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16 VECTOR CALCULUS
16.1 Vector Fields
1. F({> |) = 0= 3 i 3 0 = 4 j
All vectors in this field are identical, with length 0 = 5 and
parallel to h 3 > 34 i.
2. F({> |) = 12 { i + | j
The length of the vector 12 { i + | j is
t 1 4 {
Vectors point roughly away from the origin and vectors
farther from the origin are longer.
3. F({> |) = 3 12 i + (| 3 {) j
The length of the vector 3
1 2 i^ + (|^3 {)^ j^ is
t 1 4 + (|^3 {)
(^2). Vectors along the line | = { are
horizontal with length
1
4. F({> |) = | i + ({ + |) j
The length of the vector | i + ({ + |) j is
s |^2 + ({ + |)^2. Vectors along the {-axis are vertical,
and vectors along the line | = 3 { are horizontal with
length |||.
5. F({> |) =
| i + { j s {^2 + |^2
The length of the vector
| i + { j s {^2 + |^2
is 1.
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624 ¤^ CHAPTER 16 VECTOR CALCULUS
6. F({> |) =
| i 3 { j s {^2 + |^2
All the vectors F({> |) are unit vectors tangent to circles
centered at the origin with radius
s {^2 + |^2.
7. F({> |> }) = k
All vectors in this field are parallel to the }-axis and have
length 1.
8. F({> |> }) = 3 | k
At each point ({> |> }), F({> |> }) is a vector of length |||.
For | A 0 , all point in the direction of the negative }-axis,
while for |? 0 , all are in the direction of the positive
}-axis. In each plane | = n, all the vectors are identical.
9. F({> |> }) = { k
At each point ({> |> }), F({> |> }) is a vector of length |{|.
For { A 0 , all point in the direction of the positive }-axis,
while for {? 0 , all are in the direction of the negative
}-axis. In each plane { = n, all the vectors are identical.
10. F({> |> }) = j 3 i
All vectors in this field have length
I
2 and point in the
same direction, parallel to the {|-plane.
11. F({> |) = h{> 3 |i corresponds to graph IV. In the first quadrant all the vectors have positive {-components and negative
|-components, in the second quadrant all vectors have negative {- and |-components, in the third quadrant all vectors have
negative {-components and positive |-components, and in the fourth quadrant all vectors have positive {- and |-components.
In addition, the vectors get shorter as we approach the origin.
° c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. °°cc 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, o2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, orr duplicated, or posted to a publicly accessible website, in whole or in partduplicated, or posted to a publicly accessible website, in whole or in part..
626 ¤^ CHAPTER 16 VECTOR CALCULUS
22. i ({> |) = tan(3{ 3 4 |) i
Qi({> |) = i{({> |) i + i| ({> |) j =
sec
2 (3{ 3 4 |) · 3
i +
sec
2 (3{ 3 4 |) · ( 3 4)
j
= 3 sec
2 (3{ 3 4 |) i 3 4 sec
2 (3{ 3 4 |) j
23. Qi ({> |> }) = i{({> |> }) i + i| ({> |> }) j + i} ({> |> }) k =
s {^2 + |^2 + }^2
i +
s {^2 + |^2 + }^2
j +
s {^2 + |^2 + }^2
k
24. Qi({> |> }) = i{({> |> }) i + i| ({> |> }) j + i} ({> |> }) k = ln(| 3 2 }) i +
j +
k
= ln(| 3 2 }) i +
j 3
k
25. i ({> |) = {
2 3 | i Qi ({> |) = 2{ i 3 j.
The length of Qi({> |) is
I
4 {^2 + 1. When { 6 = 0, the vectors point away
from the |-axis in a slightly downward direction with length that increases
as the distance from the |-axis increases.
26. i ({> |) =
s {^2 + |^2 i
Qi({> |) =
1 2 ({
2
2 )
31 @ 2 (2{) i +
1 2 ({
2
2 )
31 @ 2 (2|) j
s {^2 + |^2
i +
s {^2 + |^2
j or
s {^2 + |^2
({ i + | j).
Qi ({> |) is not defined at the origin, but elsewhere all vectors have length 1
and point away from the origin.
27. We graph Qi ({> |) =
1 + {^2 + 2|^2
i +
1 + {^2 + 2|^2
j along with
a contour map of i.
The graph shows that the gradient vectors are perpendicular to the
level curves. Also, the gradient vectors point in the direction in
which i is increasing and are longer where the level curves are closer
together.
28. We graph Qi({> |) = 3 sin { i 3 2 cos | j along with a contour map
of i.
The graph shows that the gradient vectors are perpendicular to the
level curves. Also, the gradient vectors point in the direction in
which i is increasing and are longer where the level curves are closer
together.
° c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. °°cc 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, o2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, orr duplicated, or posted to a publicly accessible website, in whole or in partduplicated, or posted to a publicly accessible website, in whole or in part..
SECTION 16.1 VECTOR FIELDS ¤^ 627
29. i ({> |) = {
2
2 i Qi ({> |) = 2{ i + 2| j. Thus, each vector Qi ({> |) has the same direction and twice the length of
the position vector of the point ({> |), so the vectors all point directly away from the origin and their lengths increase as we
move away from the origin. Hence, Qi is graph III.
30. i ({> |) = {({ + |) = {
2
- {| i Qi ({> |) = (2{ + |) i + { j. The |-component of each vector is {, so the vectors
point upward in quadrants I and IV and downward in quadrants II and III. Also, the {-component of each vector is 0 along the
line | = 32 { so the vectors are vertical there. Thus, Qi is graph IV.
31. i ({> |) = ({ + |)
2 i Qi ({> |) = 2({ + |) i + 2({ + |) j. The {- and |-components of each vector are equal, so all
vectors are parallel to the line | = {. The vectors are 0 along the line | = 3 { and their length increases as the distance from
this line increases. Thus, Qi is graph II.
