ⎠ ⎝ ∂x ∂z ∂z ∂x ⎠ ⎝ ∂x ∂y ∂y ∂x ⎠
∂ ⎛ ∂f ∂g ∂f ∂g ⎞ ∂ ⎛ ∂f ∂g ∂f ∂g ⎞ ∂ ⎛ ∂f ∂g ∂f ∂g ⎞



Therefore, div(∇f × ∇g ) = ⎜
⎟− ⎜
⎟.
⎟+ ⎜
∂x ⎝ ∂y ∂z ∂z ∂y ⎠ ∂y ⎝ ∂x ∂z ∂z ∂x ⎠ ∂z ⎝ ∂x ∂y ∂y ∂x ⎠
Using the product rule for partials and some algebra will yield the result
div(∇f × ∇g ) = 0

lim

( x , y , z ) →( a , b , c )

F ( x, y, z ) = L if for each ε > 0 there is a δ > 0 such that 0 < x, y, z – a, b, c < δ implies that

F ( x, y , z ) – L < ε .

F is continuous at (a, b, c) if and only if

lim


( x , y , z ) →( a , b , c )

= F(a, b, c).

14.2 Concepts Review
1. Increasing values of t
n

2.

∑ f ( xi , yi )Δsi
i =1

3.

f ( x(t ), y (t )) [ x′(t )]2 + [ y ′(t )]2

4. F ⋅

dr
dt

Instructor’s Resource Manual
Section 14.2
861
© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


Problem Set 14.2

1.

1

∫0 (27t

3

(

)

+ t 3 )(9 + 9t 4 )1/ 2 dt = 14 2 2 – 1 ≈ 25.5980
1/ 2

t ⎞ ⎛ 1 25t 3 ⎞
2. ∫ ⎜ ⎟ (t ) ⎜ +

⎜4
0⎝ 2 ⎠
4 ⎟⎠

1⎛

⎛ 1 ⎞ 3/ 2
– 1) ≈ 0.2924
dt = ⎜
⎟ (26
⎝ 450 ⎠


3. Let x = t, y = 2t, t in [0, π ] .
π

∫C (sin x + cos y)ds = ∫0 (sin t + cos 2t )

Then

1 + 4dt = 2 5 ≈ 4.4721

4. Vector equation of the segment is
x, y = −1, 2 + t 2, −1 , t in [0, 1].
1

∫0 (–1 + 2t )e
1

5.

∫0 (2t + 9t

6.

∫0



3

2– t


(4 + 1)1/ 2 dt = 5e2 (1 – 3e –1 ) ≈ −1.7124

⎛1⎞
)(1 + 4t 2 + 9t 4 )1/ 2 dt = ⎜ ⎟ (143 / 2 – 1) ≈ 8.5639
⎝6⎠

(16 cos 2 t + 16sin 2 t + 9t 2 )(16sin t 2 + 16 cos 2 t + 9)1/ 2 dt = ∫



0

(16 + 9t 2 )(5)dt



= ⎡80t + 15t 3 ⎤ = 160π + 120π3 ≈ 4878.11

⎦0
2

7.

∫0 [(t

8.

∫0 (–1)dx + ∫–1 (4)

9.


∫C y

10.

∫–2 [(t

–1)(2) + (4t 2 )(2t )]dt =

2

4

3

3

2

100
3

dy = 60

dx + x3 dy = ∫ y 3 dx + x3 dy + ∫
C1

1

2


C2

– 3)3 (2) + (2t )3 (2t )]dt =

y3 dx + x3 dy = ∫

–2

1

2

(–4)3 dy + ∫ (–2)3 dx = 192 + (–48) = 144
–4

828
≈ 23.6571
35

11. y = –x + 2
3

∫1 ([ x + 2(– x + 2)](1) + [ x – 2(– x + 2)](–1))dx = 0
12.

1

∫0 [ x


2

1

+ ( x)2 x]dx = ∫ 3x 2 dx = 1
0

(letting x be the parameter; i.e., x = x, y = x 2 )
13.

x, y, z = 1, 2,1 + t 1, -1, 0
1

∫0 [(4 – t )(1) + (1 + t )(–1) – (2 – 3t + t
14

1

∫0 [(e

3t

2

)(–1)]dt =

17
≈ 2.8333
6


5
⎛1⎞
⎛2⎞
⎛1⎞
)(et ) + (e – t + e2t )(–3– t ) + (et )(2e2t )]dt = ⎜ ⎟ e 4 + ⎜ ⎟ e3 – e + ⎜ ⎟ e –2 –
≈ 23.9726
4
3
2
12
⎝ ⎠
⎝ ⎠
⎝ ⎠

862
Section 14.2
Instructor’s Resource Manual
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


15. On C1 : y = z = dy= dz = 0
On C2 : x = 2, z = dx = dz = 0
On C3 : x = 2, y = 3, dx = dy = 0

2
4
⎡ x2 ⎤

z2 ⎤

2 3
∫0 x dx + ∫0 (2 – 2 y)dy + ∫0 (4 + 3 – z )dz = ⎢⎢ 2 ⎥⎥ + [2 y – y ]0 + ⎢⎢7 z – 2 ⎥⎥ = 2 + (–3) + 20 = 19
⎣ ⎦0

⎦0
2

16.

3

4

x, y, z = t 2,3, 4 , t in [0, 1].
1

∫0 [(9t )(2) + (8t )(3) + (3t )(4)]dt = 27
2
⎛k⎞
17. m = ∫ k x ds = ∫ k x (1 + 4 x 2 )1/ 2 dx = ⎜ ⎟ (173 / 2 – 1) ≈ 11.6821k
–2
C
⎝6⎠

18. Let δ(x, y, z) = k (a constant).


m = k ∫ 1 ds = k ∫ 1(a 2 sin 2 t + a 2 cos 2 t + b 2 )1/ 2 dt = 3πk (a 2 + b 2 )1/ 2
0


C

M xy = k ∫ z ds = k (a 2 + b 2 )1/ 2 ∫



M xz = k ∫ y ds = k (a 2 + b 2 )1/ 2 ∫



0

C

C

M yz = k ∫ x ds =
C

Therefore, x =

19

∫C ( x

20.

∫C e

x


3

9π2 bk (a 2 + b 2 )1/ 2
2

bt dt =

a sin t dt = ak (a 2 + b 2 )1/ 2 (2) = 2ak (a 2 + b 2 )1/ 2

0

k (a 2 + b 2 )1/ 2
a cos t dt
0

M yz
m



= 0; y =

M xy 3πb
M xz 2a
=
=
; z=
.
m

m

2

0
7
≈ −0.1591
– y 3 )dx + xy 2 dy = ∫ [(t 6 – t 9 )(2t ) + (t 2 )(t 6 )(3t 2 )]dt = −
–1
44

5⎡
⎛ 3 ⎞ ⎛ 1 ⎞⎛ 1 ⎞ ⎤
dx – e – y dy = ∫ ⎢(t 3 ) ⎜ ⎟ – ⎜ ⎟⎜ ⎟ ⎥dt = 123.6
1 ⎣
⎝ t ⎠ ⎝ 2t ⎠⎝ t ⎠ ⎦

21. W = ∫ F ⋅ dr = ∫ ( x + y )dx + ( x – y )dy = ∫
C

=∫

π/ 2

0

π/ 2

0


C

[(a cos t + b sin t )(– a sin t ) + (a cos t – b sin t )(b cos t )]dt

[–(a 2 + b 2 ) sin t cos t + ab(cos 2 t – sin 2 t )]dt = ∫

π/ 2

0

π/2

⎡ (a 2 + b 2 ) cos 2t ab sin 2t ⎤
=⎢
+

4
2
⎢⎣
⎥⎦ 0

22.

= ak (a 2 + b 2 )1/ 2 (0) = 0

=

–(a 2 + b 2 ) sin 2t
+ ab cos 2t dt
2


a 2 + b2
–2

x, y, z = t 1,1,1 , t in [0, 1].
1

∫C (2 x – y)dx + 2 z dy + ( y – z)dz = ∫0 (t + 2t + 0)dt = 1.5

Instructor’s Resource Manual
Section 14.2
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© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


23.

