Austin Cholley Zhu MAT 267 ONLINE A Fall 2021
Assignment Section 12.3 due 09/19/2021 at 11:59pm MST
Problem 1. (1 point)
Using polar coordinates, evaluate the integral Z ZR
sin(x2+y2)dA
where R is the region 9 ≤x2+y2≤81.
Solution:
SOLUTION
The region Ris described in polar coordinates by
R={(r,θ)|3≤r≤9,0≤θ≤2π}.Then
Z ZR
sin(x2+y2)dA =Z9
3Z2π
0
sin(r2)r dθdr
=2πZ9
3
sin(r2)r dr
[Substituting u=r2,du =2rdr]
=πZ81
9
sin(u)du
=−π[cos(u)]81
9
=π(cos(9)−cos(81))
Answer(s) submitted:
•-5.3024
(correct)
Correct Answers:
•-5.30243111246558
Problem 2. (1 point)
By changing to polar coordinates, evaluate the integral
ZZD
(x2+y2)3/2dA where Dis the disk x2+y2≤25.
The value is .
Solution:
SOLUTION
The region Dis described in polar coordinates by
D={(r,θ)|0≤r≤5,0≤θ≤2π}.Then
Z ZD
(x2+y2)3/2dA =Z5
0Z2π
0
(r2)3/2r dθdr
=2πZ5
0
r4dr
=2π
5r55
0
=2π
555
=1250π
Answer(s) submitted:
•3926.9908
(correct)
Correct Answers:
•3926.99081698724
1
Problem 3. (1 point)
Use polar coordinates to find the volume of a sphere of radius 2.
Solution:
SOLUTION
The sphere of radius 2, centered at the origin, has cartesian equa-
tion x2+y2+z2=22and intersects the xy plane in the circle
x2+y2=22.
By symmetry,
V=2ZZx2+y2≤22p22−x2−y2dA
=2Z2π
0Z2
0p22−r2r drdθ
=2Z2π
0
dθZ2
0p22−r2r dr
=4πZ2
0p22−r2r dr
[Substituting u=22−r2,du =−2rdr]
=−4π1
2Z0
22√udu
=2π2
3u3/222
0
=4
3π(22)3/2
=4
3π·23
Answer(s) submitted:
•33.5103
(correct)
Correct Answers:
•33.5103216382911
Problem 4. (1 point)
Find the volume of the ellipsoid x2+y2+7z2=64.
Solution:
SOLUTION
The ellipsoid x2+y2+7z2=64 intersects the xy-plane in the
circle x2+y2=64.
Solving for zyields z=1
√7p64 −x2−y2.
By symmetry,
V=2ZZx2+y2≤64
1
√7p64 −x2−y2dA
=2
√7Z2π
0Z8
0p64 −r2r drdθ
=2
√7Z2π
0
dθZ8
0p64 −r2r dr
=4π
√7Z8
0p64 −r2r dr
[Substituting u=64 −r2,du =−2rdr]:
=−2π
√7Z0
64
√udu
=2π
√72
3u3/264
0
=4π
3√7(64)3/2
=4π
3√7·83
Answer(s) submitted:
•810.6055
(correct)
Correct Answers:
•810.60550773673
2
Problem 5. (1 point)
Find the volume of the solid enclosed by the paraboloids z=
4x2+y2and z=32 −4x2+y2.
Solution:
SOLUTION
The two paraboloids intersect when
4x2+y2=32 −4x2+y2or x2+y2=4. So
V=ZZx2+y2≤432 −4x2+y2−4x2+y2dA
=Z2π
0Z2
0
(32 −8r2)r dr dθ
=Z2π
0
dθZ2
0
(32r−8r3)dr
= [θ]2π
016r2−2r42
0
=2π16(2)2−2(2)4
=2π(32) = 64π
Answer(s) submitted:
•7.5921
(incorrect)
Correct Answers:
•201.062
Problem 6. (1 point)
A cylindrical drill with radius 4 is used to bore a hole through the
center of a sphere of radius 6. Find the volume of the ring shaped
solid that remains.