32. i ({> |) = sin
s {^2 + |^2 i
Qi ({> |) =
k cos
s {^2 + |^2 ·
1 2 ({
2
2 )
31 @ 2 (2{)
l i +
k cos
s {^2 + |^2 ·
1 2 ({
2
2 )
31 @ 2 (2|)
l j
cos
s {^2 + |^2 s {^2 + |^2
{ i +
cos
s {^2 + |^2 s {^2 + |^2
| j or
cos
s {^2 + |^2 s {^2 + |^2
( { i + | j)
Thus each vector is a scalar multiple of its position vector, so the vectors point toward or away from the origin with length that
changes in a periodic fashion as we move away from the origin. Qi is graph I.
33. At w = 3 the particle is at (2> 1) so its velocity is V(2> 1) = h 4 > 3 i. After 0.01 units of time, the particle’s change in
location should be approximately 0 = 01 V(2> 1) = 0= 01 h 4 > 3 i = h 0 = 04 > 0 = 03 i, so the particle should be approximately at the
point (2= 04 > 1 =03).
34. At w = 1 the particle is at (1> 3) so its velocity is F(1> 3) = h 1 > 31 i. After 0.05 units of time, the particle’s change in
location should be approximately 0 = 05 F(1> 3) = 0= 05 h 1 > 31 i = h 0 = 05 > 30 = 05 i, so the particle should be approximately at
the point (1= 05 > 2 =95).
35. (a) We sketch the vector field F({> |) = { i 3 | j along with
several approximate flow lines. The flow lines appear to
be hyperbolas with shape similar to the graph of
| = ± 1 @{, so we might guess that the flow lines have
equations | = F@{.
(b) If { = {(w) and | = |(w) are parametric equations of a flow line, then the velocity vector of the flow line at the
point ({> |) is {
0 (w) i + |
0 (w) j. Since the velocity vectors coincide with the vectors in the vector field, we have
0 (w) i + |
0 (w) j = { i 3 | j i g{@gw = {, g|@gw = 3 |. To solve these differential equations, we know
g{@gw = { i g{@{ = gw i ln |{| = w + F i { = ±h
w + F = Dh
w for some constant D, and
° c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. °°cc 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, o2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, rduplicated, or posted to a publicly accessible website, in whole or in partlicated, or posted to a publicly accessible website, in whole or in par.
SECTION 16.2 LINE INTEGRALS ¤^ 629
Integrating by parts with x = w i gx = gw, gy = sin(3 + 3w)gw i y = 3
1 3 cos(3 + 3w)^ gives
U F {^ sin^ | gv^ = 20
1 3 w^ cos(3 + 3w) +^
1 9 sin(3 + 3w)
0
1 3 cos 6 +^
1 9 sin 6 + 0^3
1 9 sin 3
20 9 (sin 6^3 3 cos 6^3 sin 3)
5. If we choose { as the parameter, parametric equations for F are { = {, | =
I
{ for 1 $ { $ 4 and
U
F
{^2 |^3
I
g| =
U 4
1
k {^2 · (
I
{ )^3
I
l 1
2
I
g{ = (^12)
U 4
1
{^3 3
g{
1 2
4 3 {
1 =^
1 2
1 4 + 1
243 8
6. Choosing | as the parameter, we have { = |
3 , | = |, 31 $ | $ 1. Then
U
F h
{ g{ =
U 1
31 h
|^3 · 3 |
2 g| = h
|^3
l 1
31
= h
1 3 h
31 = h 3
1 h.
7. F = F 1 + F 2
On F 1 : { = {, | =
1 2 {^ i^ g|^ =^
1 2 g{,^0 $^ {^ $^2.
On F 2 : { = {, | = 3 3 { i g| = 3 g{, 2 $ { $ 3.
Then
U F ({^ + 2|)^ g{^ +^ {
2 g| =
U
F 1 ({^ + 2|)^ g{^ +^ {
2 g| +
U
F 2 ({^ + 2|)^ g{^ +^ {
2 g|
U 2
0
2 ^1
2
g{ +
U 3
2
2 ( 3 1)
g{
U 2
0
1 2 {
2 ^
g{ +
U 3
2
2 ^
g{
2
1 6 {
3 ^2
0
1 2 {
2 3
1 3 {
3 ^3
2
16 3 3 0 +^
9 2 3
22 3 =^
5 2
8. F = F 1 + F 2
On F 1 : { = 2 cos w i g{ = 3 2 sin w gw, | = 2 sin w i
g| = 2 cos w gw, 0 $ w $ 2.
On F 2 : { = 4w i g{ = 4 gw, | = 2 + w i
g| = gw, 0 $ w $ 1.
Then
U F {
2 g{ + |
2 g| =
U
F 1 {
2 g{ + |
2 g| +
U
F 2 {
2 g{ + |
2 g|
U @ 2
0 (2 cos^ w)
2 ( 3 2 sin w gw) + (2 sin w)
2 (2 cos w gw) +
U 1
0 (4w)
2 (4 gw) + (2 + w)
2 gw
U @ 2
0 (^3 cos
2 w sin w + sin
2 w cos w) gw +
U 1
0 (65w
2
3 cos
3 w +
1 3 sin
3 w
0
3 w
3
2
0
1 3
65 3 + 2 + 4 =^
83 3
° c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. °°cc 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, o2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, rduplicated, or posted to a publicly accessible website, in whole or in partlicated, or posted to a publicly accessible website, in whole or in par.
