Exam 3 Math 251 Sec 519, 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? Note how the
vector field’s x-component is positive for 0 < x < 3, negative for −3 <
x < 0, roughly speaking. Note how the y-component has the opposite sign,
but same strength, across the y axis, as it has on the other side. These
behaviors, and more, match the gradient hsin y, x cos yi which points to
answer (a).
• (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, yR = t, z = −3 cos t,
starting at (0, 0, −3) and ending at (3, π/2, 0). Find C xz ds, where ds
stands for integration with
√ respect to arc length. The relation between dt
and ds here is that ds = 10dt because the velocity√vector for the parametric curve is h3 cos t, 1, 3 sin ti and that has length 10. So the parametric
√ R π/2
version of the question is 10 t=0 −9 cos t sin t dt. With the substitution
√
√ R1
u = sin t, this becomes −9 10 u=0 u2 du = −9 10/2. Choose (b).
√
• (a) 9 10/2
√
• (b) −9 10/2
1
• (c)
9π 2 /2
• (e)
9π/2
• (d) −27/2
3. Let F be the vector field given by F(x, y, z) = 2xyez i + yzex k. Find divF .
The answer is (c).
• (a) (2y + yz)ex+z
• (b) (yz + y + z)ex + 2(x + y + xy)ez
• (c) y(ex + 2ez )
• (d) h2yez + yzex , 2xez + zex , yex + 2xyez i
• (e) hzex , 2xyez − yzex , yex + 2xyez i
4. Let F be the vector field given by F(x, y, z) = 2xyez i+yzex k. Find curlF .
The answer is (e).
• (a) (2y + yz)ex+z
• (b) (yz + y + z)ex + 2(x + y + xy)ez
• (c) y(ez + 2ez )
• (d) h2yez + yzex , 2xez + zex , yex + 2xyez i
• (e) hzex , 2xyez − yzex , −2xez i
5. Let M = 2xy and N = 3xy. Let C be the path taken counterclockwise
alongRline segments from (0, 0) to (3, 0), thence to (2, 4), and back to (0, 0).
Find C M dx+N dy. Using Green’s theorem, the required integral is equal
R 4 R 3−y/4
to y=0 x=y/2 (3y − 2x) dx dy. This works out to 4. Choose (b).
• (a) 2
• (b) 4
• (c) 6
• (d) 8
• (e) 10
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
R Rchange of variable T given by (x, y) = T (u, v) =
x dA. In the parametric setting, the integral
(3u + 2v, u − v). Find
R
R3R2
becomes 5 0 0 3u + 2v dv du, since the Jacobian is 5. That works out to
195. Choose (b).
• (a) 160
• (b) 195
• (c) 220
• (d) 245
2
• (e) 270
7. Consider the region V bounded by the surface S whose spherical-coordinates
equation is ρ = cos(2φ) where φ ranges from 0 to π/4.
(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 1 at φ = 0 and 0 at φ = π/4. It decreases, slowly
at first and then more rapidly, over the interval from 0 to π/4. 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:
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 this region. The limits of integration are that θ runs 0 to
3
2π, (because it’s a region got by rotating about the z-axis), φ runs
0 to π/4 (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/4
Z
cos(2φ)
ρ2 sin φ dρ dφ dθ.
0
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 0 to π/4. 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.
(e) Find
Z
S
(curlF · dS),
where F is the vector field xi+(x+y)j+(y +z)k. Explain. There are
several ways to solve this, but all the reasonable approaches involve
using a theorem. Call I the required integral.
First, one may consider that the region is a surface to which Stokes’
theorem applies, with a boundary that has vanished to nothing, say
at the origin. The path integral around this pinpoint of a circle is 0,
so I = 0.
A second approach is to take any path that cleanly divides the surface
into two parts. One choice as good as another, so say the circle at
height z = 1/2. Taking this circle counterclockwise as seen from
above gives a path integral equal to the contribution from the top
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half of S to I. Taking this same circle counterclockwise as seen from
below is, on the one hand, the exact opposite of the first path integral
we were thinking about, but on the other hand, it’s the path integral
that yields the contribution to I from the bottom half. Thus the one
piece of I, plus the other piece, are given by two path integrals that
exactly cancel each other. Again I = 0.
Finally, there’s the divergence theorem. We already have two solutions not involving it, but there is no law against reading ahead
or attending class the day before the exam and picking up on what
was presented. The divergence of the curl of any F is 0, so the field
whose surface integral we’re taking has divergence 0. By the divergence theorem, the integral of F over the closed surface F is equal to
the integral of 0 over the interior, and that’s 0. The answer is that
I = 0.
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