Math 261 – Solutions To Sample Exam 2 Problems

1
Solutions to Sample Exam 2 Problems – Math 261
Math 261 – Solutions To Sample Exam 2 Problems
√
1. Given to the right is the graph of a portion of four curves: x = 0, y = 1, x − 3y = 0
(0,2)
and x2 + y 2 = 4. Note that these curves divide the plane into 3 separate regions, which
R1
(0,1)
have been marked on the diagram.
R2
R
(0,0)
(a) Write R1 2x dA as an iterated integral, both in the “dx dy” order and in the “dy dx”
R3
order. Then, evaluate one of the two integrals.
R
(b) Set up R2 2x dA as an iterated integral, both in the “dx dy” order and in the
(0,−2)
“dy dx” order.
R
(c) Use polar coordinates to evaluate R3 (x2 + y 2 ) dA.
R
(d) Use polar coordinates to evaluate R 12x dA, where R is the region formed by combining the regions
R1 and R2 .
R
(e) Set up, but do not evaluate, an iterated integral equivalent to S 2x dA, where S is the region formed
by combining the regions R2 and R3 . Use whatever coordinate system you think is easiest.
Solution.
(a) Integrating in the “dy dx” direction means that we will be integrating in the ydirection first, meaning that we need to find the curves at the bottom and top of
the region R1 . From the diagram, we see that the curve on the bottom is the line
y = 1, and the curve on √the top is the circle x2 + y 2 = 4. Solving for y, we have
x2 + y 2 = 4 =⇒ y = 4 − x2 , so
Z
2x dA =
R1
√
3
Z
Z
√
H0, 2L
x2 +y2 =4
!!!
H 3, 1L
4−x2
2x dy dx.
1
0
Integrating in the “dx dy” direction means that we will be integrating in the x-direction first, meaning
that we need to find the curves on the left and right of the region R1 . From the diagram above, we see
that the curve on the left is the line x =p
0, and the curve on the right is the circle x2 + y 2 = 4. Solving
for x, we have x2 + y 2 = 4 =⇒ x = 4 − y 2 , so
Z Z √ 2
Z
2
4−y
2x dx dy.
2x dA =
Calculating, we see that
Z Z √ 2
2
4−y
2x dx dy
2
Z
=
0
1
0
1
R1
1
Z 2
√4−y2
y 3 2
5
x dy =
(4 − y 2 ) dy = 4y −
= .
3
3
0
1
1
2
(b) As above, integrating in the “dy dx” direction means that we need to locate the top
and bottom curves and solve
√ for y. The top curve is the line y = 1, while for the
bottom curve we have x − 3y = 0 =⇒ y = √13 x. Therefore,
Z
2x dA =
R2
Z
√
3
0
Z
H0, 1L
!!!
x- 3 y = 0
1
√
(1/ 3)x
2x dy dx.
!!!
H 3, 1L
H0, 0L
To treat the “dx dy” direction,
√ on the left side of R2 is x = 0, while for the curve on
√ note that the curve
the right side we have x − 3y = 0 =⇒ x = 3y. Therefore,
Z 1 Z √3y
Z
2x dx dy.
2x dA =
R2
0
0
√
(c) Since the point ( 3, 1) at the upper right-hand corner of R3 is given by (2, π/6) in polar coordinates, it
follows that R3 = {(r, θ) : 0 ≤ r ≤ 2, −π/2 ≤ θ ≤ π/6} . Therefore, we have
Z 2 Z π6
Z
r2 r dθ dr (since x2 + y 2 = r2 and dA = r dθ dr)
(x2 + y 2 ) dA =
R3
0
−π
2
2
Solutions to Sample Exam 2 Problems – Math 261
π
=
6
8π
.
3
=
+
π
2
Z
2
r3 dr
0
(d) We have R = {(r, θ) : 0 ≤ r ≤ 2, π/6 ≤ θ ≤ π/2} , so
Z
Z
(12x) dA =
R
Z
=
π
2
π
6
Z
2
12r cos θ r dr dθ
(since x = r cos θ and dA = r dr dθ)
0
π
2
π
6
2
(4r ) cos θ dθ
3
0
π/2
32 sin θ
=
π/6
=
!!!
H 3, 1L
16.
(e) Since the region S is bounded
by only one curve on the left (x = 0) and only one
p
curve on the right (x = 4 − y 2 ), the integration will be the simplest in the “dx dy”
order. We therefore have
Z Z √ 2
Z
1
2x dA
2x dx dy.
=
S
x2 +y2 =4
4−y
−2
0
H0,-2L
Other equivalent answers require two integrals. Here are two alternate answers that are correct, though
they are more complicated:
Z
√
0
Z
π
6
−π
2
3
Z
1
√
− 4−x2
Z
2x dy dx
+
2
2r cos θ r dr dθ
+
0
Z
2
√
Z
3
π
2
π
6
√
Z
4−x2
√
− 4−x2
Z
2x dy dx,
or
1
sin θ
2r cos θ r dr dθ.