630 ¤^ CHAPTER 16 VECTOR CALCULUS
9. { = 2 sin w, | = w, } = 3 2 cos w, 0 $ w $ . Then by Formula 9,
U
F {|} gv^ =^
U
0 (2 sin^ w)(w)(^3 2 cos^ w)
t (^) g{ gw
(^) g| gw
(^) g} gw
gw
U
0 34 w^ sin^ w^ cos^ w^
s (2 cos w)^2 + (1)^2 + (2 sin w)^2 gw =
U
0 32 w^ sin 2w^
s 4(cos^2 w + sin^2 w) + 1 gw
I
U
0
w sin 2w gw = 32
I
3 12 w cos 2w + 14 sin 2w
0
integrate by parts with x = w, gy = sin 2w gw
I
2 3 0
I
10. Parametric equations for F are { = 3 1 + 2w, | = 5 + w, } = 4w, 0 $ w $ 1. Then
U
F {|}
2 gv =
U 1
0 (^3 1 + 2w)(5 +^ w)(4w)
2
I
22 + 1^2 + 4^2 gw =
I
U 1
0 (32w
4
3 3 80 w
2 ) gw
I
w
5
w
4
w
3
0
I
5 + 36^3
80 3
236 15
I
11. Parametric equations for F are { = w, | = 2w, } = 3w, 0 $ w $ 1. Then
U
F
{h|}^ gv =
U 1
0
wh(2w)(3w)
I
12 + 2^2 + 3^2 gw =
I
U 1
0
wh^6 w
2 gw =
I
k 1 12 h
6 w^2
l 1
0
I 14 12 (h
s (g{@gw)^2 + (g|@gw)^2 + (g}@gw)^2 =
s 12 + ( 3 2 sin 2w)^2 + (2 cos 2w)^2 =
s 1 + 4(sin^2 2 w + cos^2 2 w) =
I
- Then
U
F ({
2
2
2 ) gv =
U 2
0 (w
2
2 2 w + sin
2 2 w)
I
5 gw =
I
U 2
0 (w
2
I
3 w
3
0
I
3 ) + 2 )
I
3
U
F
{|h
|} g| =
U 1
0
(w)(w
2 )h
(w^2 )(w^3 ) · 2 w gw =
U 1
0
2 w
4 h
w^5 gw = 25 h
w^5
l 1
0
= 25 (h
1 3 h
0 ) = 25 (h 3 1)
U
F | g{^ +^ } g|^ +^ { g}^ =^
U 4
1 w^ ·^
1 2 w
31 @ 2 gw + w
2 · gw +
I
w · 2 w gw =
U 4
1
1 2 w
1 @ 2
2
3 @ 2
gw
k 1 3 w
3 w
5 w
5 @ 2
l 4
1
15. Parametric equations for F are { = 1 + 3w, | = w, } = 2w, 0 $ w $ 1. Then
U
F }
2 g{ + {
2 g| + |
2 g} =
U 1
0 (2w)
2 · 3 gw + (1 + 3w)
2 gw + w
2 · 2 gw =
U 1
0
23 w
2
gw
23 3
w^3 + 3w^2 + w
0
3
3
16. On F 1 : { = w i g{ = gw> | = 0 i
g| = 0 gw> } = w i g} = gw> 0 $ w $ 1.
On F 2 : { = 1 3 w i g{ = 3 gw> | = w i
g| = gw> } = 1 + w i g} = gw> 0 $ w $ 1.
° c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. °°cc 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, o2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, orr duplicated, or posted to a publicly accessible website, in whole or in partduplicated, or posted to a publicly accessible website, in whole or in part..
632 ¤^ CHAPTER 16 VECTOR CALCULUS
25. { = w
2 , | = w
3 , } = w
4 so by Formula 9,
U
F
{ sin(| + }) gv =
U 5
0
(w^2 ) sin(w^3 + w^4 )
s (2w)^2 + (3w^2 )^2 + (4w^3 )^2 gw
U 5
0 w
2 sin(w
3
4 )
I
4 w^2 + 9w^4 + 16w^6 gw E 15 = 0074
U
F }h
3 {| gv =
U 1
0 (h
3 w )h
3 w·w^2
s (1)^2 + (2w)^2 + ( 3 h^3 w)^2 gw =
U 1
0 h
3 w 3 w^3 I 1 + 4w^2 + h^32 w^ gw E 0 = 8208
27. We graph F({> |) = ({ 3 |) i + {| j and the curve F. We see that most of the vectors starting on F point in roughly the same
direction as F, so for these portions of F the tangential component F · T is positive. Although some vectors in the third
quadrant which start on F point in roughly the opposite direction, and hence give negative tangential components, it seems
reasonable that the effect of these portions of F is outweighed by the positive tangential components. Thus, we would expect
U
F
F · gr =
U
F
F · T gv to be positive.
To verify, we evaluate
U
F F^ ·^ gr. The curve^ F^ can be represented by^ r(w) = 2 cos^ w^ i^ + 2 sin^ w^ j,^0 $^ w^ $^
3 2 ,
so F(r(w)) = (2 cos w 3 2 sin w) i + 4 cos w sin w j and r^0 (w) = 3 2 sin w i + 2 cos w j. Then
U
F F^ ·^ gr^ =^
U 3 @ 2
0 F(r(w))^ ·^ r
0 (w) gw
U 3 @ 2
0
[ 3 2 sin w(2 cos w 3 2 sin w) + 2 cos w(4 cos w sin w)] gw
U 3 @ 2
0 (sin
2 w 3 sin w cos w + 2 sin w cos
2 w) gw
2 3 [using a CAS]
28. We graph F({> |) =
s {^2 + |^2
i +
s {^2 + |^2
j and the curve F. In the
first quadrant, each vector starting on F points in roughly the same direction
as F, so the tangential component F · T is positive. In the second quadrant,
each vector starting on F points in roughly the direction opposite to F, so
F · T is negative. Here, it appears that the tangential components in the first
and second quadrants counteract each other, so it seems reasonable to guess
that
U
F F^ ·^ gr^ =^
U
F F^ ·^ T^ gv^ is zero. To verify, we evaluate^
U
F F^ ·^ gr. The curve^ F^ can be represented by
r(w) = w i + (1 + w
2 ) j, 31 $ w $ 1 , so F(r(w)) =
w t
w^2 + (1 + w^2 )
2
i +
1 + w^2 t
w^2 + (1 + w^2 )
2
j and r
0 (w) = i + 2w j. Then
U
F F^ ·^ gr^ =^
U 1
31 F(r(w))^ ·^ r
0 (w) gw =
] 1
31
C
w t
w^2 + (1 + w^2 )
2
2 w(1 + w^2 ) t
w^2 + (1 + w^2 )
2
D (^) gw
] 1
31
w(3 + 2w
2 ) I w^4 + 3w^2 + 1
gw = 0 [since the integrand is an odd function]
° c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. °°cc 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, o2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, orr duplicated, or posted to a publicly accessible website, in whole or in partduplicated, or posted to a publicly accessible website, in whole or in part..