π ⎡⎛ π ⎞

⎛ πt ⎞

⎛ πt ⎞

⎛ πt ⎞

⎛ πt ⎞




2

∫0 ⎣⎢⎝⎜ 2 ⎠⎟ sin ⎝⎜ 2 ⎠⎟ cos ⎝⎜ 2 ⎠⎟ + πt cos ⎝⎜ 2 ⎠⎟ + sin ⎝⎜ 2 ⎠⎟ – t ⎦⎥ dt = 2 – π ≈ 1.3634
2

2

0

0

24. W = ∫ F ⋅ dr = ∫ y dx + z dy + x dz = ∫ [(t 2 )(1) + (t 3 )(2t ) + (t )(3t 2 )]dt = ∫ (2t 4 + 3t 3 + t 2 )dt
C

=

C

64
8 412
+ 12 + =
≈ 27.4667
5
3 15

25. The line integral

∫ F • dr represents the work done in moving a particle through the force field

F along the curve


Ci

Ci , i = 1, 2,3,

a. In the first quadrant, the tangential component to C1 of each force vector is in the positive y-direction , the same
direction as the object moves along C1 . Thus the line integral (work) should be positive.
b. The force vector at each point on C2 appears to be tangential to the curve, but in the opposite direction as the
object moves along C2 . Thus the line integral (work) should be negative.
c. The force vector at each point on C3 appears to be perpendicular to the curve, and hence has no component in
the direction the object is moving. Thus the line integral (work) should be zero
26. The line integral

∫ F • dr represents the work done in moving a particle through the force field

F along the curve

Ci

Ci , i = 1, 2,3,

a. In the first quadrant, the tangential component to C1 of each force vector is in the positive y-direction , the same
direction as the object moves along C1 . Thus the line integral (work) should be positive.
b. The force vector at each point on C2 appears to be perpendicular to the curve, and hence has no component in
the direction the object is moving. Thus the line integral (work) should be zero
c. The force vector at each point on C3 is along the curve, and in the same direction as the movement of the
object. Thus the line integral (work) should be positive.
27.




y⎞

Christy needs

28.

2

∫C ⎜⎝1 + 3 ⎟⎠ ds = ∫0 (1 + 10sin

∫C

3

t )[(−90cos 2 t sin t ) 2 + (90sin 2 t cos t ) 2 ]1/ 2 dt = 225

450
= 2.25 gal of paint.
200


0, 0, 1.2 ⋅ dx, dy, dz = ∫ 1.2 dz = ∫ 1.2(4) dt = 38.4π ≈ 120.64 ft-lb
C

0

Trivial way: The squirrel ends up 32π ft immediately above where it started.
( 32π ft)(1.2 lb) ≈ 120.64 ft-lb
29. C: x + y = a

Let x = t, y = a – t, t in [0, a].
Cylinder: x + y = a; ( x + y )2 = a 2 ; x 2 + 2 xy + y 2 = a 2
Sphere: x 2 + y 2 + z 2 = a 2
The curve of intersection satisfies: z 2 = 2 xy; z = 2 xy .

864
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Instructor’s Resource Manual
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currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


Area = 8∫

C

2 xyds = 8∫

a

0

2t (a – t ) (1)2 + (–1) 2 dt = 16 ∫

( )

a

0


at – t 2 dt

a

⎡ a
a 2
⎛ ⎡ ⎛ a 2 ⎞ ⎛ π ⎞ ⎤ ⎡ ⎛ a 2 ⎞ ⎛ −π ⎞ ⎤ ⎞
⎛ t – a ⎞ ⎥⎤
⎢t – 2
2
2
–1 ⎜
2⎟
⎜ ⎢0 + ⎜ ⎟ ⎜ ⎟ ⎥ − ⎢0 + ⎜ ⎟ ⎜ ⎟ ⎥ ⎟
16
= 16 ⎢
at – t +
sin
=
⎜ a ⎟⎥
⎜ ⎢ ⎜⎝ 8 ⎟⎠ ⎝ 2 ⎠ ⎥ ⎢ ⎜⎝ 8 ⎟⎠ ⎝ 2 ⎠ ⎥ ⎟
2
2
⎦ ⎣
⎦⎠
⎝ 2 ⎠ ⎥⎦
⎝⎣
⎢⎣
0
2

= 2a π
Trivial way: Each side of the cylinder is part of a plane that intersects the sphere in a circle. The radius of each
a
⎛ a ⎞⎛ a ⎞ a 2
circle is the value of z in z = 2 xy when x = y = . That is, the radius is 2 ⎜ ⎟ ⎜ ⎟ =
. Therefore, the
2
2
⎝ 2 ⎠⎝ 2 ⎠
2⎤
⎡ ⎛
a 2⎞ ⎥
2

total area of the part cut out is r π ⎜⎜
⎟ = 2a π.
⎢ ⎝ 2 ⎟⎠ ⎥



ka3
0
c
3
(using same parametric equations as in Problem 29)
I x = I y (symmetry)
a

30. I y = ∫ kx 2 ds = 4k ∫ t 2 2dt = 4 2


Iz = Ix + I y = 8 2

ka3
3

31. C: x 2 + y 2 = a 2
⎡ π⎤
Let x = a cos θ, y = a sin θ, θ in ⎢ 0, ⎥ .
⎣ 2⎦

Area = 8∫

C

= 8∫

π/2

= 8∫

π/2

0
0
2

a 2 – x 2 ds

(a sin θ ) (– a sin θ ) 2 + (a cos θ )2 dθ
(a sin θ ) a 2 dθ = 8a 2 [– cosθ ]0π / 2


= 8a

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32. Note that r = a cos θ along C.

Then (a 2 – x 2 – y 2 )1/ 2 = (a 2 – r 2 )1/ 2 = a cos θ .
⎧ x = r cosθ = (a sin θ ) cosθ ⎫
⎡ π⎤
Let ⎨
⎬ , θ in ⎢ 0, ⎥ .
=
=
y
r
sin
(
a
sin
)
sin
θ
θ
θ

⎣ 2⎦


Therefore, x′(θ ) = a cos 2θ ; y ′(θ ) = a sin 2θ .
Then Area = 4∫ (a 2 – x 2 – y 2 )1/ 2 ds = 4 ∫
C

33. a.

∫C

b.

∫C

x 2 y ds = ∫

4

π/ 2

0

π/ 2

(a cos θ )[(a sin 2θ )2 + (a cos 2θ )2 ]1/ 2 dθ = 4a 2 .

(3sin t )2 (3cos t )[(3cos t )2 + (–3sin t )2 ]1/ 2 dt = 81∫

π/2


0

0

π/ 2

⎡⎛ 1 ⎞

sin 2 t cos t dt = 81 ⎢⎜ ⎟ sin 3 t ⎥
3
⎣⎝ ⎠
⎦0

3

3

3

0

0

0

= 27

xy 2 dx + xy 2 dy = ∫ (3 – t )(5 – t )2 (–1)dt + ∫ (3 – t )(5 – t )2 (–1)dt = 2 ∫ (t 3 – 13t 2 + 55t – 75)dt = –148.5


14.3 Concepts Review

7. M y = 2e y – e x = N x so F is conservative.
f ( x, y ) = 2 xe y – ye x + C

1. f(b) – f(a)
2. gradient; ∇f (r )

8. M y = – e – x y –1 = N x , so F is conservative.
f ( x, y ) = e – x ln y + C

3. 0; 0
4. F is conservative.

9. M y = 0 = N x , M z = 0 = Px , and N z = 0 = Py ,

so F is conservative. f satisfies
f x ( x, y , z ) = 3 x 2 , f y ( x, y, z ) = 6 y 2 , and

Problem Set 14.3
1. M y = –7 = N x , so F is conservative.
f ( x, y ) = 5 x 2 – 7 xy + y 2 + C

2. M y = 6 y + 5 = N x , so F is conservative.
f ( x, y ) = 4 x3 + 3xy 2 + 5 xy – y 3 + C

3. M y = 90 x 4 y – 36 y 5 ≠ N x since
N x = 90 x 4 y – 12 y 5 , so F is not conservative.

4. M y = –12 x y + 9 y = N x , so F is

2 3

8

f z ( x, y , z ) = 9 z 2 .
Therefore, f satisfies

1. f ( x, y, z ) = x3 + C1 ( y, z ),
2. f ( x, y , z ) = 2 y 3 + C2 ( x, z ), and
3. f ( x, y, z ) = 3 z 3 + C3 ( x, y ).
A function with an arbitrary constant that
satisfies 1, 2, and 3 is
f ( x, y , z ) = x 3 + 2 y 3 + 3 z 3 + C .

10. M y = 2 x = N y , M z = 2 z = Px , and
N z = 0 = Py , so F is conservative.
f ( x, y, z ) = x 2 y + xz 2 + sin πz + C

conservative.
f ( x, y ) = 7 x – x y + xy + C
5

3 4

9

⎛ 12 ⎞
5. M y = ⎜ – ⎟ x 2 y –3 = N x , so F is conservative.
⎝ 5⎠
⎛2⎞

f ( x, y ) = ⎜ ⎟ x3 y –2 + C
⎝5⎠

6. M y = (4 y 2 )(–2 xy sin xy 2 ) + (8 y )(cos xy 2 ) ≠ N x

since N x = (8 x )(– y 2 sin xy 2 ) + (8)(cos xy 2 ), so F
is not conservative.

