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Math 2263 Summer 2014 Exam 3 7/25/14

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Math 2263 Summer 2014 Exam 3 7/25/14
Math 2263
Summer 2014
Exam 3
7/25/14
Time Limit: 50 Minutes
Name (Print):
This exam contains 7 pages (including this cover page) and 5 problems. Check to see if any pages
are missing. Enter all requested information on the top of this page, and put your initials on the
top of every page, in case the pages become separated.
You may not use your books, notes, or a graphing calculator on this exam.
You are required to show your work on each problem on this exam. The following rules apply:
• If you use a “fundamental theorem” you
must indicate this and explain why the theorem
may be applied.
• Organize your work, in a reasonably neat and
coherent way, in the space provided. Work scattered all over the page without a clear ordering
will receive very little credit.
• Mysterious or unsupported answers will not
receive full credit. A correct answer, unsupported by calculations, explanation, or algebraic
work will receive no credit; an incorrect answer
supported by substantially correct calculations and
explanations might still receive partial credit.
• If you need more space, use the back of the pages;
clearly indicate when you have done this.
Do not write in the table to the right.
Problem
Points
1
20
2
15
3
20
4
20
5
25
Total:
100
Score
Math 2263
Exam 3 - Page 2 of 7
7/25/14
1. Let E be the solid region above the plane z = 2 and below the upper hemisphere of the sphere
x2 + y 2 + z 2 = 16. Set up (but do not evaluate) an iterated integral that would calculate the
volume of E using:
(a) (10 points) Cylindrical coordinates.
Solution
Z
ZZZ
dV =
2π
16−z 2
r dr dz dθ
0
E
√
4Z
Z
2
2π
Z
0
√
Z
12 Z
√
=
16−r2
r dz dr dθ
0
0
2
(b) (10 points) Spherical coordinates.
Solution
The lower bound on ρ come from the intersection of the sphere with the z = 2 plane. So
we set z = 2 in the spherical coordinate for z and get the lower bound on ρ to be cos2 φ .
To get the upper bound on φ we solve the trig equation 4 = cos2 φ and find that φ = π3 .
ZZZ
Z
2π
Z
dV =
E
0
0
π
3
Z
4
ρ2 sin φ dρ dφ dθ
2
cos φ
Math 2263
Exam 3 - Page 3 of 7
7/25/14
√
2. (15 points) Let C be the piece
Z of the graph of the function y = 2 x − 1 between the points
1
( 41 , 0) and (1, 1). Calculate
ds.
2
C x
Solution
√
Parametrize the curve by letting t = x so that the vector equation is ~r(t) = t, 2 t − 1 for
r
1
t ∈ [1/4, 1]. The ds = 1 + . The line integral become
t
r
Z 1
Z
1
1
1
1
1
ds
=
let u = 1 + so that du = − 2 dt
1 + dt
2
1 t2
x
t
t
t
C
4
Z
√
=−
u du
2
= − u3/2
3
1
1 2
=− 1+
3
t 1/4
√
2 √
= (5 5 − 2 2)
3
= 5.567
Math 2263
Exam 3 - Page 4 of 7
7/25/14
3. Let C be the closed, positively-oriented curve formed by the right half of the circle x2 + y 2 = 9
and the line segment between the points (0, −3) and (0, 3). Calculate
I
−xy dx + 2 dy
C
in two ways:
(a) (10 points) Directly.
Solution
We need to first parametrize the curve using C. Let C1 be the circular
arc and let C2 be
the line segment. Then a parametrization
for C1is given by r~1 (t) = 3 cos t, 3 sin t for
0
t ∈ [− π2
, π2 ]. Then
r
~
(t)
=
−
3
sin
t
dt,
3
cos
for C2 is given by
1
t dt A parametrization
0
r~2 (t) = 0, 3 − 3t for t ∈ [0, 2]. Then r~2 (t) = 0, −3 dt . The line integral becomes
Z
Z
I
−xy dx + 2 dy =
−xy dx + 2 dy +
−xy dx + 2 dy
C1
C
Z
=
π
2
C2
2
27 cos t sin t + 6 cos t dt +
− π2
Z
2
−6 dt
0
= 30 − 12
= 18
Note that the line integral over the segment could be computed without parametrizing the
segment because along the line segment dx = 0 and dy = 6.