0
2. Pictured to the right is a thin metal plate in the shape of a quarter circle placed in the first
quadrant. All distances are measured in inches.
1
y
(a) Find the mass of the plate if its density is a uniform 10 grams per square inch.
p
(b) Find the mass of the plate if its density is given by σ(x, y) = 100x2 + 100y 2 grams
per square inch. Do NOT use your calculator to evaluate the integral.
1
x
Solution. First, we recall that, if σ(x, y) represents our density function, then
Mass =
Z
σ(x, y) dA,
E
which makes sense because σ has units of grams/in2 and dA has units of in2 , indicating that the above
integral will have units of grams.
(a) We could calculate an integral to find the mass, but since the density is constant for part (a), we simply
have
5π
grams.
Mass = (Density)(Area of Plate) = (10)(π/4) =
2
(b) In this case, we need to integrate since the density is not constant. Therefore, the mass is given by
Z
Z p
p
100x2 + 100y 2 dA =
10 x2 + y 2 dA =
E
E
=
Z
0
π
2
Z
0
1
10r · r dr dθ
5π
grams.
3
3
Solutions to Sample Exam 2 Problems – Math 261
R
3. Evaluate W z dV, where W is the first-octant portion of the solid sphere of radius 2 centered at the origin.
Do this in 2 ways: using cylindrical coordinates and using spherical coordinates.
Solution.
(a) In spherical coordinates, we have z = ρ cos φ, and the equation of our sphere becomes ρ = 2. Therefore,
we have
Z 2 Z π2 Z π2
Z
Z Z π
π 2 2 3
(ρ cos φ)p2 sin φ dθ dφ dρ =
ρ sin φ cos φ dφ dρ
z dV =
2 0 0
0
0
0
W
Z
π 2 3 sin2 φ π2
=
ρ
dρ
2 0
2 0
Z
π 2 3
ρ dρ
=
4 0
= π.
(b) In cylindrical coordinates, we have
x2 + y 2 + z 2 = 4
r2 + z 2 = 4
=⇒
=⇒
z=±
p
4 − r2 ,
and since z ≥ 0 for the
√ portion of the sphere we are interested in, the equation for the surface of the
sphere becomes z = 4 − r2 . Therefore, we have
Z
z dV =
W
Z
0
2
Z
0
√
4−r 2
Z
π
2
0
π
zr dθ dz dr =
2
Z
2
0
Z
√
4−r 2
zr dz dr
=
0
π
2
0
π
4
= π.
=
2
Z
Z
2
0
√
z 2 4−r2
r dr
2 0
r(4 − r2 ) dr
4. Consider the solid region in the 1st octant that lies below the plane 2y + z = 4
and inside the cylinder x2 + z 2 = 4 (see diagram to the right). Set up, but do not
evaluate, iterated integrals that give the volume of this region in two ways: in the
“dy dz dx” order and in the “dx dz dy” order.
z
y
x
Solution.
First, we tackle the “dy dz dx” order of integration. The surface on the left hand
side of the region is the plane y = 0, and the surface on the right hand side of the
region is the plane 2y + z = 4, which is equivalent to y = 12 (4 − z). To the right,
note that the xz-projection of this solid is shown, where the x- and z-intercepts
have been found by setting z = 0 and x = 0 in the equation x2 + z 2 = 4 and
observing that x and z are both positive. Therefore, we have
z
2
z=
"############2
4-x
2
Volume =
Z
0
2
Z
0
√
4−x2
Z
1
2 (4−z)
dy dz dx.
0
x
4
Solutions to Sample Exam 2 Problems – Math 261
Next, we tackle the “dx dz dy” order of integration. In this case, the surface on
z
the back side of the region is the plane x = 0, and the surface on the front
side
√
2
of the region is the cylinder x2 + z 2 = 4, which is equivalent to x = 4 − z 2
z=4-2y
since we are in the first octant. To the right, note that the yz-projection of
this solid is shown, where the various y- and z-intercepts have been found by
equating appropriate surfaces. Because there are two different “top curves” in
y
1 2
the projection picture, we will need two separate integrals to describe the integral
over the projected region shown above. Therefore, we obtain
Z 1 Z 2 Z √4−z2
Z 2 Z 4−2y Z √4−z2
Volume =
dx dz dy +
dx dz dy.
0
0
0
1
0
0
5. Find parameterizations for the following curves.
(a) The circle of radius 4, parallel to the xz-plane and centered at (0, 3, 0).
(b) The line segment starting at (2, 1, 4) and ending at (3, 5, −1).
Solution.