866
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11. Writing F in the form
F ( x, y , z ) = M ( x, y , z ) i + N ( x , y , z ) j + P ( x , y , z ) k

we have M ( x, y, z ) =
P ( x, y , z ) =

−2 x
(x + z2 )
−2 z
2

, N ( x, y, z ) = 0,

( x2 + z 2 )


so that
4 xz
∂M
∂N ∂M
∂P
,
,
=0=
=
=
2
2 2
∂y
∂x ∂z
∂x
(x + z )
∂N
∂P
. Thus F is conservative by Thm. D.
=0=
∂z
∂y

We must now find a function f ( x, y, z ) such that
∂f
−2 x
∂f
∂f
−2 z
,

=
= 0,
=
.
2
2
2
∂x ( x + z ) ∂y
∂z ( x + z 2 )
Note that F is a function of x and z alone so f
will be a function of x and z alone.
a. Applying the first condition gives
−2 x
f ( x, y , z ) = ∫
dx
2
x + z2
⎛ 1 ⎞
= ln ⎜
⎟ + C1 ( z )
⎝ x2 + z 2 ⎠
b. Applying the second condition,
−2 z
∂f
∂ ⎛ 1 ⎞ ∂C1
=
= ln ⎜
=
⎟+
2

2
( x + z ) ∂z ∂z ⎝ x 2 + z 2 ⎠ ∂z
−2 z
( x2 + z 2 )

+

∂C1
∂C1
=0
which requires
∂z
∂z

⎛ 1 ⎞
Hence f ( x, y, z ) = ln ⎜
⎟+C
⎝ x2 + z 2 ⎠
12. Writing F in the form
F ( x, y , z ) = M ( x, y , z ) i + N ( x , y , z ) j + P ( x , y , z ) k

we have M ( x, y , z ) = 0, N ( x, y, z ) = 1 + 2 yz 2 ,
P ( x, y , z ) = 1 + 2 y 2 z

so that
∂M
∂N ∂M
∂P ∂N
∂P
=0=

=0=
= 4 yz =
,
,
.
∂y
∂x ∂z
∂x ∂z
∂y
Thus F is conservative by Thm. D.
We must now find a function f ( x, y, z ) such that
∂f
∂f
∂f
= 1 + 2 yz 2 ,
= 0,
= 1 + 2 yz 2 .
∂y
∂x
∂z
Note that f is a function of y and z only.

a. Applying the first condition gives

b. Applying the second condition,
∂C
∂f

= ( y + y2 z2 ) + 1
1 + 2 y2 z =

∂z ∂z
∂z
∂C
= 2 y2 z + 1
∂z
∂C1
which requires
= 1 or C1 ( z ) = z + c .
∂z

Hence f ( x, y, z ) = y + z + y 2 z 2 + C
13. M y = 2 y + 2 x = N x , so the integral is

independent of the path. f ( x, y ) = xy 2 + x 2 y
(3,1)

∫(−1, 2) ( y

+ 2 xy )dx + ( x 2 + 2 xy )dy

2

= [ xy 2 + x 2 y ](3,1)
( −1, 2) = 14

14. M y = e x cos y = N x , so the line integral is

independent of the path.
Let f x ( x, y ) = e x sin y and f y ( x, y ) = e x cos y.
Then f ( x, y ) = e x sin y + C1 ( y ) and

f ( x, y ) = e x sin y + C2 ( x).

Choose f ( x, y ) = e x sin y.
By Theorem A,
(1, π / 2) x

∫(0, 0)

e sin y dx + e x cos y dy

π / 2)
= [e x sin y ](1,
(0, 0) = e

( Or

use line segments (0, 0) to (1, 0), then (1, 0)

⎛ π⎞ ⎞
to ⎜1, ⎟ . ⎟
⎝ 2⎠ ⎠
15. For this problem, we will restrict our
consideration to the set
D = {( x, y ) | x > 0, y > 0}
(that is, the first quadrant), which is open and
simply connected.
a. Now, F ( x, y ) = M i + N j where
M ( x, y ) =

thus


x3
( x 4 + y 4 )2

, N ( x, y ) =

y3
( x 4 + y 4 )2

;

∂M
−8 x3 y 3
∂N
=
=
so F is
4
4
3
∂y ( x + y )
∂x

conservative by Thm. D, and hence

∫ F(r) i dr

is

C


independent of path in D by Theorem C.

f ( x, y, z ) = ∫ (1 + 2 yz 2 ) dy = y + y 2 z 2 + C1 ( z )

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b. Since F is conservative, we can find a function
f ( x, y ) such that
∂f
x3
∂f
y3
=
,
=
.
∂x ( x 4 + y 4 ) 2 ∂y ( x 4 + y 4 )2
Applying the first condition gives
x3
−1
f ( x, y ) = ∫
dx =
+ C ( y)
4

4 2
4
(x + y )
4( x + y 4 )

16. For this problem, we can use the whole real plane
as D.
a. F ( x, y ) = M i + N j where
M ( x, y ) = 3x 2 − 2 xy − y 2 ,
N ( x, y ) = 3 y 2 − 2 xy − x 2
∂M
∂N
so F is
= −2 x − 2 y =
∂y
∂x
conservative by Thm. D, and hence
∫ F(r) i dr is independent of path in D by

thus

Appling the second condition gives
⎞ ∂C
y3
∂f
∂ ⎛
−1
=
= ⎜
=

⎟+
4
4 2
4
4

∂y ∂y ⎝ 4( x + y ) ⎟⎠ ∂y
(x + y )
y3
(x + y )
4

4 2

hence

(6,3)
(2,1)

∂C
;
∂y

b. Since F is conservative, we can find a function
f ( x, y ) such that

∂C
= 0 and C(y) = constant. Therefore
∂y


f ( x, y ) =



+

−1
4( x + y 4 )
4

x3
( x 4 + y 4 )2

C

Theorem C.

dx +

+ c , and so, by Thm. A,

y3
( x4 + y 4 )2

dy

1

⎞ ⎛ 1


= f (6,3) − f (2,1) = ⎜ −
+ c⎟ −⎜− + c⎟
⎝ 5508
⎠ ⎝ 68

20
=
1377

∂f
∂f
= 3x 2 − 2 xy − y 2 ,
= 3 y 2 − 2 xy − x 2 .
∂x
∂y
Applying the first condition gives
f ( x, y ) = ∫ 3x 2 − 2 xy − y 2 dx
= x3 − x 2 y − xy 2 + C ( y )
Appling the second condition gives
∂f
∂ 3
∂C
=
x − x 2 y − xy 2 +
3 y 2 − 2 xy − x 2 =
∂y ∂y
∂y
∂C
= − x 2 − 2 xy +
;

∂y

(

hence
x
, 2 ≤ x ≤ 6 in
2
D which connects the points (2, 1) and (6, 3);
1
then dy = dx . Thus
2

c. Consider the linear path C : y =

(6,3)



x3
(x + y )
4

(2,1)

4 2

dx +

y3

(x + y )
4

4 2

∫ ( x 4 + ( x )4 )2 dx +
6

6

x3

2

2

(

1
1
⎡ −4 ⎤
∫ 17 x5 dx = ⎢⎣17 x 4 ⎥⎦ = − 5508 + 68 =
2
2
16

20
1377

∂C

3
= 3 y 2 and C ( y ) = y + c. Therefore
∂y

f ( x, y ) = x3 − x 2 y − xy 2 + y 3 + c , and so, by
Thm. A,
(4,2)



(3 x 2 − 2 xy − y 2 ) dx + (3 x 2 − 2 xy − y 2 ) dy

( −1,1)

dy =

x 3
)
1
2
( dx)
x 4 2 2
4
(x + ( ) )
2

6

)


= f (4, 2) − f (−1,1) = ( 24 + c ) − ( 0 + c ) = 24

=

c. Consider the simple linear path
C : x = 5 y − 6, 1 ≤ y ≤ 2 in D which connects the

( −1,1)

points

and (4, 2); then dx = 5 dy .

Thus
(4,2)



(3x 2 − 2 xy − y 2 ) dx + (3 y 2 − 2 xy − x 2 ) dy =

( −1,1)
2

∫ (64 y

2

− 168 y + 108)(5 dy ) + (−32 y 2 + 72 y − 36) dy

1

2

= ∫ (288 y 2 − 768 y + 504) dy
1
2

= ⎡96 y 3 − 384 y 2 + 504 y ⎤

⎦1
= 240 + (−216) = 24

868
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17. M y = 18 xy 2 = N x , M z = 4 x = Px , N z = 0 = Py .

