Answers and Solutions to Section 13.3 Homework
Problems 1-23 (odd) and 29-33
S. F. Ellermeyer
1.
By looking at the picture in the book, we see that
C f  dr 50  10  40.
3.
For the vector field
Fx, y  6x  5yi  5x  4yj,
we have
P  5
y
Q
 5.
x
Since the (implied) domain of F is  2 , which is a simply–connected set, we
conclude that F is conservative.
To find f such that f  F, we begin with
f x x, y  6x  5y.
This gives us
fx, y  3x 2  5xy  hy.
Differentiation with respect to y then gives us
f y x, y  5x  h  y.
However, we must also have
f y x, y  5x  4y.
Thus
h  y  4y
which means that
hy  2y 2  C.
Since we are only looking for a single potential function, we might as well
take C  0. We thus obtain
fx, y  3x 2  5xy  2y 2 .
Let us check that this is correct:
fx, y  6x  5yi  5x  4yj  Fx, y.
5.
For the vector field
Fx, y  xe y i  ye x j,
we have
1
7.
P  xe y
y
Q
 ye x .
x
Since P/y  Q/x, we conclude that the vector field F is not conservative.
For the vector field
Fx, y  2x cosy  y cosxi  x 2 siny  sinxj,
we have
P  2x siny  cosx
y
Q
 2x siny  cosx.
x
Since the (implied) domain of F is  2 , which is a simply–connected set, we
conclude that F is conservative.
To find f such that f  F, we begin with
f x x, y  2x cosy  y cosx.
This gives us
fx, y  x 2 cosy  y sinx  hy.
Differentiation with respect to y then gives us
f y x, y  x 2 siny  sinx  h  y.
However, we must also have
f y x, y  x 2 siny  sinx.
Thus
h  y  0
which means that
hy  C.
Since we are only looking for a single potential function, we might as well
take C  0. We thus obtain
fx, y  x 2 cosy  y sinx.
Let us check that this is correct:
fx, y  2x cosy  y cosxi  x 2 siny  sinxj
 Fx, y.
9.
For the vector field
Fx, y  ye x  sinyi  e x  x cosyj,
we have
P  e x  cosy
y
Q
 e x  cosy.
x
2
Since the (implied) domain of F is  2 , which is a simply–connected set, we
conclude that F is conservative.
To find f such that f  F, we begin with
f x x, y  ye x  siny.
This gives us
fx, y  ye x  x siny  hy.
Differentiation with respect to y then gives us
f y x, y  e x  x cosy  h  y.
However, we must also have
f y x, y  e x  x cosy.
Thus
h  y  0
which means that
hy  C.
Since we are only looking for a single potential function, we might as well
take C  0. We thus obtain
fx, y  ye x  x siny.
Let us check that this is correct:
fx, y  ye x  sinyi  e x  x cosyj Fx, y.
11. The vector field Fx, y  2xyi  x 2 j is conservative, so all line integrals of F
are path independent. A potential function for F is fx, y  x 2 y so, by the
Fundamental Theorem for line integrals, the integral of F over any path
beginning at 1, 2 and ending at 3, 2 is
 F dr   f  dr
 f3, 2  f1, 2
 3 2 2  1 2 2
 16.
13. First we find a potential function, f, for the vector field
Fx, y  x 3 y 4 i  x 4 y 3 j.
The potential function must satisfy
f x x, y  x 3 y 4
and thus
fx, y  1 x 4 y 4  hy
4
and
f y x, y  x 4 y 3  h  y.
Since f must also satisfy
3
f y x, y  x 4 y 3 ,
we obtain h  y  0 and thus hy  C.
Therefore, a potential function for F is
fx, y  1 x 4 y 4 .
4
The path
rt  t i  1  t 3 j
0t1
has initial point r0  0, 1 and terminal point r1  1, 2. Thus, by the
Fundamental Theorem for Line integrals, we have
C F dr  f1, 2  f0, 1
 1 1 4 2 4  1 0 4 1 4
4
4
 4.
15. First we find a potential function, f, for the vector field
Fx, y, z  yzi  xzj  xy  2zk.
The potential function must satisfy
f x x, y, z  yz
and thus
fx, y, z  xyz  hy, z
and
f y x, y, z  xz  h .
y
Since f must also satisfy
f y x, y, z  xz
we see that h/y  0 and hence that
hy, z  gz.
We now have
fx, y, z  xyz  gz
This tells us that
f z x, y, z  xy  g  z
Since f must also satisfy
f z x, y, z  xy  2z,