SECTION 16.2 LINE INTEGRALS ¤^ 633
29. (a)
U
F F^ ·^ gr^ =^
U 1
0
G
h
w^231
w
5
H
· 2 w> 3 w
2 ^
gw =
U 1
0
2 wh
w^231
7
gw =
k h
w^231
3 8 w
8
l 1
0
11 8 3 1 @h
(b) r(0) = 0 , F(r(0)) = h
31
0
r
I^1 2
G
1 2 >^
1 2
I 2
H
, F
r
I^1 2
G
h^31 @^2 > 1 4
I 2
H
r(1) = h 1 > 1 i, F(r(1)) = h 1 > 1 i.
In order to generate the graph with Maple, we use the line command in
the plottools package to define each of the vectors. For example,
v1:=line([0,0],[exp(-1),0]):
generates the vector from the vector field at the point (0> 0) (but without an arrowhead) and gives it the name v1. To show
everything on the same screen, we use the display command. In Mathematica, we use ListPlot (with the
PlotJoined - A True option) to generate the vectors, and then Show to show everything on the same screen.
30. (a)
U
F F^ ·^ gr^ =^
U 1
31 2 w> w
2
3 w
· h 2 > 3 > 32 wi gw =
U 1
31 (4w^ + 3w
2 3 6 w
2 ) gw =
2 w
2 3 w
3
31 =^32
(b) Now F(r(w)) = 2 w> w
2
3 w
, so F(r( 3 1)) = h 32 > 1 > 33 i, F
r
1 2
1 4 >^3
3 2
, F
r
1 2
1 4 >^
3 2
and F(r(1)) = h 2 > 1 > 3 i.
31. { = h^3 w^ cos 4w, | = h^3 w^ sin 4w, } = h^3 w, 0 $ w $ 2 .
Then
g{
gw
= h
3 w ( 3 sin 4w)(4) 3 h
3 w cos 4w = 3 h
3 w (4 sin 4w + cos 4w),
g|
gw
= h
3 w (cos 4w)(4) 3 h
3 w sin 4w = 3 h
3 w ( 3 4 cos 4w + sin 4w), and
g}
gw
= 3 h
3 w , so
v g{
gw
g|
gw
g}
gw
s ( 3 h^3 w)^2 [(4 sin 4w + cos 4w)^2 + ( 3 4 cos 4w + sin 4w)^2 + 1]
= h^3 w^
s 16(sin^2 4 w + cos^2 4 w) + sin^2 4 w + cos^2 4 w + 1 = 3
I
2 h^3 w
Therefore
U
F {
3 |
2 } gv =
U 2
0 (h
3 w cos 4w)
3 (h
3 w sin 4w)
2 (h
3 w ) (
I
2 h
3 w ) gw
U 2
I
2 h
37 w cos
3 4 w sin
2 4 w gw =
172 , 704 5 , 632 , 705
I
2 (1 3 h
314 )
32. (a) We parametrize the circle F as r(w) = 2 cos w i + 2 sin w j, 0 $ w $ 2 . So F(r(w)) = 4 cos
2 w> 4 cos w sin w
r
0 (w) = h 3 2 sin w> 2 cos wi, and Z =
U
F F^ ·^ gr^ =^
U 2
0 (^3 8 cos
2 w sin w + 8 cos
2 w sin w) gw = 0.
° c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. °°cc 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, o2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, rduplicated, or posted to a publicly accessible website, in whole or in partlicated, or posted to a publicly accessible website, in whole or in par.
SECTION 16.2 LINE INTEGRALS ¤^ 635
I
2 (4 ^2 + 3)
] 2
0
I
2 cos w
(w
2
I
2 (4 ^2 + 3)
] 2
0
I
2 sin w
(w
2
- gw = 0. Hence ({> |> }) =
2
4 ^2 + 3
37. From Example 3, ({> |) = n(1 3 |), { = cos w, | = sin w, and gv = gw, 0 $ w $ i
L{ =
U
F |
2 ({> |) gv =
U
0 sin
2 w [n(1 3 sin w)] gw = n
U
0 (sin
2 w 3 sin
3 w) gw
1 2 n^
U
0 (1^3 cos 2w)^ gw^3 n^
U
0 (1^3 cos
2 w) sin w gw
Let x = cos w, gx = sin w gw in the second integral
= n
k 2 +^
U 31
1 (1^3 x
2 ) gx
l = n
2 3
4 3
L| =
U
F
2 ({> |) gv = n
U
0
cos
2 w (1 3 sin w) gw = n 2
U
0
(1 + cos 2w) gw 3 n
U
0
cos
2 w sin w gw
= n
2 3
, using the same substitution as above.
