Exam 3 Math 251 Sec 520, Fall 2015
Multiple-choice problems are worth 15 each. Each of the five parts of the
last problem is worth 6 points.
1. Consider the vector field shown:
Which of the following formulas could have given this field? The answer is
(d). Among many other reasons, the y-direction component of the vector
field in the picture is positive when x = 0, 0 when x is close to 1.5 near
π/2, and negative when x is close to 3 near π. That fits with the second
component of the gradient being cos x for formula (d).
• (a) ∇(x sin y)
• (b) ∇(y sin x)
• (c) ∇(x cos y)
• (d) ∇(y cos x)
• (e) None of the above, because the field is not conservative.
2. Consider the parametric curve C given by x = 3 sin t, y R= 4t, z = −3 cos t,
starting at (0, 0, −3) and ending at (0, 8π, −3). Find C x2 ds, where ds
stands for integration withprespect to arc length. The parametrization
traces the path at speed (3 cos t)2 + 42 + (3 sin t)2 = 5, so ds = 5 dt
R 2π
and our answer is 5 t=0 (9 sin2 t). Now the integral of sin2 t over a full
period is half the integral of 1 over the same period because sin2 oscillates
symmetrically about 1/2, going back and forth from 0 to 1. Thus the
answer is 45π. Choose (c). The integral can also be done using the
identity sin2 t = (1 − cos(2t))/2, and it can also be done by observing that
R 2π
R 2π
sin t = cos(π/2 − t) so that, using periodicity, 0 sin2 t dt = 0 cos2 t dt.
Since the sum of these integrals is the integral of 1, each must give half
that, and half of 2π is π.
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• (a) 9π
• (b) 36π
• (c) 45π
• (d) 90π
• (e) 180π
3. Let F be the vector field given by F(x, y, z) = xez i + zexy j. Find divF .
The answer is (e).
• (a) xzez−1 + xyzexy−1
• (b) ez + xzexy + exy
• (c) xez i + xyzexy j
• (d) ez i + xzexy j
• (e) ez + xzexy
4. Let F be the vector field given by F(x, y, z) = xez i + zexy j. Find curlF .
The answer is (d).
• (a) hxey , xez , zexy i
• (b) hxyzexy−1 , xzez−1 , xyzexy−1 i
• (c) hez , xzexy , 0i
• (d) h−exy , xez , yzexy i
• (e) h−exy , −xez , yzexy i
5. Let M = 3xy and N = 2xy. Let C be the path taken counterclockwise
along Rline segments from (0, 0) to (3, 0), thence to (3,
R 6), and back to (0, 0).
Find C M dx + N dy. (Also could be written as C (M i + N j) · dr.) We
do best to use Green’s theorem, with ∂N/∂y = 2x and ∂M/∂x = 3y.
Our double integral giving the same answer as the answer to the original
question is then
Z Z
2x
3
(2x − 3y) dy dx.
0
0
That can be evaluated by straightforward plugging but it’s more fun to
use what we know about x and y. These are the centroid of the triangle,
which for triangles is given by the average of the corner coordinates. Thus
x = 2 and y = 2. The area of the triangle is 9 so the contribution to the
answer from integrating 2x is 9 ∗ 2 ∗ 2 and the contribution from −3y is
−9 ∗ 3 ∗ 2. Since 36 − 54 = −18, the answer is −18. Choose (a).
• (a) -18
• (b) -9
• (c) 0
• (d) 9
2
• (e) 18
6. Let S and R be regions in the plane. S is the region [0, 3] × [0, 2], and
R is the image of S under the change
of variable T given by (x, y) =
RR
x
dA.
In the parametrized domain,
T (u, v) = (2u + 3v, u − v). Find
R
the region of integration
is
0
to
3,
0
to
2.
The
Jacobian is 5. So since
R3R2
R3 x =
2u + 3v, the answer is 0 0 (2u + 3v) dv du. That evaluates to 5 0 (2uv +
R3
(3/2)v 2 )|v=2
v=0 du = 5 0 (4u + 6) du = 180. Choose (a).
