Daniel Mirandoli Ruedemann MAT 267 ONLINE B Summer 2019
Assignment Section 12.7 due 08/12/2019 at 11:59pm MST
1. (1 point) What are the rectangular coordinates of the point
whose spherical coordinates are 2,3π
4,π
6?
x=
y=
z=
Solution:
SOLUTION
x=ρcosθsinφ=2cos3π
4sinπ
6=2−√2
21
2
y=ρsinθsinφ=2sin3π
4sinπ
6=2√2
21
2
z=ρcosφ=2cosπ
6=2√3
2
So the point is 2−√2
21
2,2√2
21
2,2√3
2in rectan-
gular coordinates.
Correct Answers:
•-0.707107
•0.707107
•1.73205
2. (1 point) What are the spherical coordinates of the point
whose rectangular coordinates are
(1,3,4)?
ρ=
θ=
φ=
Solution:
SOLUTION
ρ=p(1)2+ (3)2+ (4)2=√26
cosφ=z
ρ=4
√26 ⇒φ=arccos4
√26 .
The point (1,3)is in the first quadrant of the xy-plane, so
θ=arctan3
1.
Thus spherical coordinates are
√26,arctan3
1,arccos4
√26
Correct Answers:
•5.09901951359278
•1.24904577239825
•0.668964074268407
3. (1 point) Express the point given in Cartesian coordinates
in spherical coordinates (ρ,θ,φ). Note: you really only have
to do the work for one, if you use a little geometry and your
knowledge of the trig functions.
A) 9
4√6,9
4√2,−9
2√2=
B)−9
4√6,9
4√2,−9
2√2=
C)9
4√6,−9
4√2,+9
2√2=
D)−9
4√6,−9
4√2,+9
2√2=
Solution:
SOLUTION
A) ρ2=px2+y2+z2=r9
4√62+9
4√22+−9
2√22=
√92=9
The point 9
4√6,9
4√2is in the first quadrant of the xy-plane,
so θ=arctany
x=arctan9
4√2/9
4√6=arctan1
√3=π
6
cos(φ) = z
ρ=−9
2√2
9=−1
2√2⇒φ=3π
4
Thus spherical coordinates are 9,π
6,3π
4.
B) ρ2=px2+y2+z2=r−9
4√62+9
4√22+−9
2√22=
√92=9
The point −9
4√6,9
4√2is in the second quadrant of the xy-
plane, so θ=arctan y
x+π=arctan9
4√2/−9
4√6+π=
arctan−1
√3+π=−π
6+π=5π
6
cos(φ) = z
ρ=−9
2√2
9=−1
2√2⇒φ=3π
4
Thus spherical coordinates are 9,5π
6,3π
4.
C) ρ2=px2+y2+z2=r9
4√62+−9
4√22++9
2√22=
√92=9
The point 9
4√6,−9
4√2is in the fourth quadrant of the xy-
plane, so θ=arctan y
x+2π=arctan−9
4√2/9
4√6+
2π=arctan−1
√3+2π=−π
6+2π=11π
6
cos(φ) = z
ρ=+9
2√2
9= +1
2√2⇒φ=π
4
Thus spherical coordinates are 9,11π
6,π
4.
D) ρ2=px2+y2+z2=r−9
4√62+−9
4√22+−9
2√22=
√92=9
The point −9
4√6,−9
4√2is in the third quadrant of the xy-
plane, so θ=arctan y
x+π=arctan−9
4√2/−9
4√6+π=
arctan1
√3+π=π
6+π=7π
6
1
cos(φ) = z
ρ=+9
2√2
9= +1
2√2⇒φ=π
4
Thus spherical coordinates are 9,7π
6,π
4.
Correct Answers:
•(9,0.523599,2.35619)
•(9,2.61799,2.35619)
•(9,5.75959,0.785398)
•(9,3.66519,0.785398)
4. (1 point) What are the cylindrical coordinates of the point
whose spherical coordinates are
(ρ,θ,φ) = (5,5,π
2)?
r=
θ=
z=
Solution:
SOLUTION
r=ρsin(φ) = 5sinπ
2=5(1) = 5
θ=5
z=ρcos(φ) = 5cosπ
2=5(0) = 0
Thus cylindrical coordinates are (5,5,0)
Correct Answers:
•5
•5
•0
5. (1 point)
Match the given equation with the verbal description of the
surface:
A. Cone
B. Elliptic or Circular Paraboloid
C. Circular Cylinder
D. Half plane
E. Plane
F. Sphere
1. ρcos(φ) = 4
2. r=2cos(θ)
3. z=r2
4. ρ=2cos(φ)
5. r=4
6. ρ=4
7. θ=π
3
8. φ=π
3
9. r2+z2=16
Solution: 1. ρcos(φ) = 4 is equivalent to z=4 and the sur-
face is a horizontal plane.
Therefore the answer is E.
2. r=2cos(θ)⇒r2=2rcosθ⇔x2+y2=2x⇔
(x−1)2+y2=1. Thus the surface is a right circular cylinder of
radius 1 with vertical axis through the point (1,0,0).