(a) x = 4 cos t, y = 3, z = 4 sin t, for 0 ≤ t ≤ 2π.
(b) First, we note that the vector ~v = (3 − 2)~i + (5 − 1)~j + (−1 − 4)~k = ~i + 4~j − 5~k is parallel to the line
segment. Therefore, a parameterization for this segment is given by
x = 2 + t, y = 1 + 4t, z = 4 − 5t, where 0 ≤ t ≤ 1.
6. The National Park Service has conducted a cactus
census in Arizona. In order to collect data, the service
divided the desert into a grid whose lines are 10 miles
apart in each direction. The entries in the table to
the right represent thousands of cacti per square mile
at grid points in Saguaro East National Park, where
the visitor center is located at (0, 0).
Distance, y
(miles south)
0
10
20
30
Distance, x
0
10
8.5 8.2
9.5 10.6
9.3 10.5
8.3 8.6
(miles
20
7.9
10.5
10.4
9.3
east)
30
8.1
10.1
9.5
9.1
(a) Let f (x, y) be the number of cacti (in thousands) per square mile x miles east and y miles south of the
Z 30 Z 30
visitor center. Explain what
f (x, y) dy dx represents in the context of this problem.
0
0
(b) Give the best estimate you can based on the data of the number of cacti in the 30 by 30 test area.
Solution.
(a) First, we examine the units of the various pieces of our integral. The function f (x, y) is measured in
thousands of cacti per square mile, and dx and dy are both measured in miles. Therefore, the units of
the integral are given by
thousands of cacti
· mi · mi = thousands of cacti.
mi2
Therefore, this integral represents the population of cacti (in thousands) in the 30 by 30 region of the
desert described by the provided table.
(b) To estimate the number of cacti, we need to estimate the population density in each
subsquare represented by the table and multiply by the area. For example, in the
9.2
9.3
subsquare described by 0 ≤ x ≤ 10 and 0 ≤ y ≤ 10 (the upper left square described by
the table), there are four different population densities we could use: 8.5, 8.2, 9.5, or
9.975 10.5
10.6. These are just the numbers that are given at the four corners of the subsquare.
Therefore, a reasonable estimate of the population density for this subsquare would be
9.175
8.5 + 8.2 + 9.5 + 10.6
= 9.2 thousand cacti per square mile.
4
9.7
9.15
10.125
9.575
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Solutions to Sample Exam 2 Problems – Math 261
Doing this for each of the 9 subsquares described from the table, we obtain the population densities
described by the diagram above.
To get a population, we need to take each of these population densities and multiply by the area of the
corresponding subsquare, which is 100 square miles for each of them. Therefore, our best estimate of the
cactus population is given by
100(9.2 + 9.975 + 9.175 + 9.3 + 10.5 + 9.7 + 9.15 + 10.125 + 9.575) = 8670 thousand cacti,
or about 8.67 million cacti.
7. Given below are three vector fields, F~1 , F~2 , and F~3 . Let C1 and C2 denote circles of radius 1 and 2, respectively,
both centered at (0, 0) and oriented counterclockwise. Also, let C3 denote the line segment starting at
(−3, −3) and ending at (2, 2), and let C4 denote the line segment starting at (2, 2) and ending at (2, −2).
F~1 (x, y)
F~2 (x, y)
F~3 (x, y)
3
3
3
2
2
2
1
1
1
0
0
0
-1
-1
-1
-2
-2
-2
-3
-3 -2 -1 0
1
2
-3
-3 -2 -1 0
3
1
2
-3
-3 -2 -1 0
3
1
2
3
(a) For each of the four curves C1 , C2 , C3 , and C4 , and for each of the vector fields F~1 , F~2 , and F~3 , decide
R
whether C F~ · d~r appears to be positive, negative, or zero. Note that there are 12 different integrals
to be examined.
R
R
F~1 · d~r ? Explain.
F~1 · d~r or
(b) Which is larger,
C2
C1
Solution.
(a) Vector Field F~1 :
Z
F~1 · d~r > 0,
Z
F~2 · d~r < 0,
Z
F~3 · d~r = 0,
Z
F~2 · d~r < 0,
Z
F~3 · d~r = 0,
C2
C1
Vector Field F~3 :
F~1 · d~r > 0,
C2
C1
Vector Field F~2 :
Z
C1
C2
Z
Z
Z
F~1 · d~r = 0,
F~1 · d~r < 0.
Z
F~2 · d~r > 0.
Z
F~3 · d~r = 0.
C4
C3
F~2 · d~r > 0,
C4
C3
F~3 · d~r < 0,
C3
Z
C4
(b) All the vectors in the vector field appear to have the same length. Therefore, since both curves go
R flow of the vector field but C2 is a longer path than C1 , we conclude that
Rdirectly with the
~
~1 · d~r <
r.