Solution:
SOLUTION
The given solid is the region outside the cylinder x2+y2=16 be-
tween the surfaces z=p36 −x2−y2and z=−p36 −x2−y2.
The region in the xy-plane is the annular region 16 ≤x2+y2≤36.
So
V=ZZ16≤x2+y2≤36 hp36 −x2−y2−−p36 −x2−y2idA
=ZZ16≤x2+y2≤36
2p36 −x2−y2dA
=2Z2π
0Z6
4p36 −r2r dr dθ
=2[θ]2π
0Z6
4p36 −r2r dr
=4πZ6
4p36 −r2r dr
[Using the substitution u=36 −r2,du =−2rdr]
=−2πZ0
20
√udu
=2π2
3u3/220
0
=4
3π203/2
Answer(s) submitted:
•
(incorrect)
Correct Answers:
•374.656785655505
3
Problem 7. (1 point)
A volcano fills the volume between the graphs z=0 and z=
1
(x2+y2)18 , and outside the cylinder x2+y2=1. Find the vol-
ume of this volcano.
Solution:
SOLUTION
The given solid is the region outside the cylinder x2+y2=1 be-
tween the surfaces z=1
(x2+y2)18 and z=0 . The region in the
xy-plane is the annular region 1 ≤x2+y2<∞. So
V=ZZ1≤x2+y2<∞
1
(x2+y2)18 dA
=Z2π
0Z∞
1
1
(r2)18 r dr dθ
= [θ]2π
0Z∞
1
1
r35 dr
=2πZ∞
1
1
r35 dr
=2πlim
b→∞Zb
1
1
r35 dr
=2πlim
b→∞−1
34r34 b
1
=πlim
b→∞1
17 −1
17b34
=π
17
Answer(s) submitted:
•(pi /(17))
(correct)
Correct Answers:
•0.184799567858223
Problem 8. (1 point)
For the region Rbelow, write RRRf dA as an iterated integral in
polar coordinates.
With a=,b=,c=, and d=,
RRRf dA =Rb
aRd
cf dA, where dA =d d
Note: Use tfor θin your expressions.
Solution:
SOLUTION
ZZR
f dA =Z3π/2
π/2Z2
1
f·r dr dθ.
Answer(s) submitted:
•
•((3pi )/2)
•-1
•-2
•r
•r
•t
(score 0.428571)
Correct Answers:
•pi/2
•3*pi/2
•1
•2
•r
•r
•t
4
Problem 9. (1 point)
(a) Graph r=1/(3cosθ)for −π/2<θ<π/2 and r=1. Then
write an iterated integral in polar coordinates representing the area
inside the curve r=1 and to the right of r=1/(3cosθ). (Use t
for θin your work.)
With a=,b=,
c=, and d=,
area = Rb
aRd
cd d
(b) Evaluate your integral to find the area.
area =
Note: You must complete part (a) in order to receive any partial
credit.
Solution:
SOLUTION
(a) The curve r=1/(3cosθ), or rcosθ=1/3, is the line x=1/3.
The curve r=1 is the circle of radius 1 centered at the origin.
This gives the graph
The line intersects the circle where 3cosθ=1, so θ=
±arccos(1/3). Thus
Area =Zarccos(1/3)
−arccos(1/3)Z1
1/(3cosθ)
r dr dθ.
Evaluating gives
Area =Zarccos(1/3)
−arccos(1/3) r2
2
1
1/(3cosθ)!dθ
=1
2Zarccos(1/3)
−arccos(1/3)1−1
9cos2θdθ
=1
2θ−tanθ
9
arccos(1/3)
−arccos(1/3)
=1
22(arccos(1/3)) −2tan(arccos(1/3))
9
=arccos(1
3)−1
9tan(arccos(1
3)).