we obtain g z  2z and thus gz  z 2  C.
Therefore, a potential function for F is
fx, y, z  xyz  z 2 .
By the Fundamental Theorem for Line integrals, we have
4
C F dr  f4, 6, 3  f1, 0, 2
 463  3 2  102  2 2
 77.
17. First we find a potential function, f, for the vector field
Fx, y, z  y 2 coszi  2xy coszj  xy 2 sinzk.
The potential function must satisfy
f x x, y, z  y 2 cosz
and thus
fx, y, z  xy 2 cosz  hy, z
and
f y x, y, z  2xy cosz  h .
y
Since f must also satisfy
f y x, y, z  2xy cosz
we see that h/y  0 and hence that
hy, z  gz.
We now have
fx, y, z  xy 2 cosz  gz
This tells us that
f z x, y, z  xy 2 sinz  g  z
Since f must also satisfy
f z x, y, z  xy 2 sinz,
we obtain g  z  0 and thus gz  C.
Therefore, a potential function for F is
fx, y, z  xy 2 cosz.
The path
rt  t 2 i  sintj tk
0t
has initial point r0  0, 0, 0 and terminal point r   2 , 0, .
By the Fundamental Theorem for Line integrals, we have
C F dr  f 2 , 0,   f0, 0, 0
  2 0 2 cos  00 2 cos0
 0.
19. For the vector field
Fx, y  2x sinyi  x 2 cosy  3y 2 j,
5
we have
P  2x cosy
y
Q
 2x cosy.
x
Since the (implied) domain of F is  2 , which is a simply–connected set, we
conclude that F is conservative (and hence that all line integrals of F are
path–independent).
To find f such that f  F, we begin with
f x x, y  2x siny.
This gives us
fx, y  x 2 siny  hy.
Differentiation with respect to y then gives us
f y x, y  x 2 cosy  h  y.
However, we must also have
f y x, y  x 2 cosy  3y 2 .
Thus
h  y  3y 2
which means that
hy  y 3  C.
Since we are only looking for a single potential function, we might as well
take C  0. We thus obtain
fx, y  x 2 siny  y 3 .
Now, by the Fundamental Theorem for Line Integrals, for a path with initial
point 1, 0 and terminal point 5, 1, we have
C 2x siny dx  x 2 cosy  3y 2  dy  C F dr
f5, 1  f1, 0
 25 sin1  1.
21. The force field
Fx, y  x 2 y 3 i  x 3 y 2 j
is conservative and has potential function
fx, y  1 x 3 y 3 .
3
Thus, the work done by this force field in moving an object from the point
0, 0 to the point 2, 1 is
6
 F dr  f2, 1  f0, 0
 1 2 3 1 3  1 0 3 0 3
3
3
 8.
3
23. The vector field, F, shown in the picture (page 944) is not conservative. If it
were, then it would have to satisfy
P x, y  Q x, y
y
x
at all points x, y. However, the picture shows that this is not true.
For example, look at the vertical column of vectors that is the second to the
left from the y axis. (Hence x  x 0 is constant for all vectors in this column.) If
we move from bottom to top (in the direction of increasing y) along this
column, we see that the vectors make a transition from pointing to the left to
pointing to the right. There is some point x 0 , y 0  in the third quadrant at
which Fx 0 , y 0   0. Also, since Px 0 , y transitions from negative to positive
as y increases, we see that
P x 0 , y 0   0.
y
However, if we now look at the horizontal row of vectors with y  y 0 , we see
that these vectors transition from pointing up to pointing down as x increases.
This means that
Q
x 0 , y 0   0.
x
Therefore it is not true that
P x , y   Q x , y .
0 0
0 0
y
x
x, y | x  0 and y  0 is open, connected, and
simply–connected.
The set x, y | x  0 is open, but not connected or simply–connected.
The set x, y | 1  x 2  y 2  4 is open and connected, but not
simply–connected.
The set x, y | x 2  y 2  1 or 4  x 2  y 2  9 is neither open, nor
connected, nor simply–connected.
Consider the vector field
y
i 2 x 2 j
Fx, y  2
x y
x  y2
with domain D  x, y | x, y  0, 0 . The vector field F satisfies
29. The set
30.
31.
32.
33.
7
2
2
P  x  y 1  y2y  y 2  x 2
2
2
y
x 2  y 2 
x 2  y 2 
x 2  y 2 1  x2x
y2  x2
Q


2
2
x
x 2  y 2 
x 2  y 2 
and thus
P  Q .
y
x
However, since the domain D is not a simply–connected set, we are not
guaranteed that the vector field F is conservative. In fact, F is not
conservative as we will show by computing line integrals of F over two
different paths joining that begin at the point 1, 0 and end at the point
1, 0.
First, we use the path
rt  costi  sintj
0  t  .
(This path traces out the top half of the unit circle counterclockwise.)
For this path, we have

C F dr   0 Frt  r  t dt


 0  sinti  costj   sinti  costj dt

 0 1 dt

 .
Next, we use the path
rt  costi  sintj
0  t  .
(This path traces out the bottom half of the unit circle clockwise.)
For this path, we have

C F dr   0 Frt  r  t dt


 0 sinti  costj   sinti  costj dt

 0 1 dt

 .
8