Math 251
Chapter 14 Sample Exam Problems
Answers
May 6, 2011
1. B
2. A
3. B There is a typo on the worksheet, and there are two ‘C’ answers. The correct
answer is what would be ‘B’ or π + πe . The potential function you want to find is
f (x, y, z) = x + y + z + xyz.
4. A This is a problem using the Divergence Theorem. We see that div(F) = 4. If we
let E denote the solid region surrounded by S and the unit disk D in the xy-plane,
then by the Divergence Theorem
ZZ
ZZZ
ZZ
F · n dS =
div(F) dV −
F · n dS
S
E
D
ZZZ
ZZ
=
4 dV −
hx, y, 2zi · h0, 0, −1i dS.
E
D
RRR
The triple integral evaluates to 4 times the volume of E: check that
4 dV = 2π.
E
The surface integral over the disk D is 0, because at every point in D we have z = 0,
so the dot product in the integrand is always 0 on D. Thus the answer is 2π .
q 2
2
dx 2
+ dy
dt, we see
5. (a) Using that ds =
+ dz
dt
dt
dt
Z
2
2
2
Z
(x + y + z ) ds =
C
2π
(16 cos2 t + 16 sin2 t + 9t2 )
p
16 sin2 t + 16 cos2 t + 9 dt.
0
Evaluating the integral on the right, we obtain an answer of 160π + 120π 3 .
(b) Let R be the triangular region bounded by the points (0, 0), (2, 0), and (2, 4) in
the xy-plane. Then Green’s Theorem says that
Z
ZZ
2
x2
x
3 2
3x − 3ex dA,
3ye + 2e dx + 2 x + sin y =
C
Z 2RZ 2x
2
=
3x − 3ex dy dx.
0
0
Evaluating the iterated integral, we obtain an answer of 19 − 3e4 .
Z
1
6. (a)
−1
√
Z
1−x2
√
− 1−x2
−4 dy dx
1
(b) Use the usual parametrization of C : x = cos t, y = sin t, 0 ≤ t ≤ 2π. So
dx = − sin t dt, dy = cos t dt. Then
I
Z
Z 2π
F · dr = (2y dx − 2x dy) =
(−2 sin2 t − 2 cos2 t) dt.
C
0
This integral evaluates to −4π .
(c) The integral in (a) is simply −4 times the area of the region of integration, which
is a disk of radius 1. Thus,
Z
1
−1
√
Z
1−x2
√
− 1−x2
−4 dy dx = −4π
as expected from (b).
7. By Stokes’ Theorem, the surface integral in question is the same as a line integral along
the boundary of S. More to the point, if we let C circle of radius 3 in the xy-plane
defined by x = 3 cos t, y = 3 sin t, z = 0, 0 ≤ t ≤ 2π, then
ZZ
I
Z 2π
(curl F) · n dS =
F · dr =
(−9 sin2 t − 9 cos2 t) dt = −18π .
S
C
0
8. In this problem we use Stokes’ Theorem to evaluate the desired line integral in terms
of a surface integral. Given that F = hy, −x, yzi, we calculate that
curl F = hz, 0, −2i.
Now if we let n be the upward unit normal vector on S, then
I
ZZ
(curl F) · n dS.
F · dr =
S
When calculating surface integrals on surfaces defined by the graph of a function, in
this case z = x2 + y 2 , we have some relatively nice formulas. For example, see formula
(8) on p. 922 of Stewart. Then if we let D be the unit disk in the xy-plane,
ZZ
ZZ
ZZ
(curl F) · n dS =
hz, 0, −2i · n dS =
(−z(2x) − 0 · (2y) − 2) dA.
S
S
D
The double integral on the right then evaluates to −2π .
2