23. F( x, y, z ) = k r

By paths (0, 0, 0) to (1, 0, 0); (1, 0, 0) to (1, 1, 0);
(1, 1, 0) to (1, 1, 1)
1

1

∫0 0 dx + ∫0 9 y


2

⎛k⎞
f ( x, y, z ) = ⎜ ⎟ ( x 2 + y 2 + z 2 ) works.
⎝2⎠

1

dy + ∫ (4 z + 1)dz = 6
0
2 3

(Or use f ( x, y ) = 3x y + 2 xz 2 + z.)
18. M y = z = N x , M z = y = Px , N z = x = Py .

f(x, y) = xyz + x + y + z
Thus, the integral equals
1, 1)
[ xyz + x + y + z ](1,
(0.1, 0) = 3.

⎛1⎞
24. Let f = ⎜ ⎟ h(u ) where u = x 2 + y 2 + z 2 .
⎝2⎠
⎛1⎞
⎛1⎞
Then f x = ⎜ ⎟ h′(u )u x = ⎜ ⎟ g (u )(2 x ) = xg (u ).
2
⎝ ⎠
⎝2⎠

Similarly, f y = yg (u ) and f z = zg (u ).

Therefore, f ( x, y, z ) = g (u ) x, y, z
= g ( x 2 + y 2 + z 2 ) x, y, z = F( x, y, z ).

19. M y = 1 = N x , M z = 1 = Px , N z = 1 = Py (so path

independent). From inspection observe that
f ( x, y, z ) = xy + xz + yz satisfies

25.

f = y + z , x + z , x + y , so the integral equals
[ xy + xz +

0, π)
yz ](–1,
(0, 0, 0)

b

= – π. (Or use line

a

b

⎡ ( x ′) 2 ( y ′) 2 ( z ′)2 ⎤
= m⎢
+

+

2
2 ⎥⎦
⎣⎢ 2
a
m⎡
m⎡
2 ⎤b
2
2
= ⎢ r ′(t ) ⎥ = ⎢ r ′(b) – r ′(a) ⎤⎥




2
2
a

20. M y = 2 z = N x , M z = 2 y = Px , N z = 2 x = Py by

paths (0, 0, 0) to (π , 0, 0) , (π , 0, 0) to (π , π , 0) .

∫0

π

cos x dx + ∫ sin y dy = 2
0


Or use f ( x, y, z ) = sin x + 2 xyz – cos y +
21.

z2
.
2

fx = M , f y = N, fz = P
f xy = M y , and f yx = N x , so M y = N x .
f xz = M z and f zx = Px , so M z = Px .
f yz = N z and f zy = Py , so N z = Py .

22.

f x ( x, y , z ) =

– kx
x + y2 + z2
2

26. The force exerted by Matt is not the only force
acting on the object. There is also an equal but
opposite force due to friction. The work done by
the sum of the (equal but opposite) forces is zero
since the sum of the forces is zero.
27.

f ( x, y, z ) = - gmz satisfies
∇f ( x, y, z ) = 0, 0, – gm = F. Then, assuming


the path is piecewise smooth,
(x , y , z )
Work = ∫ F ⋅ dr = [– gmz ]( x2, y 2, z 2)
C
1 1 1
= – gm( z2 – z1 ) = gm( z1 – z2 ).

, so

–k
ln( x 2 + y 2 + z 2 ) + C1 ( y, z ).
f ( x, y , z ) =
2
Similarly,
–k
ln( x 2 + y 2 + z 2 ) + C2 ( y, z ),
f ( x, y , z ) =
2
using f y ; and
f ( x, y , z ) =

b

∫C F ⋅ dr = ∫a (mr ′′ ⋅ r ′)dt
= m ∫ ( x′′x ′ + y ′′y ′ + z ′′z ′)dt

segments (0,1, 0) to (1,1, 0) , then (1,1, 0) to
(1,1,1) .)


π

r
= kr = k x, y, z
r

–k
ln( x 2 + y 2 + z 2 ) + C3 ( y , z ),
2

using f z .
Thus, one potential function for F is
–k
ln( x 2 + y 2 + z 2 ).
f ( x, y , z ) =
2

28. a.

Place the earth at the origin.
GMm ≈ 7.92(1044 )
f (r ) =

–GMm
is a potential function of
r

F(r). (See Example 1.)
147.1(109 )


⎡ – GMm ⎤
Work = ∫ F (r ) ⋅ dr = ⎢

C
⎢⎣ r ⎥⎦ r =152.1(109 )
≈ –1.77(1032 ) joules

b. Zero

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29. a.

M =

y
(x + y )
2

N=–

b.

M =


2

x
( x2 + y2 )
y

(x + y )
2

N=–

2

x
(x + y )
2

2

(x + y )
2

; Nx =

=

2 2

1.


( x2 – y 2 )
( x 2 + y 2 )2

(sin t )
(cos t + sin 2 t )
2

=

Problem Set 14.4

( x2 – y2 )

;My =

= sin t

(– cos t )
(cos t + sin t )
2

2

∫ C 2 xy dx + y

= − cos t

=∫

∫C F ⋅ dr = ∫C Mdx + Ndy

=∫



0

= –∫





0 y2

[(sin t )(– sin t ) + (– cos t )(cos t )]dt

0

2 2y

2

dy = ∫∫ (0 − 2 x)dA
S

–2 x dx dy = –

64
≈ –4.2667
15


2.

1 dt = –2π ≠ 0

30. f is not continuously differentiable on C since f is
undefined at two points of C (where x is 0).
31. Assume the basic hypotheses of Theorem C are
satisfied and assume ∫ F(r ) i dr = 0 for every
C

closed path in D . Choose any two distinct points
A and B in D and let C1, C2 be arbitrary

∫C

positively oriented paths from A to B in D. We
must show that
∫ F(r) i dr = ∫ F(r) i dr

⎛1⎞ 2 x
= ⎜ ⎟∫ ∫
⎝2⎠ 0 0

C1

=–

C2


Let −C2 be the curve C2 with opposite
orientation; then −C2 is a positively oriented path
from B to A in D. Thus the curve C = C1 ∪ −C2
is a closed path (in D) between A and B and so,
by our assumption,
0 = ∫ F (r ) i dr =
C

∫ F(r) i dr + ∫
C1

y dx + x dy = ∫∫

S

2

/2

(

)

1 −1/ 2
– y −1/ 2 dA
x
2

( x –1/ 2 – y –1/ 2 )dy dx


3 2
≈ –0.8485
5

3.

F (r ) i dr =

− C2

∫ F(r) i dr − ∫ F(r) i dr
C1

C2

Thus we have

∫ F(r) i dr = ∫ F(r) i dr
C1

proves independence of path.

14.4 Concepts Review
1.

∂N ∂M

∂x ∂y

C2


which

∫ C (2 x + y

2

)dx + ( x 2 + 2 y )dy = ∫∫ (2 x – 2 y )dA
S

2⎡x
x6 ⎤
(2
x

2
y
)
dy
dx
=


∫0 ⎢ 2 16 ⎥⎥ dx
0 ∫0


16 8 72
= – =
≈ 2.0571

5 7 35

=∫

3

2 x /4

4

2. –2; –2
3. source; sink
4. rotate; irrotational
870
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4.

8.

∫ C xy dx + ( x + y)dy = ∫∫S (1 – x)dA
=∫

1 –2 y + 2




0 0

(1 – x)dx dy =

⎛1⎞
A( S ) = ⎜ ⎟ ∫ x dy – y dx
⎝2⎠ C
⎛ 1 ⎞ 2 ⎡⎛ 3 ⎞
⎛1⎞ ⎤
⎛1⎞ 0
= ⎜ ⎟ ∫ ⎢⎜ ⎟ x3 – ⎜ ⎟ x3 ⎥ dx − ⎜ ⎟ ∫ [2 x 2 – x 2 ]dx
0
⎝ 2 ⎠ ⎣⎝ 2 ⎠
⎝2⎠ ⎦
⎝2⎠ 2

1
3

5.

=

2
3

9.

∫C (x


2

+ 4 xy )dx + (2 x 2 + 3 y )dy = ∫∫ (4 x – 4 x)dA
S

=0
6.

a.

∫∫S div F dA = ∫∫S ( M x + N y )dA
= ∫∫ (0 + 0)dA = 0
S

b.

∫∫S (curl F) ⋅ k dA = ∫∫S ( N x – M y )dA
= ∫∫ (2 x – 2 y )dA = ∫
S
1

∫C

∫1 ∫2

4

(2 x – 2 y )dx dy


0

S

(2 x – 2)dx dy = ∫ 24dy = 24(3) = 72



= ∫ (1 – 2 y )dy = 0

(e3 x + 2 y )dx + ( x 2 + sin y )dy = ∫∫ (2 x – 2)dA

4 6

1 1

0 0

10. a.