Answer(s) submitted:
•-(acos((1/3)))
•acos((1/3))
•(1/(3cost))
•1
•r
•r
•t
•arccos((1/3))-(1/9)tan(arccos((1/3)))
(correct)
Correct Answers:
•-acos(1/3)
•acos(1/3)
•1/[3*cos(t)]
•1
•r
•r
•t
•acos(1/3)-[tan(acos(1/3))]/(3ˆ2)
5
Problem 10. (1 point)
For each of the following, set up the integral of an arbitrary func-
tion f(x,y)over the region in whichever of rectangular or polar
coordinates is most appropriate. (Use tfor θin your expressions.)
(a) The region
With a=,b=,
c=, and d=,
integral = Rb
aRd
cd d
(b) The region
With a=,b=,
c=, and d=,
integral = Rb
aRd
cd d
Solution:
SOLUTION
(a) Since this is a rectangular region, we use Cartesian coordi-
nates. This gives
Z5
2Z6
2
f(x,y)dy dx.
(b) Since this is a partially-circular region, we use polar coordi-
nates. This gives
Zπ/2
π/4Z1
0
f(rcos(θ),rsin(θ))r dr dθ.
Answer(s) submitted:
•2
•2
•
•
•
•
•
•0
•0
•0
•0
•
•
•
(incorrect)
Correct Answers:
•2
•5
•2
•6
•f(x,y)
•y
•x
•0.785398
•1.5708
•0
•1
•r*f(r*cos(t),r*sin(t))
•r
•t
Problem 11. (1 point)
Find the volume of the region between the graph of f(x,y) =
1−x2−y2and the xyplane.
volume =
Solution:
SOLUTION
The graph of f(x,y) = 1−x2−y2is an upside down bowl, and
the region whose volume we want is contained between the bowl
(above) and the xy-plane (below). We must first find the region
in the xy-plane where f(x,y)is positive. To do that, we set
f(x,y)≥0 and get x2+y2≤1. The disk x2+y2≤1 is the re-
gion Rover which we integrate.
Volume =ZZR
(1−x2−y2)dA =Z2π
0Z1
0
(1−r2)r dr dθ
=Z2π
01
2r2−1
4r4
1
0
dθ
=1
4Z2π
0
dθ=1
2π.
Answer(s) submitted:
•(pi /2)
(correct)
Correct Answers:
•pi*1/2
6
Problem 12. (1 point)
Consider the solid shaped like an ice cream cone that is bounded
by the functions z=px2+y2and z=p32 −x2−y2.Set up an
integral in polar coordinates to find the volume of this ice cream
cone.
Instructions: Please enter the integrand in the first answer box,
typing theta for θ. Depending on the order of integration you
choose, enter dr and dtheta in either order into the second and
third answer boxes with only one dr or dtheta in each box. Then,
enter the limits of integration and evaluate the integral to find the
volume.
ZB
AZD
C
A =
B =
C =
D =
Volume =
Solution:
SOLUTION
The functions z=px2+y2and z=p32 −x2−y2intersect when
px2+y2=p32 −x2−y2⇒x2+y2=32−x2−y2⇒x2+y2=
16.
Thus the region of integration is the disk
D=(x,y)|x2+y2≤16
={(r,θ)|0≤θ≤2π,0≤r≤4}.
In polar coordinates, the cone z=px2+y2has equation
z=r, while the hemisphere z=p32 −x2−y2has equation
z=√32 −r2, so the volume is given by
V=Z2π
0Z4
0p32 −r2−rr dr dθ
=2πZ4
0p32 −r2r−r2dr
=2πZ4
0p32 −r2r dr −2πZ4
0
r2dr [Substitution :u=32 −r2,du =−2rdr]
=−πZ16
32
u1/2du −2πr3
34
0
=2
3πhu3/2i32
16 −2π64
3
=2
3π323/2−163/2−2π64
3
=128
3π23/2−2
Answer(s) submitted:
•sqrt(32-rˆ(2)-r)(r)
•
(score 0.235)
Correct Answers:
•[sqrt(32-rˆ2)-r]*r; dr; dtheta; 0; 2*pi; 0; 4
•0.828427*64*2*pi/3
7
Problem 13. (1 point)
Match each double integral in polar with the graph of the
region of integration.