38. The wire is given as { = 2 sin w, | = 2 cos w, } = 3w, 0 $ w $ 2 with ({> |> }) = n. Then
gv =
s (2 cos w)^2 + ( 3 2 sin w)^2 + 3^2 gw =
s 4(cos^2 w + sin
2 w) + 9 gw =
I
13 gw and
L{ =
U
F (|
2
2 ) ({> |> }) gv =
U 2
0 (4 cos
2 w + 9w
2 )(n)
I
13 gw =
I
13 n
1 2 w^ +^
1 4 sin 2w
3
0
I
13 n(4 + 24 ^3 ) = 4
I
13 n(1 + 6 ^2 )
L| =
U
F ({
2
2 ) ({> |> }) gv =
U 2
0
4 sin
2 w + 9w
2
(n)
I
13 gw =
I
13 n
1 2 w^3
1 4 sin 2w
3
0
I
13 n(4 + 24 ^3 ) = 4
I
13 n(1 + 6 ^2 )
L} =
U
F ({
2
2 ) ({> |> }) gv =
U 2
0 (4 sin
2 w + 4 cos
2 w)(n)
I
13 gw = 4
I
13 n
U 2
0 gw^ = 8 ^
I
13 n
39. Z =
U
F
F · gr =
U 2
0
hw 3 sin w> 3 3 cos wi · h 1 3 cos w> sin wi gw
U 2
0 (w^3 w^ cos^ w^3 sin^ w^ + sin^ w^ cos^ w^ + 3 sin^ w^3 sin^ w^ cos^ w)^ gw
U 2
0
(w 3 w cos w + 2 sin w) gw =
1 2 w
(^2 3) (w sin w + cos w) 3 2 cos w
0
integrate by parts in the second term
2
40. Choosing | as the parameter, the curve F is parametrized by { = |
2
- 1, | = |, 0 $ | $ 1. Then
Z =
U
F F^ ·^ gr^ =^
U 1
0
G
2
|h
|^2 +
H
· h 2 |> 1 i g| =
U 1
0
k 2 |
2
|^2 +
l g|
k 1 3
2
1 2 h
|^2 +
l 1
0
8 3 +^
1 2 h
2 3
1 3 3
1 2 h^ =^
1 2 h
2 3
1 2 h^ +^
7 3
41. r(w) = h 2 w> w> 1 3 wi, 0 $ w $ 1.
Z =
U
F F^ ·^ gr^ =^
U 1
0 2 w^3 w
2
w 3 (1 3 w)
2
1 3 w 3 (2w)
2 ^
· h 2 > 1 > 31 i gw
U 1
0 (4w^3 2 w
2
2 3 1 + w + 4w
2 ) gw =
U 1
0 (w
2
3 w
3
2 3 2 w
0
7 3
42. r(w) = 2 i + w j + 5w k, 0 $ w $ 1. Therefore
Z =
U
F F^ ·^ gr^ =
] 1
0
Nh 2 > w> 5 wi
(4 + 26w^2 )^3 @^2
· h 0 > 1 > 5 i gw = N
] 1
0
26 w
(4 + 26w^2 )^3 @^2
gw = N
k 3 (4 + 26w
2 )
31 @ 2
l 1
0
= N
1 2 3
I^1 30
° c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. °°cc 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, o2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, rduplicated, or posted to a publicly accessible website, in whole or in partlicated, or posted to a publicly accessible website, in whole or in par.
636 ¤^ CHAPTER 16 VECTOR CALCULUS
43. (a) r(w) = dw
2 i + ew
3 j i v(w) = r
0 (w) = 2dw i + 3ew
2 j i a(w) = v
0 (w) = 2d i + 6ew j, and force is mass times
acceleration: F(w) = p a(w) = 2pd i + 6pew j.
(b) Z =
U
F F^ ·^ gr^ =^
U 1
0 (2pd^ i^ + 6pew^ j)^ ·^ (2dw^ i^ + 3ew
2 j) gw =
U 1
0 (4pd
2 w + 18pe
2 w
3 ) gw
2 pd
2 w
2
9 2 pe
2 w
4 ^1
0
= 2pd
2
9 2 pe
2
44. r(w) = d sin w i + e cos w j + fw k i v(w) = r
0 (w) = d cos w i 3 e sin w j + f k i a(w) = v
0 (w) = 3 d sin w i 3 e cos w j
and F(w) = p a(w) = 3 pd sin w i 3 pe cos w j. Thus
Z =
U
F F^ ·^ g^ r^ =^
U @ 2
0 (^3 pd^ sin^ w^ i^3 pe^ cos^ w^ j)^ ·^ (d^ cos^ w^ i^3 e^ sin^ w^ j^ +^ f^ k)^ gw
U @ 2
0 (^3 pd
2 sin w cos w + pe
2 sin w cos w) gw = p(e
2 3 d
2 )
2 sin
2 w
0
1 2 p(e
2 3 d
2 )
45. Let F = 185 k. To parametrize the staircase, let { = 20 cos w, | = 20 sin w, } =
90 6 w^ =^
15 w,^0 $^ w^ $^6 ^ i
Z =
U
F F^ ·^ gr^ =^
U 6
0 h^0 >^0 >^185 i ·^3 20 sin^ w>^ 20 cos^ w>^
15
gw = (185)
15
U 6
0 gw^ = (185)(90)^ E^1 =^67 ×^10
4 ft-lb
46. This time p is a function of w: p = 185 3 9 6
w = 185 3 3 2
w. So let F =
2
w
k. To parametrize the staircase,
let { = 20 cos w, | = 20 sin w, } =
90 6 w^ =^
15 w,^0 $^ w^ $^6 . Therefore
Z =
U
F F^ ·^ gr^ =^
U 6
0 0 >^0 >^185
3 2 w
· 3 20 sin w> 20 cos w>
15
gw =
15
U 6
0
3 2 w
gw
15
185 w 3
3 4 w
2 ^6
0
9 2
E 1 = 62 × 10
4 ft-lb
47. (a) r(w) = hcos w> sin wi, 0 $ w $ 2 , and let F = hd> ei. Then
Z =
U
F
F · g r =
U 2
0
hd> ei · h 3 sin w> cos wi gw =
U 2
0
( 3 d sin w + e cos w) gw =
d cos w + e sin w
0
= d + 0 3 d + 0 = 0
(b) Yes. F ({> |) = n x = hn{> n|i and
Z =
U
F F^ ·^ g^ r^ =^
U 2
0 hn^ cos^ w> n^ sin^ wi · h^3 sin^ w>^ cos^ wi^ gw^ =^
U 2
0 (^3 n^ sin^ w^ cos^ w^ +^ n^ sin^ w^ cos^ w)^ gw^ =^
U 2
0 0 gw^ = 0.