∫∫S (0 + 0)dA = 0

1

b.

7.

11. a.
b.


1 1

∫∫S (b – a)dA = ∫0 ∫0 (b – a)dx dy = b – a

∫∫S (0 + 0)d A = 0
∫∫S (3x

2

– 3 y 2 )dA = 0, since for the

integrand, f(y, x) = –f(x, y).

⎛1⎞
A( S ) = ⎜ ⎟ ∫ x dy – y dx
⎝2⎠ C
8
⎛1⎞ 2
⎛1⎞ 0
= ⎜ ⎟ ∫ [4 x 2 – 2 x 2 ]dx + ⎜ ⎟ ∫ [4 x – 4 x]dx =
0
2
3
⎝2⎠
⎝2⎠

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12.
15.

a.

b.

13.

∫∫S div F dA = ∫∫S ( M x + N y )dA
= ∫∫ (1 + 1)dA = 2[ A( S )] = 2π
S

= ∫∫ (–2 y – 2 y )dA = ∫

∫∫S (curl F) ⋅ k dA = ∫∫S ( N x – M y )dA
= ∫∫ (0 – 0)dA = 0
S

∫∫S (curl F) ⋅ kdA = ∫C F ⋅ Tds – ∫C
1

S
1

1 1




0 0

–4 ydxdy

= ∫ –4 y dy = –2
0

16.

F ⋅ Tds

∫ C F ⋅ Tds = ∫∫S (2 y – 2 y)dA = 0

2

17. F is a constant, so N x = M y = 0.

= 30 – (–20) = 50
14.

∫ C F ⋅ T ds = ∫∫S ( N x – M y )dA

W=

∫ C F ⋅ Tds = ∫∫S ( N x – M y )dA = 0

1 1


∫ C F ⋅ n ds = ∫∫S (2 x + 2 x)dA = ∫0 ∫0 4 x dx dy = 2
18.

∫ M dx + N dy = ∫∫S ( N x – M y )dA = 0

Therefore,

∫C F ⋅ dr

∫C F ⋅ dr – ∫C
1

2

is independent of path since

F ⋅ d r = ∫ F ⋅ dr = 0
C

(Where C is the loop C1 followed by –C2 . )
Therefore,

∫C F ⋅ dr = ∫C
1

F ⋅ dr, so F is
2

conservative.


19. a.
b.
c.
20. a.

Each equals ( x 2 – y 2 )( x 2 + y 2 ) –2 .

∫ C y( x

2

+ y 2 ) –1 dx – x ( x 2 + y 2 ) –1 dy = ∫



0

(– sin 2 t – cos 2 t )dt = ∫



0

–1dt

M and N are discontinuous at (0, 0).
Parameterization of the ellipse: x = 3 cos t, y = 2 sin t, t in [0, 2π ].
2π ⎡


∫0

2sin t
3cos t

(–3sin t ) –
(2 cos t ) ⎥ dt = –2π

2
2
2
2
9 cos t + 4sin t
⎣ 9 cos t + 4sin t


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1

) dy + ∫

–1

1


∫–1 –(1 + y

c.

Green’s Theorem applies here. The integral is 0 since N x – M y .

2 –1

1

( x 2 + 1) –1 dx + ∫

–1

b.

1

(1 + y 2 ) –1 dy + ∫ –( x 2 + 1) –1 dx = –2π
–1

21. Use Green’s Theorem with M(x, y) = –y and N(x, y) = 0.
∫ (– y)dx = ∫∫ [0 – (–1)]dA = A(S )
C

S

Now use Green’s Theorem with M(x, y) = 0 and N(x, y) = x.
∫ x dy = ∫∫ (1 – 0)dA = A( S )

C

22.

S

⎛ 1⎞ 2
⎛1⎞
dx = ∫∫ (0 + y )dA = M x ⋅ ∫ ⎜ ⎟ x 2 dy =
S
C⎝ 2⎠

∫ C ⎜⎝ – 2 ⎟⎠ y

∫∫S ( x – 0)dA = M y

⎛ 1 ⎞ 2π
⎛1⎞
⎛ 3⎞
23. A( S ) = ⎜ ⎟ ∫ x dy – y dx = ⎜ ⎟ ∫ [(a cos3 t )(3a sin 2 t )(cos t ) – (a sin 3 t )(3a cos 2 t )(– sin t )]dt = ⎜ ⎟ a 2 π
C
0
2
2
⎝ ⎠
⎝ ⎠
⎝8⎠

24. W =


⎛ 3a 2 π ⎞
15a 2 π
, using
⎟=–

15
⎝ 8 ⎠

∫ C F ⋅ Tds = ∫∫S (curl F) ⋅ kdA = ∫∫S ( N x – M y )dA = ∫∫S (–3 – 2)dA = –5[ A(S )] = –5 ⎜⎜

the result of Problem 23.
25. a.

F ⋅n =

x2 + y2
(x + y )
2

2 3/ 2

div F =

c.

M =

1
(x + y )
2


2 1/ 2

1

1
a

=

1

∫C F ⋅ n ds = a ∫C 1ds = a (2πa) = 2π.

Therefore,

b.

=

( x 2 + y 2 )(1) – ( x)(2 x)
( x 2 + y 2 )2
x

(x + y2 )
2

+

( x 2 + y 2 )(1) – ( y )(2 y )

( x 2 + y 2 )2

=0

is not defined at (0, 0) which is inside C.

d. If origin is outside C, then

∫ C F ⋅ n ds = ∫∫S div F dA = ∫∫S 0 dA = 0.

If origin is inside C, let C ′ be a circle (centered at the origin) inside C and oriented clockwise. Let S be the
region between C and C ′. Then 0 = ∫∫ div F dA (by “origin outide C” case)
S

= ∫ F ⋅ n ds – ∫ F ⋅ n ds (by Green’s Theorem) = ∫ F ⋅ n ds – 2π (by part a), so
C′

C

26. a.

Equation of C:
x, y = x0 , y0 + t x1 – x0 , y1 – y0 ,

C

b.

∫C F ⋅ n ds = 2π.


Area( P ) = ∫ x dy where
C

C = C1 ∪ C2 ∪…∪ Cn and Ci is the ith
edge. (by Problem 21)

t in [0, 1].
Thus;
1

∫C x dy = ∫0 [ x0 + t ( x1 – x0 )]( y1 – y0 )dt ,

= ∫ x dy + ∫

which equals the desired result.

=∑

C1

C2

x dy +…+ ∫

Cn

x dy

( xi – xi –1 )( yi – yi –1 )
(by part a)

2
i =1

c.

Immediate result of part b if each xi and
each yi is an integer.

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d. Formula gives 40 which is correct for the
polygon in the figure below.

b.
29. a.

27. a.

1 3 3 ⎡
⎛ x⎞
⎛ y ⎞⎤
– sec2 ⎜ ⎟ + sec2 ⎜ ⎟ ⎥ dy dx = 0




–3
–3
9
⎝3⎠
⎝ 3 ⎠⎦

div F = –2 sin x sin y
div F < 0 in quadrants I and III
div F > 0 in quadrants II and IV

div F = 4
sin(x)sin(y)

x2 + y 2

b. Flux across boundary of S is 0.

Flux across boundary T is –2(1 – cos 3) 2 .
30. div F =
b. 4(36) = 144
28. a.

1
⎛ x⎞ 1
⎛ y⎞
div F = – sec2 ⎜ ⎟ + sec 2 ⎜ ⎟
9
⎝3⎠ 9
⎝3⎠




⎛ x ⎞⎞
⎛ y ⎞⎞
ln ⎜ cos ⎜ ⎟ ⎟ – ln ⎜ cos ⎜ ⎟ ⎟
3
⎝ ⎠⎠
⎝ 3 ⎠⎠



1 –( x 2 + y 2 ) / 4 2
e
( x + y 2 – 4)
4

so div F < 0 when x 2 + y 2 < 4 and div F > 0
when x 2 + y 2 > 4.

⎛ ( x2 + y2 ) ⎞
exp ⎜ –



4



874
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∫∫R y(4 y

14.5 Concepts Review

=∫

1. surface integral

2

+ 1)1/ 2 dA = ∫

3 (173 / 2

0

– 1)

12

dx =

17

3 2




0 0
3/ 2

(4 y 2 + 1)1/ 2 y dy dx

–1

4

≈ 17.2732

n

2.

∑ g ( xi , yi , zi )ΔSi

7.

i =1

+

f x2

3.