• (a) 180
• (b) 210
• (c) 240
• (d) 270
• (e) 300
7. Consider the surface S whose spherical-coordinates equation is ρ = − cos(2φ)
where φ ranges from π/4 to π/2.
(a) Sketch S. When there is no mention of θ, the region or surface will
be rotationally symmetric about the z-axis. It’s a surface of rotation,
and the way to think about these is to look at a cross section. Say
we pick the y-z plane to look at. Our situation is now just like polar
coordinates, but with ρ in place of r and with the z-axis taking the
place of the x-axis.
Now − cos(2φ) is 0 at φ = π/4 and 1 at φ = π/2. It increases, faster
at first and then more slowly, over the interval from π/4 to π/2. The
resulting two-dimensional graph thus has connected-dots in a pattern
like this:
and if we rotate that, we get sort of a skeleton for the figure:
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Put it all together and you get something like this:
.
(b) Set up, but do not evaluate, a spherical-coordinates integral for the
volume of the region below S but above the x-y plane. The limits
of integration are that θ runs 0 to 2π, (because it’s a region got
by rotating about the z-axis), φ runs π/4 to π/2 (because that’s
specified in the question), and ρ runs 0 to − cos(2φ) (because the
surface bounding the region has that equation). The Jacobian for
spherical coordinates is ρ2 sin φ. The upshot is that the answer is
that the volume is given by
V =
Z
0
2π
Z
π/2
π/4
Z
− cos(2φ)
ρ2 sin φ dρ dφ dθ.
0
(c) The surface S is given by these parametric equations over a parametric domain R. Give some reasons why these equations are appropriate. The quickest reason is that the equations converting spherical
to rectangular give exactly this, if we use u and v in place of φ and
θ and take into account that ρ = − cos(2φ). Other reasons would be
that the formula captures the rotation about the z axis by attaching
to
p x and y a cos v and sin v, respectively, and that if we work out
x2 + y 2 + z 2 from the expressions for x, y, and z, we get − cos(2u),
which is consistent with the defining formula for the surface.
x = − cos(2u) sin(u) cos(v)
y = − cos(2u) sin(u) sin(v)
z = − cos(2u) cos(u)
(d) Find the simplest domain R so that the parametric mapping from R
(u-v coordinates) to S (x, y, and z-coordinates) is one-to-one. There
are two reasonable choices here. In both, u runs π/4 to π/2. We can
continue by taking v to range over any interval of length 2π. Of these,
[0, 2π) and [−π, π) stand out as particularly simple. Technically, one
should use the half-open interval, but since the domains are used for
integration, it won’t matter if we double-map a few points because
v = 0 gives the same thing as v = 2π.
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(e) Find
Z
(curlF · dS),
S
where dS points more or less up, and where F is the vector field xi +
(x + y)j + (y + z)k. Explain. The head-on integration is ugly. Best to
use Stokes’ theorem. That says that we can replace the given integral
with the path integral of F · dr over the boundary of S. Here, that
boundary is just the unit circle in the x-y plane, so r = cos ti + sin tj
and dr = − sin ti + cos tj. The vector field F = cos ti + (cos t + sin t)j
when
R 2πwe use our parametrization, so the technical integral boils down
to t=0 (− sin t cos t + cos t(cos t + sin t)) dt. Two of the terms cancel
R 2π
leaving 0 cos2 t dt = π.
That’s probably the most easily found solution, but here’s another
one: imagine that we put a floor under our surface, by adding the
disk D that it covers in the x-y plane. Now we have a closed surface, and the divergence theorem applies. The divergence of the
curl of anything
is 0, so our surface integral I, plus a correspondRR
ing JRR= D curlF · (−k) dA, gives 0. But curlF · k = h1, 0, 1i, so
J = D (−1) = −π. Thus I = π.
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