48. Consider the base of the fence in the {|-plane, centered at the origin, with the
height given by } = k ({> |). The fence can be graphed using the parametric
equations { = 10 cos x, | = 10 sin x,
} = y
4 + 0=01((10 cos x)^2 3 (10 sin x)
2 )
= y(4 + cos^2 x 3 sin^2 x)
= y(4 + cos 2x), 0 $ x $ 2 , 0 $ y $ 1.
The area of the fence is
U
F
k({> |) gv where F, the base of the fence, is given by { = 10 cos w, | = 10 sin w, 0 $ w $ 2 .
Then U
F
k({> |) gv =
U 2
0
4 + 0=01((10 cos w)
2 3 (10 sin w)
2 )
s ( 3 10 sin w)^2 + (10 cos w)^2 gw
U 2
0
(4 + cos 2w)
I
100 gw = 10
4 w + 12 sin 2w
0
= 10(8 ) = 80 m^2
If we paint both sides of the fence, the total surface area to cover is 160 m
2 , and since 1 L of paint covers 100 m
2 , we require
160 100 = 1=^6 ^ E^5 =^03 L of paint.
° c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. °°cc 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, o2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, orr duplicated, or posted to a publicly accessible website, in whole or in partduplicated, or posted to a publicly accessible website, in whole or in part..
638 ¤^ CHAPTER 16 VECTOR CALCULUS
3. C(2{ 3 3 |)@C| = 3 3 = C( 33 { + 4| 3 8)@C{ and the domain of F is R
2 which is open and simply-connected, so by
Theorem 6 F is conservative. Thus, there exists a function i such that Qi = F, that is, i{({> |) = 2{ 3 3 | and
i| ({> |) = 33 { + 4| 3 8. But i{({> |) = 2{ 3 3 | implies i({> |) = {
2 3 3 {| + j(|) and differentiating both sides of this
equation with respect to | gives i| ({> |) = 33 { + j
0 (|). Thus 33 { + 4| 3 8 = 33 { + j
0 (|) so j
0 (|) = 4| 3 8 and
j(|) = 2|
2 3 8 | + N where N is a constant. Hence i ({> |) = {
2 3 3 {| + 2|
2 3 8 | + N is a potential function for F.
4. C(h
{ sin |)@C| = h
{ cos | = C(h
{ cos |)@C{ and the domain of F is R
2
. Hence F is conservative so there exists a function i
such that Qi = F. Then i{({> |) = h
{ sin | implies i ({> |) = h
{ sin | + j(|) and i| ({> |) = h
{ cos | + j
0 (|). But
i| ({> |) = h{^ cos | so j^0 (|) = 0 i j(|) = N. Then i ({> |) = h{^ sin | + N is a potential function for F.
5. C(h{^ cos |)@C| = 3 h{^ sin |, C(h{^ sin |)@C{ = h{^ sin |. Since these are not equal, F is not conservative.
6. C(3{
2 3 2 |
2 )@C| = 34 |, C(4{| + 3)@C{ = 4|. Since these are not equal, F is not conservative.
7. C(|h
{
{
{
- { cos |)@C{ and the domain of F is R
2
. Hence F is conservative so there
exists a function i such that Qi = F. Then i{({> |) = |h
{
- sin | implies i ({> |) = |h
{
i| ({> |) = h{^ + { cos | + j^0 (|). But i| ({> |) = h{^ + { cos | so j(|) = N and i ({> |) = |h{^ + { sin | + N is a potential
function for F.
8. C(2{| + |^32 )@C| = 2{ 3 2 |^33 = C({^2 3 2 {|^33 )@C{ and the domain of F is {({> |) | | A 0 } which is open and
simply-connected. Hence F is conservative, so there exists a function i such that Qi = F. Then i{({> |) = 2{| + |
32
implies i ({> |) = {
2 | + {|
32
2 3 2 {|
33
0 (|). But i| ({> |) = {
2 3 2 {|
33 so
j
0 (|) = 0 i j(|) = N. Then i ({> |) = {
2 | + {|
32
- N is a potential function for F.
9. C(ln | + 2{|
3 )@C| = 1@| + 6{|
2 = C(3{
2 |
2
- {@|)@C{ and the domain of F is {({> |) | | A 0 } which is open and simply
connected. Hence F is conservative so there exists a function i such that Qi = F. Then i{({> |) = ln | + 2{|
3 implies
i ({> |) = { ln | + {
2 |
3
- j(|) and i| ({> |) = {@| + 3{
2 |
2
0 (|). But i| ({> |) = 3{
2 |
2
0 (|) = 0 i
j(|) = N and i ({> |) = { ln | + {^2 |^3 + N is a potential function for F.
C({| cosh {| + sinh {|)
C|
2 | sinh {| + { cosh {| + { cosh {| = {
2 | sinh {| + 2{ cosh {| =
C({^2 cosh {|)
C{
and the domain of F is R
2
. Thus F is conservative, so there exists a function i such that Qi = F. Then
i{({> |) = {| cosh {| + sinh {| implies i({> |) = { sinh {| + j(|) i i| ({> |) = {
2 cosh {| + j
0 (|). But
i| ({> |) = {
2 cosh {| so j(|) = N and i ({> |) = { sinh {| + N is a potential function for F.