? 1. Z2π
0Z5
4
f(r,θ)r dr dθ
? 2. Z4
0Z3π/4
−π/2
f(r,θ)r dθdr
? 3. Z5
4Z7π/4
3π/4
f(r,θ)r dθdr
? 4. Z4
0Z3π/4
0
f(r,θ)r dθdr
? 5. Z3π/4
−π/4Z5
4
f(r,θ)r dr dθ
? 6. Z2π
3π/2Z5
0
f(r,θ)r dr dθ
? 7. Z3π/2
3π/4Z4
0
f(r,θ)r dr dθ
? 8. Z2π
0Z4
0
f(r,θ)r dr dθ
A B C
D E F
G H
Answer(s) submitted:
•E
•B
•C
•A
•D
•G
•F
•H
(correct)
Correct Answers:
•E
•B
•C
•A
•D
•G
•F
•H
8
Problem 14. (1 point)
Convert the integral below to polar coordinates and evaluate the
integral.
Z4/√2
0Z√16−y2
y
xy dx dy
Instructions: Please enter the integrand in the first answer box,
typing theta for θ. Depending on the order of integration you
choose, enter dr and dtheta in either order into the second and
third answer boxes with only one dr or dtheta in each box. Then,
enter the limits of integration and evaluate the integral to find the
volume.
ZB
AZD
C
A =
B =
C =
D =
Volume =
Solution:
SOLUTION
(Click on graph to enlarge)
The graph of the region of integration is shown above.
The point Ahas coordinates 4
√2,4
√2. The region is bounded on
the left by the line y=xand on the right by the circle of radius 4
centered at the origin.
In polar coordinates the region is described by
(r,θ)|0≤θ≤π
4,0≤r≤4
Thus
Z4/√2
0Z√16−y2
y
xy dx dy =Zπ/4
0Z4
0
(rcosθ)(rsinθ)r dr dθ
=Z4
0
r3dr Zπ/4
0
cosθsinθdθ
=r4
44
0sin2θ
2π/4
0
=44
4
1
2 √2
2!2
=16
Answer(s) submitted:
•rˆ(2)cos(theta)*sin(theta)
•
(score 0.235)
Correct Answers:
•rˆ2*cos(theta)*sin(theta)*r; dr; dtheta; 0; pi/4; 0; 4
•16
9
Problem 15. (1 point)
Find the volume of the wedge-shaped region (Figure 1) contained
in the cylinder x2+y2=49 and bounded above by the plane z=x
and below by the xy-plane.
V=
Solution:
Solution:
[Step 1.] Express Win cylindrical coordinates. Wis bounded
above by the plane z=xand below by z=0,
therefore 0 ≤z≤x, in particular x≥0. Hence, Wprojects onto
the semicircle Din the xy-plane of radius 7, where x≥0.
r=7
In polar coordinates,
D:−π
2≤θ≤π
2,0≤r≤7
The upper surface is z=x=rcosθand the lower surface is z=0.
Therefore,
W:−π
2≤θ≤π
2,0≤r≤7,0≤z≤rcosθ
[Step 2.] Set up an integral in cylindrical coordinates and evalu-
ate. The volume of Wis the triple integral RRRW1dV .
Using change of variables in cylindrical coordinates gives
ZZZW
1dV =Zπ/2
−π/2Z7
0Zrcosθ
0
r dz dr dθ=Zπ/2
−π/2Z7
0
rz
rcosθ
z=0
dr dθ=
Zπ/2
−π/2Z7
0
r2cosθdr dθ=Zπ/2
−π/2
r3
3cosθ
7
r=0
dθ=Zπ/2
−π/2
73
3cosθdθ=
73
3sinθ
π/2
−π/2
=73
3sin π
2−sin−π
2=228.667
Answer(s) submitted:
•228.667
(correct)
Correct Answers:
•228.667
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