Right side (y = 1):

1

1 1

3

8 3
= ∫ ∫ 3( x + y + x + y + 1)dx dy =
0 0
3

Front (x = 1):

∫0 ∫0 (1 + y)dy dz = 2

∫∫R

2

2

⎛1 1 ⎞
∫∫R x ⎜⎝ 4 + 4 + 1⎟⎠

dA = ∫

∫∫R ( x + y)
=∫


3 1

=∫

3

Therefore, the integral equals
1 3 1 3
1 + 1 + + + + = 6.
2 2 2 2

1 1⎛



6⎞

⎟⎟ x dx dy
⎝ 2 ⎠

0 0⎜

6
4
[– x(4 – x 2 ) –1/ 2 ]2 + 0 + 1 dA

8. Bottom (z = 0): The integrand is 0 so the integral
is 0.
4 8–2 x

128
Left face (y = 0): ∫ ∫
z 1dz dx =
0 0
3
Right face (z = 8 – 2x – 4y):

2( x + y )

2 4–2 y

∫0 ∫0

∫0 (4 – x2 )1/ 2 dy dx

0

0

2x +1
(4 – x 2 )1/ 2

(8 – 2 x – 4 y )(4 + 16 + 1)1/ 2 dx dy

⎛ 32 ⎞
= ⎜ ⎟ 21
⎝ 3 ⎠

dx



⎛ x ⎞⎤
= ⎢ –2(4 – x 2 )1/ 2 + sin –1 ⎜ ⎟ ⎥
⎝ 2 ⎠⎦0

π+6
π
=
= 2 + ≈ 3.0472
3
3

6.

1 1

∫0 ∫0 (0 + y)dy dz = 2

=

5.

3

1 1

∫0 ∫0 ( x + 1)dx dz = 2

Back (x = 0):


1/ 2

4.

1

1 1

∫0 ∫0 ( x + 0)dx dz = 2

[ x 2 + y 2 + ( x + y + 1)](1 + 1 + 1)1/ 2 dA

1 1

3

1 1

∫0 ∫0 ( x + y)dx dy = 1

Left side (y = 0):

Problem Set 14.5

2.

dA

Top (z = 1): Same integral


4. 2; 18π

1.

1/ 2

Bottom (z = 0):

+1

f y2

∫∫R ( x + y)(0 + 0 + 1)

Back face (x = 0):

3

2 8–4 y

∫0 ∫0

z 1 dz dy =

64
3

⎛ 32 ⎞
Therefore, integral = 64 + ⎜ ⎟ 21 ≈ 112.88.
⎝ 3 ⎠


9.

⎛ π⎞
(4r 2 + 1)1/ 2 r dr dθ = ⎜ ⎟ (25 5 + 1)
⎝ 60 ⎠
≈ 2.9794
2π 1 2

∫0 ∫0 r

π sin θ

∫0 ∫0

⎛5⎞
(4r 2 + 1)r dr dθ = ⎜ ⎟ π ≈ 1.9635
⎝8⎠

∫∫G F ⋅ n ds = ∫∫R (– Mf x – Nf y + P)dA
=∫

1 1– y



0 0
1

(8 y + 4 x + 0)dx dy


= ∫ [8(1 – y ) y + 2(1 – y ) 2 ]dy
0
1

= ∫ (–6 y 2 + 4 y + 2)dy = 2
0

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10.

11.

3 (6–2 x ) / 3

∫0 ∫0

M xy = ∫∫ z ds = ∫∫ (a – x – y ) 3 dA

⎛ 1⎞
( x 2 – 9) ⎜ – ⎟ dy dx = 11.25
⎝ 2⎠

5 1


∫0 ∫–1[– xy(1 – y

= 3∫

)

+ 2]dy dx = 20

12.

∫∫R [– Mf x – Nf y + P]dA
= ∫∫ [–2 x( x 2 + y 2 ) –1/ 2 – 5 y ( x 2 + y 2 ) –1/ 2 + 3]dA
R
2π 1

–1

0 0


1

0

0

∫ [(–2r cosθ – 5r sin θ )r

=∫


+ 3]r dr dθ

(–2 cos θ – 5sin θ + 3)dθ ∫ r dr

13. m = ∫∫ kx ds = ∫∫ kx
2

G

R

2




a⎡
(a – y )2
= 3 ∫ ⎢ a(a – y ) –
– y (a – y ) ⎥ dy
0
2
⎢⎣
⎥⎦
a ⎛ a2
y2 ⎞
a3 3
= 3∫ ⎜
– ay +

⎟ dy =
0⎜ 2
2 ⎟⎠
6

M xy a
a
z=
= ; then x = y = (by symmetry).
3
m
3

16. By using the points (a, 0, 0), (0, b, 0), (0, 0, c) we
can conclude that the triangular surface is a
x y z
portion of the plane
+ + = 1 , or
a b c
⎛ x y⎞
z = f ( x, y ) = c ⎜ 1 − − ⎟ , over the region
⎝ a b⎠
Rxy = {( x, y ) | 0 ≤ x ≤ a, 0 ≤ y ≤ b

3dA

( )} . Since
1−

x

a

we are assuming a homogeneous surface, we will
assume δ ( x, y, z ) = 1 .
f x2 + f y2 + 1 dA

R
c2

= ∫∫

a2

R

+

c2 c2
+
b2 c2

b 2 c 2 + a 2 c 2 + a 2b 2
ab

=

dA

∫∫ 1 dA
R


14. m = ∫∫ kxy ds = ∫∫ kxy ( x 2 + y 2 + 1)1/ 2 dA
0 0

(a – x – y )dx dy

S

⎛ 3k ⎞ 4
a a– x 2
= 3k ∫ ∫
x dy dx = ⎜⎜
⎟⎟ a
0 0
⎝ 12 ⎠
G
1 1

R

a. m = ∫∫ 1 dS = ∫∫

⎛1⎞
= (6π) ⎜ ⎟ = 3π ≈ 9.4248
⎝2⎠

=∫




0 0

2 –1/ 2

(In the inside integral, note that the first term is
odd in y.)

=∫

S
a a– y

=(

R

kxy ( x 2 + y 2 + 1)1/ 2 dx dy

ab
)
2

Let w =

⎛k ⎞
= ⎜ ⎟ (9 3 – 8 2 + 1) ≈ 0.3516k
⎝ 15 ⎠

m=


b 2 c 2 + a 2 c 2 + a 2b 2
ab

b c + a 2 c 2 + a 2b2
; then
ab
2 2

abw
2

b. M xy = ∫∫ z dS = ∫∫ zw dA

15.

S
a vx

= w∫

R



c

c ⎞

∫ ⎜⎝ c − a x − b y ⎟⎠ dy dx


0 0

vx =

ab − x
a

a⎛
cb cb
cbx 2 ⎞
wabc
= w∫ ⎜ − x +
⎟ dx =
2 ⎟
⎜ 2 a
6
2a ⎠
0⎝

Let δ = 1.
m = ∫∫ 1 ds = ∫∫ (1 + 1 + 1)1/ 2 dA
S

= 3∫

R

a a– y




0 0

a

dx dy = 3 ∫ (a – y )dy =
0

a2 3
2

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c. Thus z =

M xy
m

=

2( wabc) c
= . In a like
6(abw) 3

19. r ( u , v ) = 2 cos v i + 3sin v j + u k


⎛ x z⎞
manner, using y = g ( x, z ) = b ⎜ 1 − − ⎟ over
⎝ a c⎠
the region
Rxz = {( x, z ) | 0 ≤ x ≤ a, 0 ≤ z ≤ c

( )} and
1−

x
a

y z⎞

x = h( y, z ) = a ⎜1 − − ⎟ over the region
b
c⎠

R yz = {( y, z ) | 0 ≤ y ≤ b, 0 ≤ z ≤ c

( )} ,
1−

y
b

a
b
and y =

so the center
3
3
⎛a b c⎞
of mass is ⎜ , , ⎟ .
⎝ 3 3 3⎠

we can show x =

(

20. r ( u , v ) = u i + 3sin v j + 5cos v k

)

17. r ( u , v ) = u i + 3v j + 4 − u 2 − v 2 k

21. ru (u, v) = sin v i + cos v j + 0 k ,
rv (u, v) = u cos v i − u sin v j + 1k
ru × rv = cos v i − sin v j − u k
ru × rv = cos2 v + sin 2 v + u 2 = 1 + u 2
Using integration formula 44 in the back of the
book we get
6 π

A=

(

)


