HOMEWORK 9, MATH 175 - FALL 2009
This homework assignment covers Sections 17.1-17.4 in the book.
1. Sketch the vector field F(x, y) =
1
x
i + yj.
2. Find the gradient vector field for f(x, y) = x
2
− y and sketch it.
The gradient vector field is just ∇f(x, y) = 2xi − j.
3. Evaluate the line integral
R
C
x sin y ds where C is the line segment from (0, 3) to (4, 6).
The curve C can be parametrized by r(t) = (0, 3) + t(4, 3) where 0 ≤ t ≤ 1, and then we have kr
0
(t)k =
√
4
2
+ 3
2
= 5. Hence
Z
C
x sin y ds =
Z
1
0
20t sin(3 + 3t) dt,
integration by parts (u = t and dv = sin(3 + 3t)dt) then gives
Z
1
0
20t sin(3 + 3t) dt = 20[−
1
3
t cos(3 + 3t) +
1
9
sin(3 + 3t)]
1
0
=
20
9
(sin 6 − 3 cos 6 − sin 3).
4. Evaluate the line integral
R
C
sin x dx + cos y dy, where C consists of the top half of the circle x
2
+ y
2
= 1
from (1, 0) to (−1, 0) and the line segment from (−1, 0) to (−2, 3).
If we split the curve into two parts we can find a parameterization for each part and then continue as in
3. Let’s instead use the Fundamental Theorem of Line Integrals.
Note that F (x, y) = sin xi+cos yj is a conservative vector field. Indeed if f
x
= sin x then f = −cos x+g(y)
where g is a function of y. Then we have cos y = f
y
= g
0
(y) so that g(y) = sin y + K where K is some
constant.
In particular we have F = ∇(−cos x + sin y) and hence we have
Z
C
sin x dx + cos y dy =
Z
C
∇f · dr
= f(−2, 3) − f (1, 0) = −cos 2 + sin 3 + cos 1.
5. Evaluate the line integral
R
C
F · dr where F (x, y, z) = (x + y)i + (y − z)j + z
2
k and C is given by the
vector function r(t) = t
2
i + t
3
j + t
2
k, 0 ≤ t ≤ 1.
F is not a conservative vector field and so we cannot use the Fundamental Theorem of Line Integrals. We
will have to compute this directly. Since r(t) = t
2
i + t
3
j + t
2
k we have r
0
(t) = 2ti + 3t
2
j + 2tk. Therefore
Z
C
F · dr =
Z
1
0
F (r(t)) · r
0
(t)dt
=
Z
1
0
(2t
3
+ 2t
4
+ 3t
5
− 3t
4
+ 2t
5
)dt =
Z
1
0
(5t
5
− t
4
+ 2t
3
)dt
=
5
6
−
1
5
+
1
2
=
17
15
.
1
2 HOMEWORK 9, MATH 175 - FALL 2009
6. Evaluate the line integral
R
C
F ·dr where F (x, y, z) = (2xz + y
2
)i + 2xyj + (x
2
+ 3z
2
)k and C is given by
x = t
2
, y = t + 1, z = 2t − 1, 0 ≤ t ≤ 1.
If this is a conservative vector field then we have
f
x
= 2xz + y
2
,
hence f = x
2
z + xy
2
+ g(y, z) where g is some function. Therefore we have
2xy = f
y
= 2xy + ∂g/∂y,
and so g(y, z) = h(z) for some function h. We then have f = x
2
z + xy
2
+ h(z) and so
x
2
+ 3z
2
= f
z
= x
2
+ h
0
(z),
hence h(z) = z
3
+ K for some constant K.
In particular we have shown that F = ∇(x
2
z + xy
2
+ z
3
) and so by the Fundamental Theorem of Line
Integrals we have
Z
C
F · dr = f(1, 2, 1) − f (0, 1, −1) = (1 + 4 + 1) − (0 + 0 − 1) = 7.
7. Evaluate the line integral
R
C
F · dr where F (x, y, z) = e
y
i + xe
y
j + (z + 1)e
z
k, and C is given by
r(t) = ti + t
2
j + t
3
k, 0 ≤ t ≤ 1.
Just as above, if F is a conservative vector field then we have
f
x
= e
y
,
hence f = xe
y
+ g(y, z). Therefore
xe
y
= f
y
= xe
y
+ ∂g/∂y,
and so g(y, z) = h(z). We have then f = xe
y
+ h(z) and so
(z + 1)e
z
= f
z
= h
0
(z),
therefore h(z) = ze
z
+ K and in particular we have F = ∇(xe
y
+ ze
z
) and so by the Fundamental Theorem
of Line Integrals we have
Z
C
F · dr = f(1, 1, 1) − f (0, 0, 0) = 2e.
8. Evaluate the line integral
R
C
cos y dx + x
2
sin y dy, where C is the rectangle with vertices (0, 0), (5, 0),
(5, 2), and (0, 2) oriented positively.
Let D be the region enclosed by the curve C. Using Green’s Theorem we have that
Z
C
cos y dx + x
2
sin y dy =
ZZ
D
(2x sin y + sin y) dA
=
Z
5
0
Z
2
0
(2x + 1) sin y dy dx = [x
2
+ x]
5
0
[−cos y]
2
0
= 30(1 − cos 2).
9. Evaluate the line integral
R
C
sin y dx+x cos y dy, where C is given by the ellipse x
2
+xy +y
2
= 1, oriented
positively.
Let D be the region enclosed by the curve C. Using Green’s Theorem we have that
Z
C
sin y dx + x cos y dy =
ZZ
D
(cos y − cos y)dA = 0.
