Problem 2 (20 points)
Which of the following vector fields
F
are given by
F
=
∇
f
for some function
f
? Here
f
is
assumed to have the same domain as
F
. If such an
f
exists, find it. Otherwise, show that
no such
f
can exist.
1. (
7pts
)
F
(
x, y, z
) = (
ye
y
, xe
y
+
yxe
y
, e
z
+
ze
z
)
2. (
7pts
)
F
(
x, y, z
) = (
√
xy
+
zx, x
+
z,
√
xz
+
yz
)
3. (
6pts
)
F
(
x, y, z
) =
x

y
x
2
+
y
2
,
x
+
y
x
2
+
y
2
, z
2
Solution
1. One can check that
F
is curl free, and hence is the gradient of some function.
R
F
1
dx
=
xye
y
, so call this function
f
0
. Then
∂f
∂x
=
F
1
and
∂f
∂y
=
F
2
, but
∂f
∂z
6
=
F
3
. Noticing that
R
e
z
+
ze
z
dz
=
ze
z
, we find consider the new function
f
=
f
0
+
ze
z
. Then
F
=
∇
f
.
2.
∂F
2
∂x

∂F
1
∂y
= 1

x
2
√
xy
+
zx
6
= 0, so
F
has nonzero curl and hence is not a conservative
vector field.
3. Away from the
z
axis,
∇×
F
= 0. However,
F
is not the gradient of a function because
it has nonzero circulation around, say, the unit circle
C
in the
xy
plane:
Z
C
F
·
ds
=
Z
C
x

y
x
2
+
y
2
dx
+
x
+
y
x
2
+
y
2
dy
=
Z
C
(
x

y
)
dx
+ (
x
+
y
)
dy
=
Z
C
x dx
+
y dy
=
Z
2
π
0
(cos
2
(
t
) + sin
2
(
t
))
dt
=2
π
4