11. (a) F has continuous first-order partial derivatives and
C
C|
C
C{
2 ) on R
2 , which is open and simply-connected.
Thus, F is conservative by Theorem 6. Then we know that the line integral of F is independent of path; in particular, the
value of
U
F F^ ·^ gr^ depends only on the endpoints of^ F. Since all three curves have the same initial and terminal points, U
F F^ ·^ gr^ will have the same value for each curve.
° c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. °°cc 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, o2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, orr duplicated, or posted to a publicly accessible website, in whole or in partduplicated, or posted to a publicly accessible website, in whole or in part..
SECTION 16.3 THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS ¤^ 639
(b) We first find a potential function i , so that Qi = F. We know i{({> |) = 2{| and i| ({> |) = {
2
. Integrating
i{({> |) with respect to {, we have i ({> |) = {
2 | + j(|). Differentiating both sides with respect to | gives
i| ({> |) = {
2
0 (|), so we must have {
2
0 (|) = {
2 i j
0 (|) = 0 i j(|) = N, a constant.
Thus i ({> |) = {
2 | + N. All three curves start at (1> 2) and end at (3> 2), so by Theorem 2,
U F F^ ·^ gr^ =^ i^ (3>^ 2)^3 i^ (1>^ 2) = 18^3 2 = 16^ for each curve.
12. (a) i{({> |) = {
2 implies i ({> |) =
1 3 {
3
- j(|) and i| ({> |) = 0 + j
0 (|). But i| ({> |) = |
2 so
j^0 (|) = |^2 i j(|) = 1 3
|^3 + N. We can take N = 0, so i ({> |) = 1 3
{^3 + 1
3
|^3.
(b)
U
F F^ ·^ gr^ =^ i^ (2>^ 8)^3 i^ (^31 >^ 2) =^
3 +^
512 3
1 3 +^
8 3
13. (a) i{({> |) = {|
2 implies i ({> |) =
1 2 {
2 |
2
2 | + j
0 (|). But i| ({> |) = {
2 | so j
0 (|) = 0 i
j(|) = N, a constant. We can take N = 0, so i({> |) =
1 2 {
2 |
2 .
(b) The initial point of F is r(0) = (0> 1) and the terminal point is r(1) = (2> 1), so
U F F^ ·^ gr^ =^ i^ (2>^ 1)^3 i^ (0>^ 1) = 2^3 0 = 2.
14. (a) i| ({> |) = {
2 h
{| implies i({> |) = {h
{|
{|
{|
0 ({) = (1 + {|)h
{|
0 ({). But
i{({> |) = (1 + {|)h
{| so j
0 ({) = 0 i j({) = N. We can take N = 0, so i ({> |) = {h
{| .
(b) The initial point of F is r(0) = (1> 0) and the terminal point is r( @2) = (0> 2), so
U F F^ ·^ gr^ =^ i^ (0>^ 2)^3 i^ (1>^ 0) = 0^3 h
0 = 31.
15. (a) i{({> |> }) = |} implies i ({> |> }) = {|} + j(|> }) and so i| ({> |> }) = {} + j| (|> }). But i| ({> |> }) = {} so
j| (|> }) = 0 i j(|> }) = k(}). Thus i ({> |> }) = {|} + k(}) and i} ({> |> }) = {| + k
0 (}). But
i} ({> |> }) = {| + 2}, so k
0 (}) = 2} i k(}) = }
2
- N. Hence i ({> |> }) = {|} + }
2 (taking N = 0).
(b)
U
F
F · gr = i(4> 6 > 3) 3 i (1> 0 > 3 2) = 81 3 4 = 77.
16. (a) i{({> |> }) = |
2 } + 2{}
2 implies i ({> |> }) = {|
2 } + {
2 }
2
- j(|> }) and so i| ({> |> }) = 2{|} + j| (|> }). But
i| ({> |> }) = 2{|} so j| (|> }) = 0 i j(|> }) = k(}). Thus i({> |> }) = {|
2 } + {
2 }
2
i} ({> |> }) = {|
2
2 } + k
0 (}). But i} ({> |> }) = {|
2
2 }, so k
0 (}) = 0 i k(}) = N. Hence
i ({> |> }) = {|
2 } + {
2 }
2 (taking N = 0).
(b) w = 0 corresponds to the point (0> 1 > 0) and w = 1 corresponds to (1> 2 > 1), so
U F F^ ·^ gr^ =^ i(1>^2 >^ 1)^3 i^ (0>^1 >^ 0) = 5^3 0 = 5.
17. (a) i{({> |> }) = |}h
{} implies i({> |> }) = |h
{}
- j(|> }) and so i| ({> |> }) = h
{}
- j| (|> }). But i| ({> |> }) = h
{} so
j| (|> }) = 0 i j(|> }) = k(}). Thus i ({> |> }) = |h{}^ + k(}) and i} ({> |> }) = {|h{}^ + k^0 (}). But
i} ({> |> }) = {|h
{} , so k
0 (}) = 0 i k(}) = N. Hence i ({> |> }) = |h
{} (taking N = 0).
(b) r(0) = h 1 > 31 > 0 i, r(2) = h 5 > 3 > 0 i so
U
F F^ ·^ gr^ =^ i^ (5>^3 >^ 0)^3 i^ (1>^31 >^ 0) = 3h
0
0 = 4.
° c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. °°cc 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, o2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, rduplicated, or posted to a publicly accessible website, in whole or in partlicated, or posted to a publicly accessible website, in whole or in par.
