Solutions to the 2nd Practice Midterm
1.
Let
C
be the curve in the plane given by the graph of
y
=
x
2
from the point
(1
,
1)
to the point
(2
,
4)
.
(a)
Integrate
f
(
x, y
) = 2
y/x
along
C
(with respect to arc length).
Since
C
is a graph, we can parametrize it by
x
=
t
,
y
=
t
2
for
t
∈
[1
,
2]. Then we see
that
r
(
t
) =
ı
+ 2
t
and the arclength element is
ds
=
√
1 + 4
t
2
dt
.
Further, we have
f
(
x
(
t
)
, y
(
t
)) = 2
t
, so that
C
f ds
=
2
1
2
t
1 + 4
t
2
dt
=
1
6
(
1 + 4
t
2
)
3
/
2
2
1
=
1
6
17
3
/
2

5
3
/
2
.
(b)
Let
F
(
x, y
) =
ı
+
xy
. Integrate
F
along
C
.
We use the same parametrization for
C
as in the last part. Now we have
F
(
x
(
t
)
, y
(
t
)) =
ı
+
t
3
, so that
C
F
·
dr
=
2
1
F
(
x
(
t
)
, y
(
t
))
·
r
(
t
)
dt
=
2
1
(
1 + 2
t
4
)
dt
=
t
+
2
5
t
5
2
1
=
67
5
.
2.
Let
F
(
x, y, z
) = 2
xzı
+
e
z
+ (
x
2
+
ye
z
)
k
.
(a)
If
F
has a potential, find it. If not, say why not.
We compute
curl
F
=
ı
k
∂
∂x
∂
∂y
∂
∂z
2
xz
e
z
x
2
+
ye
z
= (
e
z

e
z
)
ı

(2
x

2
x
)
+ (0

0)
k
= 0
.
Since
F
is defined everywhere on
R
3
, which is simplyconnected, this means that
F
is
conservative.
To find its potential, we integrate with respect to
x
to find
f
(
x, y, z
) =
x
2
z
+
g
(
y, z
).
Differentiating with respect to
y
gives
g
y
(
y, z
) =
e
z
, and thus
g
(
y, z
) =
ye
z
+
h
(
z
). Differentiating with respect to
z
gives
x
2
+
ye
z
+
h
(
z
) =
x
2
+
ye
z
, so that
h
(
z
)
is constant. Putting this together, we see that the potential is
f
(
x, y, z
) =
x
2
z
+
ye
z
+
K
for any constant
K
.
(b)
Compute
C
F
·
dr
where
C
is the curve
x
(
t
) = cos
2
(
πt
)
,
y
(
t
) = 1 +
t
3
,
z
(
t
) =
1
4
t
2
,
for
0
≤
t
≤
2
.
1
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The fundamental theorem for line integrals tells us that, because
f
is a potential for
F
, the integral of
F
along any curve is equal to
f
at the the end point minus
f
at the
starting point. The above curve goes from (1
,
1
,
0) to (1
,
9
,
1), and thus (we set
K
= 0 in
our definition of
f
)
C
F
·
dr
=
f
(1
,
9
,
1)

f
(1
,
1
,
0) = 1 + 9
e

1 = 9
e.
3.
Let
F
(
x, y
) = (
yx
+
yx
2
)
ı
+
(
1
3
x
3
)
. Let
C
be the curve consisting of the line segment
from
(0
,

1)
to
(0
,
1)
followed by the semicircle from
(0
,
1)
through
(1
,
0)
to
(0
,

1)
.
Compute
C
F
·
dr
.
The curve
C
is a simple, closed curve oriented clockwise. Thus we can use Green’s theorem
to evaluate the integral. The region
D
bounded by
C
is the right half of the unit disk, that is
the set of points where
x
2
+
y
2
≤
1 with
x
≥
0. The integrand for Green’s theorem is
∂Q
∂x

∂P
∂y
=
x
2

(
x
+
x
2
) =

x.
Using Green’s theorem, with a minus sign because the curve is oriented clockwise, and polar
coordinates to write the integral over the halfdisk, we have
C
F
·
dr
=

D
∂Q
∂x

∂P
∂y
dA
=

π/
2

π/
2
1
0

r
cos
θ
·
r dr dθ
=
π/
2

π/
2
cos
θ dθ
1
0
r
2
dr
= [sin
θ
]
π/
2

π/
2
1
3
r
3
1
0
= (1

(

1))
1
3

0
=
2
3
.
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 Spring '07
 NEEL
 Calculus, Arc Length, Vector Calculus, Cos, STOKE

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