Solutions for the second midterm
1.
Let
C
be the helix given by
x
(
t
) = cos
t,
y
(
t
) = sin
t,
z
(
t
) =
t
for
0
≤
t
≤
2
π
.
(a)
Integrate
f
(
x, y, z
) =
x
+ 3
z
2
along
C
with respect to arc length.
We have been given a parametrization; using it, we have
~
r
0
(
t
) = sin
t~
ı
+ cos
t~
+
~
k
and

~
r
0
(
t
)

=
p
sin
2
t
+ cos
2
t
+ 1 =
√
2
.
Writing the function in terms of this parametrization gives
f
(
x
(
t
)
, y
(
t
)
, z
(
t
)) = cos
t
+3
t
2
.
We have
Z
C
f ds
=
Z
2
π
0
(
cos
t
+ 3
t
2
)
√
2
dt
=
√
2
±
sin
t
+
t
3
²
2
π
0
= 8
√
2
π
3
.
(b)
Compute
R
C
~
F
·
d~
r
where
~
F
(
x, y, z
) =
z
2
~
ı
+
z~
+(2
xz
+
y
)
~
k
. (Hint:
~
F
is conservative.)
Computing the integral directly doesn’t look promising, so instead we follow the hint and
ﬁnd the potential function. Integrating with respect to
x
gives
f
=
xz
2
+
g
(
y, z
). Diﬀer
entiating with respect to
y
gives
∂g/∂y
=
z
and thus
g
(
y, z
) =
yz
+
h
(
z
). Diﬀerentiating
with respect to
z
gives 2
xz
+
y
+
h
0
(
z
) = 2
xz
+
y
, so that
h
(
z
) =
K
for some constant
K
.
We conclude that the potential is
f
=
xz
2
+
yz
+
K
for any constant
K
.
Note that the curve
C
begins at (1
,
0
,
0) and ends at (1
,
0
,
2
π
). We have
Z
C
~
F
·
d~
r
=
f
(1
,
0
,
2
π
)

f
(1
,
0
,
0) = 4
π
2

0 = 4
π
2
.
2.
Let
C
be the curve consisting of the following four line segments: from
(1
,
0
,
0)
to
(1
,
1
,
0)
, followed by
(1
,
1
,
0)
to
(0
,
1
,
1)
, followed by
(0
,
1
,
1)
to
(0
,
0
,
1)
, followed by
(0
,
0
,
1)
to
(1
,
0
,
0)
. In particular,
C
is a rectangle which lies in the plane