F
C2 F1 · d~
C1
~ = (y + x)~i + (y − x) ~j, where C1 is the curve consisting of the
8. Let F~ = 2xy 2 ~i + (2yx2 + 2y) ~j, and let G
circle of radius 2, centered at the origin and oriented counterclockwise, and where C2 is the curve consisting
of the line segment from (0, 0) to (1, 1) followed by the line segment from (1, 1) to (3, 1).
~ over C1 .
(a) Calculate the line integral of F~ over C1 .
(b) Calculate the line integral of G
~
~
(c) Calculate the line integral of F over C2 .
(d) Calculate the line integral of G over C2 .
~ are conservative vector fields. Since
Solution. Before we begin, let us check to determine whether F~ or G
∂
2
∂y (2xy )
= 4xy =
∂
2
∂x (2yx
+ 2y), we see that F~ is a conservative vector field. To find a potential function f
6
Solutions to Sample Exam 2 Problems – Math 261
for F~ , we integrate as follows:
Z
f (x, y) = 2xy 2 dx
Z
f (x, y) = (2yx2 + 2y) dy
= x2 y 2 + P (y)
= x2 y 2 + y 2 + Q(x)
Therefore, f (x, y) = x2 y 2 + y 2 is a potential function for F~ . On the other hand, since
~ is not a conservative vector field.
equal to ∂ (y − x) = −1, we see that G
∂
∂y (y
+ x) = 1 is not
∂x
R
(a) Since C1 is a closed curve and F~ is a conservative vector field, we know that C1 F~ · d~r = 0 without
doing any calculations. So our final answer is 0.
~ is not a conservative vector field, we must do this integral by parameterizing C1 . We can
(b) Since G
represent C1 by the parametric curve ~r(t) = 2 cos t~i + 2 sin t ~j, where 0 ≤ t ≤ 2π. Therefore, we have
Z 2π
Z
~ · d~r =
~ cos t, 2 sin t) · (−2 sin t~i + 2 cos t ~j) dt
G
G(2
0
C1
=
Z
2π
Z
2π
Z
2π
0
=
0
=
(2 sin t + 2 cos t)~i + (2 sin t − 2 cos t) ~j) · (−2 sin t~i + 2 cos t ~j) dt
[−4(sin2 t + cos2 t) − 4 sin t cos t + 4 sin t cos t] dt
(−4) dt,
0
so our final answer is −8π.
(c) Since F~ is conservative, we can use the potential function f (x, y) = x2 y 2 + y 2 that we calculated above,
and the Fundamental Theorem of Calculus for Line Integrals as follows:
Z
F~ · d~r = f (3, 1) − f (0, 0) = 10 − 0 = 10
C2
Therefore, our final answer is 10.
~ is not conservative, we must parameterize the two line segments comprising C2 and calculate
(d) Since G
this line integral using brute force. The line segment from (0, 0) to (1, 1) can be parameterized as
~r1 (t) = t~i + t ~j, where 0 ≤ t ≤ 1, and the line segment from (1, 1) to (3, 1) can be parameterized as
~r2 (t) = (1 + 2t)~i + ~j, where 0 ≤ t ≤ 1. Therefore, we have
Z 1
Z 1
Z
~ · d~r =
~ r1 (t)) · ~r1′ (t) dt +
~ r2 (t)) · ~r2′ (t) dt
G
G(~
G(~
0
C2
=
0
Z
1
Z
1
Z
1
(2t~i + 0~j) · (~i + ~j) dt +
0
=
1
((2 + 2t)~i + 2t ~j) · 2~i dt
0
(2t · 1 + 2(2 + 2t)) dt
0
=
Z
(6t + 4) dt
=
7,
0
so our final answer is 7.
R
9. Let F~ = x2 ~i+z sin(yz) ~j +y sin(yz) ~k. Calculate C F~ · d~r,
where C is the path from A = (0, 0, 1) to B = (3, 1, 2)
shown in the figure to the right.
B
x
z
A
y
7
Solutions to Sample Exam 2 Problems – Math 261
Solution. Calculating, we observe that curl F~ = ~0, so the vector field F~ is conservative and we can use the
Fundamental Theorem of Calculus for Line Integrals, provided
we do below:
Z
f (x, y, z) =
x2 dx
=
Z
f (x, y, z) =
z sin(yz) dy =
Z
f (x, y, z) =
y sin(yz) dz =
that we can find a potential function, which
x3
+ p(y, z)
3
− cos(yz) + q(x, z)
− cos(yz) + r(x, y)
3
Therefore, f (x, y, z) = x3 − cos(yz) is a potential function for F~ , so by the Fundamental Theorem of Calculus
for Line Integrals, we have
Z
F~ · d~r = f (3, 1, 2) − f (0, 0, 1) = (9 − cos 2) − (0 − cos 0) = 10 − cos 2.
C