Math 2263
Midterm 3
July 22, 2011
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your test booklet.
Make sure that you have 6 problems on your test. If any exam material
is missing, or if any questions are unclear, raise your hand or bring your
exam to the proctor.
Good luck!
1
1. Consider the vector field F (x, y) = h−xy, x2 i.
R
(a) Evaluate the line integral C F · dr, where C is the line from (0, 2)
to (2, 0).
Solution: Parametrize the line by
r(t) = h2t, 2 − 2ti, 0 ≤ t ≤ 1.
Then we compute the integral as follows.
Z 1
Z
2
− (2t)(2 − 2t)(2) + (2t) (−2) dt
F · dr =
0
C
Z
1
2
− 8t + 8t − 8t
=
2
dt
0
1
= −4t 2
0
= −4
R
(b) Evaluate the line integral C F ·dr, where C is the circle x2 +y 2 = 4
traversed clockwise from (0, 2) to (2, 0).
Solution: Parametrize the curve by
r(t) = h2 sin t, 2 cos ti, 0 ≤ t ≤ π/2.
Z π/2 Z
2
F · dr =
− (2 sin t)(2 cos t)(2 cos t) + (2 sin t) (−2 sin t) dt
C
0
Z
=
π/2
−8 sin t(cos2 t + sin2 t) dt
0
Z
π/2
−8 sin t dt
=
0
π/2
= 8 cos t
0
= −8
(c) Based on your work in parts (a) and (b), conclude that the vector
field F is either conservative or not conservative.
Solution: The vector field is not conservative because its integral is
1
2
path dependent. Also, we can see that ∂F
6= ∂F
.
∂y
∂x
2
2. A stationary charge at point (0, 0, 0) creates an electric field so that
the force experienced by a charge q at any point in R3 is given by
F (x, y, z) =
K
hx, y, zi
(x2 + y 2 + z 2 )3/2
for some constant K. Find the work done as the change q moves
along the path
r(t) = h2t, t, 2ti, 1 ≤ t ≤ 2.
Solution:
Z
Z
2
F · dr =
C
1
(4t2
Z
=
1
2
K
h2t, t, 2ti · h2, 1, 2i dt
+ + 4t2 )3/2
t2
K
(4t + t + 4t) dt
(9t2 )3/2
Z 2
K
=
(9t) dt
3
1 27t
Z 2
K
dt
=
2
1 3t
2
K =− 3t 1
K K
−
=−
6
3
=
3
K
6
R
3. Evaluate the integral C y 2 ds, where C is the curve y = e2x from the
point (0, 1) to the point (2, e4 ).
Solution: Parametrize the curve C by
r(t) = ht, e2t i, 0 ≤ t ≤ 2.
Z 2
Z
p
2
(e2t )2 (1)2 + (2e2t )2 dt
y ds =
0
C
Z
=
2
√
e4t 1 + 4e4t dt
0
Z
1+4e8
1 1/2
u du
16
5
1+4e8
1 2 3/2 =
· u 16 3
5
1
8 3/2
3/2
=
(1 + 4e ) − 5
24
=
4
4. Let F (x, y, z) = hxy cos(xy) + sin(xy), x3 cos(xy) +
1
, − lnz2y i.
yz
(a) Find a function f : R3 → R such that ∇f = F .
Solution: f (x, y, z) = x sin(xy) +
ln y
z
(b) Use the
R Fundamental Theorem of Line Integrals to evaluate the
integral C F · dr, where C is the curve parametrized by
r(t) = hπt, t2 + 1, 2t3 + 3i with 0 ≤ t ≤ 1.
Solution:
Z
Z
∇f · dr
F · dr =
C
C
= f (r(1)) − f (r(0))
= f (π, 2, 5) − f (0, 1, 3)
ln 2
ln 1
= π sin(2π) +
− 0 sin(0) +
5
3
=
5
ln 2
5
5. Find the value of the line integral
F (x, y) = hy 2 sin x + xe
x3 +3
R
C
F · dr, where
sin(y 3 + 4) ln
cos(x3 + 3),
y
p
y3 + 4
+ xi
and C consists of the line segment from (−π/2, 0) to (π/2, 0) and the
curve y = cos x from (π/2, 0) to (−π/2, 0).
Solution: We may use Green’s Theorem, since C is a positively oriented closed curve.
p
ZZ Z
∂
∂ sin(y 3 + 4) ln y 3 + 4
2
x3 +3
3
+x −
y sin x+xe
cos(x +3) dA
F ·dr =
y
∂y
D ∂x
C
Z
π/2
cos x
Z
=
−π/2
Z
0
π/2
=
−π/2
Z
1 − 2y sin x dy dx
y=cos x
(y − y 2 sin x)
dx
y=0
π/2
(cos x − cos2 x sin x) dx
=
−π/2
=
π/2
1
3
sin x + cos x 3
−π/2
= sin(π/2) − sin(−π/2)
=2
6