18. r ( u , v ) = 2u i + 3v j + u 2 + v 2 k

∫∫

u 2 + 1 dv du = π

−6 0

6



u 2 + 1 du =

−6

6

1
⎡u 2

u + 1 + ln u + u 2 + 1 ⎥ =
2
⎣2
⎦ −6

π⎢



π ⎢6 37 + ln
⎢⎣

37 + 6 ⎤
⎥ ≈ 122.49
37 − 6 ⎥⎦

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22. ru (u, v) = cos u sin v i − sin u sin v j + 0 k,
rv (u, v) = sin u cos v i + cos u cos v j + cos v k
ru × rv = − sin u sin v cos v i − cos u sin v cos v j

Using integration formula 48 in the back of the
book we get
A=

+ sin v cos v k
= sin v cos v [ − sin u i − cos u j + 1k ]

=

ru × rv = sin v cos v sin u + cos u + 1
2


2

π ⎡w

4 ⎣8

π ⎡π

(

2 w2 + 25

)

w2 + 25 −

)

(



625
ln w + w2 + 25 ⎥
8
⎦0

625
2
2

ln 4π + 16π 2 + 25
⎢ 32π + 25 16π + 25 −
4 ⎣2
8

+

2
= 2 sin v cos v =
sin 2v
2

625

ln 5⎥ ≈ 5585.42
8


Thus
2
2

A=

2π 2π

∫ ∫
0

sin 2v du dv = 2π




0



sin 2v dv =

0

⎡ π2

π 2


2π ⎢ 4 ∫ sin 2v dv ⎥ = 2 2π [ − cos 2v ]0 =
⎢⎣ 0
⎥⎦
4 2π ≈ 17.77

24. ru (u, v) = − sin u cos v i − sin u sin v j − sin u k,

1

rv (u, v) = − cos u sin v i + cos u cos v j + 0 k
ru × rv = sin u cos u cos v i + sin u cos u sin v j

0.5


− sin u cos u k

0

= sin u cos u [ cos v i + sin v j − 1k ]

1

-0.5

0.5
-1
-1

ru × rv = sin u cos u cos2 v + sin 2 v + 1

0

= 2 sin u cos u =

-0.5
-0.5

0
0.5

Thus (see problem 22)

1 -1


A=

23. ru (u, v) = 2u cos v i + 2u sin v j + 5 k ,
rv (u, v) = −u 2 sin v i + u 2 cos v j + 0 k
ru × rv = −5u cos v i − 5u sin v j + 2u k
2

2

2
sin 2u
2

3

=

2
2

π 2 2π

∫ ∫
0

0

π 2

sin 2u dv du = 2π




sin 2u du

0

2
π 2
π [ − cos 2u ]0 = 2π ≈ 4.443
2

= −u 2 [5cos v i + 5sin v j − 2u k ]
ru × rv = u 2 25 + 4u 2
Thus
2π 2π

A=
= 2π

∫ ∫u

2

0 0

2

∫u


4u 2 + 25 dv du

4u 2 + 25 du

0

=

w = 2u



4


0

dw= 2 du

π



w2
4

1
2

w2 + (5)2 ( dw) =




∫w

2

w2 + (5) 2 dw .

0

878
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25. δ ( x, y, z ) = k z = 5ku (k > 0) . Thus
2π 2π

m=

⎛ 2
∫ ∫ ( 5ku ) ⎜⎝ u

0

0


4u 2 + 25 ⎞⎟ dv du =




5k



u 3 4u 2 + 25 du

0

5k
8

16π 2 + 25



25

5k ⎡ 2
t
32 ⎢⎣ 5

5

2




50
t
3

t = 4u 2 + 25
dt =8u du

28. m = ∫∫ z ds = ∫∫ 3 dA
G

R
2

( = 3 A( R ) = 3π(3) = 27 π, ignoring the subtlety)
2 π 3−ε

∫ ∫
ε →0 0 0

= lim




⎦ 25

= −25sin v (cos u sin v)2 + (sin u sin v)2 + cos 2 v


= 25 sin v sin 2 v + cos 2 v = 25 sin v

16π 2 + 25

2⎤

ru × rv

= 25 sin v [(cos 2 u + sin 2 u ) sin 2 v + cos 2 v

=

⎛ t − 25 ⎞

⎟ t dt =
⎝ 4 ⎠
3

Thus:

5k
[139760 + 833] ≈ 21968 k
32

29. a.

26. a. δ ( x, y, z ) = k x 2 + y 2 = k cos u

3r dr dθ = lim 3(3 − ε )2 π = 27π
ε →0


0 (By symmetry, since
g(x, y, –z) = –g(x, y, z).)

b. 0 (By symmetry, since
g(x, y, –z) = –g(x, y, z).)

Thus
π 2 2π

A=

∫ ∫
0

(k cos u )( 2 sin u cos u ) dv du

c.



sin u cos 2 u du

G

2

∫∫G ( x + y + z )dS = ∫∫G x
= ∫∫ z 2 dS = 3∫∫ x 2 dS
G

G
2

0

− 2 2π k ∫ t 2 dt
1

2

2

2

dS + ∫∫ y 2 dS
G

(due to symmetry of the sphere with respect
to the origin.)
Therefore,
⎛1⎞
2
2
2
2
∫∫G x dS = ⎜⎝ 3 ⎟⎠ ∫∫G ( x + y + z )dS

2 2π k
=
≈ 2.962 k

3

b. δ ( x, y, z ) = k z = k cos u

4πa 4
⎛1⎞
= ⎜ ⎟ 4πa 4 =
.
3
⎝ 3⎠

Thus the density function is the same as in
part a. and hence so is the mass: ≈ 2.962 k
27. ru = −5sin u sin v i + 5cos u sin v j + 0 k and
rv = 5cos u cos v i + 5sin u cos v j + −5sin v k .
Thus,
i
j
k
ru × rv = −5sin u sin v 5cos u sin v
0
5cos u cos v 5sin u cos v −5sin v
= (−25cos u sin 2 v) i + (−25sin u sin 2 v) j +
( −25sin 2 u sin v cos v − 25cos 2 u sin v cos v) k
= (−25cos u sin v) i + (−25sin u sin v) j +
2

2

d. Note:


0

t =cos u
dt =− sin u du

+ y 2 + z 2 )dS = ∫∫ a 2 dS

= a Area(G ) = a (4πa ) = 4πa 4

π 2

=

2

2

0

= 2 2π k

∫∫G ( x

2

+ (−25sin v cos v) k
= (−25sin v) [ cos u sin v i + sin u sin v j + cos v k ]

e.

30. a.

⎛2⎞
2
2
4 8πa
∫∫G ( x + y )dS = ⎜⎝ 3 ⎟⎠ 4πa = 3

4

Let the diameter be along the z-axis.
I z = ∫∫ k ( x 2 + y 2 )dS

1.

G
2

∫∫G x

dS = ∫∫ y 2 dS = ∫∫ z 2 dS (by
G

G

symmetry of the sphere)
2.

∫∫G ( x


2

+ y 2 + z 2 )dS = ∫∫ a 2 dS
G

= a (Area of sphere)=a (4πa 2 ) = 4πa 4
Thus,
2
8πa 4 k
I z = ∫∫ k ( x 2 + y 2 )dS = k (4πa 4 ) =
.
G
3
3
(using 1 and 2)
2

2

Instructor’s Resource Manual
Section 14.5
879
© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


b. Let the tangent line be parallel to the z-axis.
8πa 4 k
Then I = I z + ma 2 =
+ [k (4πa 2 )]a 2

3
=

31. a.

32. x = y = 0
Now let G ′ be the 1st octant part of G.
⎛a⎞
M xy = ∫∫ k dS = 4∫∫ kz dS = 4k ∫∫ z ⎜ ⎟ dA
G
G′
R′ ⎝ z ⎠
(See Problem 19b.)
= 4ak [Area ( R′) ]

20πa k
.
3
4

Place center of sphere at the origin.
F = ∫∫ k (a – z )dS = ka ∫∫ 1 dS – k ∫∫ z dS
G

G

⎡ (a 2 – h12 ) (a 2 – h22 ) ⎤

= 4ak π ⎢


4
4
⎣⎢
⎦⎥

G

= ka (4πa ) – 0 = 4πa k
2

3

= ak π(h22 – h12 )

b. Place hemisphere above xy-plane with center
at origin and circular base in xy-plane.
F = Force on hemisphere + Force on circular
base
= ∫∫ k (a – z )dS + ka (πa 2 )
G

= ka ∫∫ 1 dS – k ∫∫ z dS + πa3 k
G

G

= ka (2πa 2 ) – k ∫∫ z
R

a2

a2 – x2 – y 2

dA + πa3 k

m(G ) = ∫∫ k dS = k[Area(G )]
G

= k[2πa(h2 – h1 )] = 2πak (h2 – h1 )

a
= 3πa3 k – k ∫∫ z dA
R z

Therefore, z =

= 3πa k – ka (πa ) = 2πa k
3

c.