(b) (10 points) Using Green’s Theorem.
Solution
ZZ ZZ
∂
∂
(2) −
(−xy) dA =
x dA
∂y
D ∂x
D
Z 3 Z √9−y2
=
x dx dy
−3
Z
3Z
=
0
Z
=
π
2
− π2
= 18
0
√
9−x2
√
− 9−x2
Z 3
2
x dy dx
r cos θ dr dθ
0
Math 2263
Exam 3 - Page 5 of 7
7/25/14
4. (20 points) Theorem: Let F~ (x, y) = P (x,Z y), Q(x, y) be a continuous vector field on an open
and connected region D and suppose that
F~ · d~r is path independent in D.
C
Then there exists a scalar function f (x, y) such that ∇f = F~ .
Prove half of the theorem by showing fx = P . Clearly indicate where you used the each hypothesis of the theorem. A picture would also be nice, but this is just a suggestion.
Since the line integral is path independent we can choose to travel from an arbitrary fixed point
(a, b) to any other point (x, y) in the region D as long as the path stays inside the region. Since
the region is connected such a path exists. Choose a path C that is the union of two paths C1
and C2 where C1 is an arbitrary path from the point (a, b) to the point (xo , y) for some xo in
an open disk with center (x, y) that is totally contained in D. Note that there is such a disk
because the region is open. Let C2 be the horizontal line segment from (xo , y) to (x, y).
Then we can construct the potential function like so,
Z
f (x, y) =
F~ · d~r =
C
Z
F~ · d~r +
C1
Z
C2
F~ · d~r =
Z
(xo ,y)
F~ · d~r +
(a,b)
Z
(x,y)
F~ · d~r.
(xo ,y)
Now we take the derivative of the equation above with respect to x. The integral over C1 is a
function of y only because xo is a constant. Thus the derivative of that integral with respect
to x vanishes. On the other hand C2 is a horizontal line segment, so dy = 0 and we get that
Z (x,y) D
Z x
E ∂
∂
~
F · d~r =
P (u, y), Q(u, y) · du, dy =
P (u, y) du
∂x (xo ,y)
∂x xo
(xo ,y)
Z
Letting W (u) = P (u, y) du we finally get the result we want;
∂
fx (x, y) =
∂x
Z
(x,y)
∂ W (x, y) − W (xo , y)
∂x
= P (x, y)
fx (x, y) =
Math 2263
Exam 3 - Page 6 of 7
7/25/14
5. Consider the vector field F~ (x, y, z) = y 3 z − 2e2x , 3xy 2 z + 2y, xy 3 .
(a) (10 points) Find curl(F~ ) and div(F~ ).
Solution
The divergence of the vector field is given by
∂
∂
∂ 3
(y z − 2e2x ) +
(3xy 2 z + 2y) +
(xy 3 )
∂x
∂y
∂z
= −4e2x + 6xyz + 2
∇ · F~ =
The curl of the vector field is given by

~i
~j
~k 
∂ 
∂
∂
∇ × F~ = Det 
∂x
∂y
∂z
y 3 z − 2e2x 3xy 2 z + 2y xy 3
= ~0
Math 2263
Exam 3 - Page 7 of 7
7/25/14
(b) (10 points) Is F~ conservative? If so, find a potential function for F~ ; if not, explain why.
Solution
The vector field is conservative because the curl is the zero vector and the vector field is
defined on all of R3 and because the partials exist and are continuous. A potential function
of the vector field is
f (x, y, z) = xy 3 z − e2x + y 2 + K,
where K ∈ R.
Z
F~ · d~r, where C is the curve described by the parametric equations
(c) (5 points) Calculate
C
x=
√
t,
y = t cos(πt),
z=
t−1
t2 + 1
(0 ≤ t ≤ 1)
Solution
Since the vector field is conservative we can use the Fundamental Theorem for Line Integrals to evaluate the given line integral.
Z
Z
~
F · d~r =
∇f · d~r
C
C
= f (~r(1)) − f (~r(0))
= f (1, −1, 0) − f (0, 0, −1)
= 2 − e2
' −5.38
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