Name:
MATH 208
Solutions
Exam # 3
This test is worth 100 points. Show all of your work to receive full and partial credit, making
sure that all work and answers are legible. Books, notes, formula sheets, etc. are not allowed.
1. (8 points) Let C be the upper half of the unit circle, traveled counter clockwise. For each of the vector
R
~ is positive, negative, or zero. (Hint: draw C on the vector fields.)
fields below, say if C F~ · dr
R
C
R
C
F~1
F~2
F~3
F~4
~ is : zero
F~1 · dr
R
~ is : positive
F~3 · dr
C
~ is : posititve
F~2 · dr
R
C
~ is : negative
F~4 · dr
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MATH 208
Solutions
2. (12 points) Set up a double integral in polar coordinates to integrate f (x, y) = 4(x2 + y 2 ) over the region
shown below.
π
Z
2
Z
0
4r3 drdθ
1
3. (a) (4 points) Sketch the region above the cone z =
p
x2 + y 2 and below the sphere x2 + y 2 + z 2 = 14.
(b) (12 points) Set up a triple integral in spherical coordinates to calculate the volume of the region in
part a).
Z
0
π/4
Z
0
2π
√
Z
14
ρ2 sin φdρdθdφ
0
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MATH 208
Solutions
4. The parameterization x = 2 cos(t), y = 2 sin(t), z = t3 describes the motion of an object. Distances are
measured in feet, and t is measured in minutes.
(a) (4 points) Where is the object when t = 0?
~r(0) = 2 cos(0)~i + 2 sin(0)~j + 03~k = 2~i. So the object is at (2, 0, 0), ie. two feet from the origin in
the positive x direction.
(b) (5 points) Find the velocity vector.
~v (t) = ~r0 (t) = −2 sin(t)~i + 2 cos(t)~j + 3t2~k
(c) (4 points) The object hits the plane z = 8. When does this happen?
When t3 = 8, so after 2 minutes.
(d) (6 points) How fast is the object traveling when it hits z = 8? (If you give your answer as a decimal
approximation, round it to three decimal places.)
~v (2) = −2 sin(2)~i + 2 cos(2)~j + 3(2)2~k, so
q
p
||~v (2)|| = (−2 sin(2))2 + (2 cos(2))2 + 122 = 4 sin2 (2) + 4 cos2 (2) + 122
q
√
= 4(sin2 (2) + cos2 (2)) + 144 = 148 ≈ 12.165 ft/min.
(e) (5 points) What is the acceleration vector of the object?
~a(t) = ~v 0 (t) = −2 cos(t)~i − 2 sin(t)~j + 6t~k
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MATH 208
Solutions
5. Consider the vector field F~ = x~j.
(a) (6 points) Sketch the vector field.
(b) (6 points) Sketch the flow of F~ .
(c) (8 points) Consider the parameterization x(t) = a, y(t) = at. Is ~r(t) = x(t)~i + y(t)~j = a~i + at~j a
flow line of F~ ? Show why or why not.
~r0 (t) = 0~i + a~j, F~ (t) = x(t)~j = a~j, so ~r0 (t) = F~ (t). Thus ~r(t) is a flow line of F~ .
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MATH 208
Solutions
6. (a) (8 points) Find a parametric equation for the line segment from (8, 8) to (7, 10).
The displacement vector between (8, 8) and (7, 10) is: ~v = (7 − 8)~i + (10 − 8)~j = −~i + 2~j.
So the line segment is parameterized by
~r(t) = (8 − t)~i + (8 + 2t)~j with 0 ≤ t ≤ 1.
(b) (12 points) Compute
R
C
F~ · d~r where F~ = (x − y)~i + y~j and C is the line segment from part a).
~r0 (t) = −~i + 2~j
F~ (~r(t)) = (x(t) − y(t))~i + y(t)~j = ((8 − t) − (8 + 2t))~i + (8 + 2t)~j = −3t~i + (8 + 2t)~j
Z 1
Z
~
~
F · dr =
(−3t~i + (8 + 2t)~j) · (−~i + 2~j)dt
0
C
Z
=
1
Z
3t + 16 + 4tdt =
0
1
16 + 7tdt
0
7
39
7
.
= 16t + t2 |10 = 16 + =
2
2
2
Have a good break!
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