SECTION 16.3 THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS ¤^ 641
25. We know that if the vector field (call it F) is conservative, then around any closed path F,
U
F F^ ·^ gr^ = 0. But take^ F^ to be a
circle centered at the origin, oriented counterclockwise. All of the field vectors that start on F are roughly in the direction of
motion along F, so the integral around F will be positive. Therefore the field is not conservative.
26. If a vector field F is conservative, then around any closed path F,
U
F
F · gr = 0. For any closed path we draw in the field, it
appears that some vectors on the curve point in approximately the same direction as the curve and a similar number point in
roughly the opposite direction. (Some appear perpendicular to the curve as well.) Therefore it is plausible that
U
F F^ ·^ gr^ = 0
for every closed curve F which means F is conservative.
27. From the graph, it appears that F is conservative, since around all closed
paths, the number and size of the field vectors pointing in directions similar
to that of the path seem to be roughly the same as the number and size of the
vectors pointing in the opposite direction. To check, we calculate
C
C|
(sin |) = cos | =
C
C{
(1 + { cos |). Thus F is conservative, by
Theorem 6.
28. Qi ({> |) = cos({ 3 2 |) i 3 2 cos({ 3 2 |) j
(a) We use Theorem 2:
U
F 1
F · gr =
U
F 1
Qi · gr = i (r(e)) 3 i (r(d)) where F 1 starts at w = d and ends at w = e. So
because i (0> 0) = sin 0 = 0 and i ( > ) = sin( 3 2 ) = 0, one possible curve F 1 is the straight line from (0> 0) to
( > ); that is, r(w) = w i + w j, 0 $ w $ 1.
(b) From (a),
U
F 2 F^ ·^ gr^ =^ i^ (r(e))^3 i^ (r(d)). So because^ i^ (0>^ 0) = sin 0 = 0^ and^ i
2 >^0
= 1, one possible curve F 2 is
r(w) =
2 w^ i,^0 $^ w^ $^1 , the straight line from^ (0>^ 0)^ to^
2 >^0
29. Since F is conservative, there exists a function i such that F = Qi , that is, S = i{, T = i| , and U = i}. Since S ,
T, and U have continuous first order partial derivatives, Clairaut’s Theorem says that CS@C| = i{| = i|{ = CT@C{,
CS@C} = i{} = i}{ = CU@C{, and CT@C} = i|} = i}| = CU@C|.
30. Here F({> |> }) = | i + { j + {|} k. Then using the notation of Exercise 29 , CS@C} = 0 while CU@C{ = |}. Since these
aren’t equal, F is not conservative. Thus by Theorem 4, the line integral of F is not independent of path.
31. G = {({> |) | 0? |? 3 } consists of those points between, but not
on, the horizontal lines | = 0 and | = 3.
(a) Since G does not include any of its boundary points, it is open. More
formally, at any point in G there is a disk centered at that point that
lies entirely in G.
(b) Any two points chosen in G can always be joined by a path that lies
entirely in G, so G is connected. (G consists of just one “piece.”)
° c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. °°cc 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, o2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, rduplicated, or posted to a publicly accessible website, in whole or in partlicated, or posted to a publicly accessible website, in whole or in par.
642 ¤^ CHAPTER 16 VECTOR CALCULUS
(c) G is connected and it has no holes, so it’s simply-connected. (Every simple closed curve in G encloses only points that are
in G.)
32. G = {({> |) | 1? |{|? 2 } consists of those points between, but
not on, the vertical lines { = 1 and { = 2, together with the points
between the vertical lines { = 31 and { = 32.
(a) The region does not include any of its boundary points, so it is open.
(b) G consists of two separate pieces, so it is not connected. [For
instance, both the points ( 31 = 5 > 0) and (1= 5 > 0) lie in G but they
cannot be joined by a path that lies entirely in G.]
(c) Because G is not connected, it’s not simply-connected.
33. G = ({> |) | 1 $ {
2
2 $ 4 > | D 0 is the semiannular region
in the upper half-plane between circles centered at the origin of radii
1 and 2 (including all boundary points).
(a) G includes boundary points, so it is not open. [Note that at any
boundary point, (1> 0) for instance, any disk centered there cannot lie
entirely in G.]
(b) The region consists of one piece, so it’s connected.
(c) G is connected and has no holes, so it’s simply-connected.
34. G = {({> |) | ({> |) 6 = (2> 3)} consists of all points in the {|-plane
except for (2> 3).
(a) G has only one boundary point, namely (2> 3), which is not included,
so the region is open.
(b) G is connected, as it consists of only one piece.
(c) G is not simply-connected, as it has a hole at (2> 3). Thus any simple
closed curve that encloses (2> 3) lies in G but includes a point that is
not in G.
35. (a) S = 3
{^2 + |^2
CS
C|
2 3 {
2
({^2 + |^2 )
2 and^ T^ =^
{^2 + |^2
CT
C{
2 3 {
2
({^2 + |^2 )
- Thus^
CS
C|
CT
C{
(b) F 1 : { = cos w, | = sin w, 0 $ w $ , F 2 : { = cos w, | = sin w, w = 2 to w = . Then
]
F 1
F · gr =
]
0
( 3 sin w)( 3 sin w) + (cos w)(cos w)
cos^2 w + sin
2 w
gw =
]
0
gw = and
]
F 2
F · gr =
]
2
gw = 3
Since these aren’t equal, the line integral of F isn’t independent of path. (Or notice that
U
F 3 F^ ·^ gr^ =^
U 2
0 gw^ = 2 ^ where
F 3 is the circle {
2
2 = 1, and apply the contrapositive of Theorem 3.) This doesn’t contradict Theorem 6, since the
domain of F, which is R
2 except the origin, isn’t simply-connected.
° c 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. °°cc 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, o2012 Cengage Learning. All Rights Reserved. May not be scanned, copied, orr duplicated, or posted to a publicly accessible website, in whole or in partduplicated, or posted to a publicly accessible website, in whole or in part..
