MAT 267
Complete Solutions to Review Chapter 13
1.
d) 1.
b) 2.
a) 3. e) 4.
f) 5.
c) 6.
2.
(a)
∇×
F
=
0
,
thus the vector field is conservative.
(b)
The potential function is
f
(
x
,
y
,
z
) =
x
2
y
+
zy
+
e
z
+ C
(c)
The initial point of the curve is
P
0
=
r
0
=
0,1,0
and the terminal point is
P
1
=
r
/
2
=/
2, 0,1
.
Using the Fundamental
Theorem of Line Integrals we have
∫
C
F
⋅
d
r
=
f
P
1
−
f
P
0
=
e
−
1
3.
∇⋅
F
=
∂
∂
x
2
x y
∂
∂
y
x
2
z
∂
∂
z
y
e
z
=
2
y
0
e
z
=
2
y
e
z
4.
We can parametrize the curve using
x
=
t
,
y
=
t
2
with
0
≤
t
≤
1.
Then
r
t
=
<
t ,t
2
>
,
r
'
t
=
<1,2
t
>
and
∣
r
'
t
∣=
1
4
t
2
.
Thus
∫
C
x ds
=
∫
0
1
t
1
4t
2
d t
=
1
12
5
5
−
1
5.
Parametrize the circle using
r
t
=
<2cos
t ,
2sin
t
>
,
−/
2
≤
t
≤/
2.
Then
r
'
t
=
<
−
2sin
t ,
2cos
t
>
,
∣
r
'
t
∣=
2
and
∫
C
x , y
ds
=
∫
−/
2
/
2
r
t
∣
r
'
t
∣
dt
=
∫
− /
2
/
2
2cos
t
4sin
2
t
2
dt
=
32
3
.
6.
Let A(1,0,0) and B(0,1,1), then
AB
=
<
−
1,1,1>
and the line through A and B can be parametrized by
r
t
=
<1
−
t ,t ,t
>
,
0
≤
t
≤
1.
Then Work
=
∫
0
1
<1
−
t ,t
1
−
t
,t
1
−
t
>
⋅
<
−
1,1
,
1>
dt
=−
1
6
.
7.
C
1
can be parametrized by
r
t
=
<
t ,t
2
>
,
0
≤
t
≤
2.
∫
C
1
F
⋅
d
r
=
∫
0
2
F
r
t
⋅
r
'
t
dt
=
∫
0
2
< 2
t
3
t
2,
3
t
>
⋅
<1
,
2
t
>
dt
=
∫
0
2
2
t
9
t
2
dt
=
28
For C
2
:
•
The line from (0,0) to (1,1) has equation
y
=
x
and can be parametrized by
r
t
=
<
t ,t
>
,
0
≤
t
≤
1.
•
The line from (1,1) to (2,4) has equation
y
= 3
x
– 2 and can be parametrized
by
r
t
=
<
t ,
3
t
−
2>
,
1
≤
t
≤
2.
Then
∫
C
2
F
⋅
d
r
=
∫
0
1
<2
t
3
t ,
3
t
>
⋅
<1,1 >
dt
∫
1
2
< 2
t
3
3
t
−
2
,
3
t
>
⋅
< 1,3>
dt
=
∫
0
1
8
t dt
∫
1
2
20
t
−
6
dt
=
28.
Since
∇×
F
=
0
the vector field is conservative and the line integral is independent of path.
Note that C
1
and C
2
have the same initial and terminal point, thus the line integral has the same value for
both paths.
8.