2

3

Place the cylinder above xy-plane with
circular base in xy-plane with the center at
the origin.
F = Force on top + Force on cylindrical side
+ Force on base


πak (h22 – h12 ) h1 + h2
.
=
2πak (h2 – h1 )
2

14.6 Concepts Review
1. boundary; ∂ S

= 0 + ∫∫ k (h − z )dS + kh(πa 2 )

2. F ⋅ n

= kh ∫∫ 1 dS – k ∫∫ z dS + πa 2 hk

3. div F

G

G

G

a2

= kh(2πah) – 4k ∫∫ z

+ 0 + 1 dA + πa 2 hk

a2 – y2

(where R is a region in the yz-plane:
0 ≤ y = a, 0 ≤ z ≤ h )
R

= 2πah 2 k + πa 2 hk – 4k ∫

az

a h



0 0

a2 – y2

4. flux; the shape

Problem Set 14.6
1.

∫∫∫S (0 + 0 + 0)dV = 0

2.

∫∫∫S (1 + 2 + 3)dV = 6V (S ) = 6

3.

∫∫∂ S F ⋅ n dS = ∫∫∫S ( M x + N y + Pz )dV


dz dy

= 2πah 2 k + πa 2 hk – πkah 2
= πah 2 k + πa 2 hk = πahk (h + a)

=∫

1

1

∫ ∫

1

−1 −1 −1

4.

(0 + 1 + 0)dx dy dz = 8

∫∫∂ S F ⋅ n dS = ∫∫∫S ( M x + N y + Pz )dV
= 3∫∫∫ ( x 2 + y 2 + z 2 ) dx dy dz
S
Converting to spherical coordinates we have

880
Section 14.6
Instructor's Resource Manual

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


3∫∫∫ ( x 2 + y 2 + z 2 ) dx dy dz
S

= 3∫∫∫ ρ 2 ( ρ 2 sin φ ) d ρ dθ dφ
S

π 2π a

= 3∫
=
=

5.

4

sin φ ) d ρ dθ dφ

0 0 0
5 π 2π

3a
5

∫ ∫ sin φ dθ dφ


6π a
5

0 0


∫ sin φ dφ =
0

12π a5
5

c b a

c b

∫∫∂ S F ⋅ n dS = ∫∫∫S ( M x + N y + Pz )dV = ∫0 ∫0 ∫0 (2 xyz + 2 xyz + 2 xyz)dx dy dz = ∫0 ∫0 3a
=

6.

∫ ∫ (ρ

2

yz dy dz = ∫

c 3a 2 b 2 z

0


2

dz

3a 2 b 2 c 2
4

⎡⎛ 4 ⎞

∫∫∫S (3 – 2 + 4)dV = 5V (S ) = 5 ⎢⎣⎜⎝ 3 ⎟⎠ π(3)

7. 2∫∫∫ ( x + y + z )dV = 2 ∫
S

2 π 2 4– r 2

0

∫0 ∫0

3⎤

⎥ = 180π = 565.49


(r cosθ + r sin θ + z )r dz dr dθ =

64π
≈ 67.02

3

8.

∫∫∫S ( M x + N y + Pz )dV = ∫∫∫S (2 x + 1 + 2 z )dV
=∫

2π 2

0

=∫

2π ⎛

0

9.

∫0 [(2r

2

=∫

2 π 2 2– r cos θ

0

∫0 ∫0


cos θ + r )(2 – r cos θ ) + r (2 – r cos θ )2 ]dr dθ = ∫

2
⎜ 12 – 4 cos θ –


2π 2

0

8cos θ
3

(2r cos θ + 1 + 2 z )r dz dr dθ

∫0 (6r – r

3

cos 2 θ – r 2 cos θ )dr dθ


⎟ dθ = 20π


∫∫∫S (1 + 1 + 0)dV = 2(volume of cylinder)
4 4– x 4– x – y

= 2π(1)2 (2) = 4π ≈ 12.5664


10.

∫∫∫S (2 x + 2 y + 2 z )dV = ∫0 ∫0 ∫0

11.

∫∫∫S (M x + N y + Pz )dV = ∫∫∫S (2 + 3 + 4)dV
= 9(Volume of spherical shell)

(2 x + 2 y + 2 z )dz dy dx = 64

12.

2π 2 2

∫∫∫S (0 + 0 + 2 z )dV = ∫0 ∫1 ∫0 2 zr dz dr dθ
= 12π ≈ 37.6991

⎛ 4π ⎞
= 9 ⎜ ⎟ (53 – 33 ) = 1176π ≈ 3694.51
⎝ 3 ⎠

Instructor’s Resource Manual
Section 14.6
881
© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.



13.

∫∫ F i n dS = ∫∫∫ (M x + N y + Pz ) dS =

∂S

S

J=

∫∫∫ (0 + 2 y + 0) dV = 2∫∫∫ y dx dy dz
S

∂x
∂ρ
∂x
∂θ
∂x
∂φ

S

J (r , y, θ ) =

∂y
∂r
∂y
∂y
∂y
∂θ


∂z
∂r
∂z
∂y
∂z
∂θ

=

cos θ

0

sin θ

0

1

0

− r sin θ

= ρ sin φ cos θ

ρ cos φ sin θ

cos φ


sin φ cos θ

0

− ρ sin φ sin θ

− ρ sin φ

ρ cos φ cos θ

= − ρ 2 sin3 φ sin 2 θ − ρ 2 sin φ cos 2 φ cos 2 θ −

ρ 2 cos2 φ sin φ sin 2 θ − ρ 2 sin3 φ cos2 θ =

=









r cos θ + r sin θ = r . Further, the region S is

now defined by r 2 ≤ 1, 0 ≤ y ≤ 10 . Hence, by
the change of variable formula in Section 13.9,

3∫∫∫ ( x 2 + y 2 + z 2 ) dx dy dz


2

S
π

2π 10 1

2∫∫∫ y dx dy dz = 2 ∫

∫ ∫ yr dr dy dθ =



0 0

0

4 2π 1

=3∫

0 0 0

S
2π 10

∫ ∫ρ

0


0 0

π


3 4
6π 4
= ∫ ∫ sin φ dθ dφ =
sin φ dφ
5
5 ∫
0

∫∫ F i n dS = ∫∫∫ (M x + N y + Pz ) dV =

=

S

3∫∫∫ ( x 2 + y 2 + z 2 ) dx dy dz
S

Use the change of variable (basically spherical
coordinates with the role of z and y interchanged
and maintaining a right handed system):
x = ρ sin φ sin θ , y = ρ cos φ , z = ρ sin φ cos θ
Then the region S becomes




2 ( ρ 2 sin φ ) d ρ dθ dφ

π

∫ ∫ y dy dθ = ∫ 50 dθ = 100π ≈ 314.16

∂S


⎠⎦

− ρ 2 sin φ ⎢sin 2 φ ⎛⎜ sin 2 θ + cos2 θ ⎞⎟ + cos2 φ ⎛⎜ sin 2 θ + cos2 θ ⎞⎟ ⎥

0 r cos θ

= − ρ 2 sin φ .
Thus,

2

14.

∂z
∂ρ
∂z
∂θ
∂z
∂φ


sin φ sin θ

Using the change of variable (from ( x, y, z ) to
(r , y, θ ) ) defined by
x = r cos θ , y = y, z = r sin θ yields the
Jacobian
∂x
∂r
∂x
∂y
∂x
∂θ

∂y
∂ρ
∂y
∂θ
∂y
∂φ


5

0

0

⎛ 2− 2 ⎞
⎜⎜
⎟⎟ ≈ 1.104

⎝ 2 ⎠

⎛1⎞
15. ⎜ ⎟ ∫∫∫ (1 + 1 + 1)dV = V ( S )
⎝3⎠ S

16.

ρ 2 ≤ 1 ( x 2 + y 2 + z 2 ≤ 1)
0 ≤φ ≤

π

sin 2 φ ≤

1
2

2

( y ≥ 0)
( x2 + z 2 ≤ y 2 )

so that

π

ρ ∈ [0,1], φ ∈ [0, 4 ], θ ∈ [0, 2π ] . The
Jacobian of the transformation is


V (S ) =
=

1
F ⋅ n dS for F = x, y, z
3 ∫∫∂ S

2π a h
1
3 dV = ∫ ∫ ∫ r dz dr dθ
∫∫∫
S
0
0 0
3

=∫

2π a

∫0

0

rh dr dθ = ∫

2π a 2 h

0


2

dθ = 2π

a2h
2

= πa h
2

882
Section 14.6
Instructor’s Resource Manual
© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


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