Therefore the answer is C.
3. z=r2is equivalento to z=x2+y2and the surface is a
circular paraboloid with vertex at the origin.
Therefore the answer is B.
4. ρ=2cos(φ)⇒ρ2=2ρcosφ⇔x2+y2+z2=2z⇔
x2+y2+ (z−1)2=1. Thus the surface is a sphere of radius 1
centered at (0,0,1).
Therefore the answer is F.
5. r=4 is equivalent to x2+y2=16 and the surface is a
circular cylinder of radius 4 and axis the z-axis.
Therefore the answer is C.
6. ρ=4 is equivalent to x2+y2+z2=16 and the surface is
a sphere with center at the origin and radius 4 .
Therefore the answer is F.
7. θ=π
3represents a vertical plane that forms an angle of π
3
with the positive x-axis.
Therefore the answer is D.
8. φ=π
3represents the top half of the right circular cone with
vertex at the origin and axis the positive z-axis.
Therefore the answer is A.
9. r2+z2=16 ⇔x2+y2+z2=16 and the surface is a sphere
of radius 4 centered at the origin.
Therefore the answer is F.
Correct Answers:
•E
•C
•B
•F
•C
•F
•D
•A
•F
6. (1 point) Match the integrals with the type of coordinates
which make them the easiest to do. Put the letter of the coordi-
nate system to the left of the number of the integral.
1. RRREz dV where E is: 1 ≤x≤2,3≤y≤4,5≤z≤6
2. RRREz2dV where E is: −2≤z≤2,1≤x2+y2≤2
3. R1
0Ry2
0
1
xdx dy
4. RRREdV where E is: x2+y2+z2≤4,x≥0,y≥0,z≥0
5. RRD
1
x2+y2dA where D is: x2+y2≤4
A. cartesian coordinates
B. spherical coordinates
C. cylindrical coordinates
D. polar coordinates
Solution: 1. Since the region of integration does not involve
spheres or cylinders, it is best solved using cartesian coordinates
2. Since Eis the region between the cylinders of radius 1 and
√2, the integral is best solved using cylindrical coordinates
3. Since the region of integration of the double integral does
not involve circles, it is best solved in cartesian coordinates
4. Since the region of integration is bounded by the sphere of
radius 2, the integral is best solved using spherical coordinates
2
5. Since the region of integration of the double integral in-
volves circles, it is best solved using polar coordinates
Correct Answers:
•A
•C
•A
•B
•D
7. (1 point)
Use spherical coordinates to evaluate the triple integral
RRRE(x2+y2+z2)dV , where Eis the ball: x2+y2+z2≤1.
Solution:
SOLUTION
The region of integration is given in spherical coordinates by
E={(ρ,θ,φ)|0≤ρ≤1,0≤θ≤2π,0≤φ≤π}
ZZZEx2+y2+z2dV =Z2π
0Zπ
0Z1
0ρ2ρ2sinφdρdφdθ
=Z2π
0
dθZπ
0
sinφdφZ1
0
ρ4dρ
= [θ]2π
0[−cosφ]π
0ρ5
51
0
=4π15
5
Correct Answers:
•2.51327412287183
8. (1 point)
Use spherical coordinates to evaluate the triple integral
ZZZE
e−(x2+y2+z2)
px2+y2+z2dV ,
where Eis the region bounded by the spheres x2+y2+z2=1
and x2+y2+z2=9.
Solution:
SOLUTION
The region of integration is given in spherical coordinates by
E={(ρ,θ,φ)|1≤ρ≤3,0≤θ≤2π,0≤φ≤π}
This represents the region between the spheres ρ=1 and ρ=3.
ZZZE
e−(x2+y2+z2)
px2+y2+z2dV =Z2π
0Zπ
0Z3
1
e−ρ2
pρ2ρ2sinφdρdφdθ
=Z2π
0
dθZπ
0
sinφdφZ3
1
ρe−ρ2dρ
= [θ]2π
0[−cosφ]π
0−1
2e−ρ23
1
=2π(2)−1
2e−9+1
2e−1
=2πe−1−e−9
Correct Answers:
•2.31067929291404
9. (1 point)
Suppose the solid Win the figure is a cone centered
about the positive z-axis with its vertex at the origin, a
90◦angle at its vertex, and topped by a sphere radius 8.
Find the limits of integration for an iterated integral of
the form
ZZZ
W
dV =ZB
AZD
CZF
E
ρ2sin(φ)dρdφdθ.
A =
B =
C =
D =
E =
F =
If necessary, enter ρas rho, φas phi, and θas theta.
(Drag to rotate)
Solution:
SOLUTION
In spherical coordiantes, the cone has equation φ=π
4, while
the sphere has equation ρ=8. Thus
ZZZ
W
dV =Z2π
0Zπ/4
0Z8
0
ρ2sin(φ)dρdφdθ..
Correct Answers:
•0
•2*pi
•0
•pi/4
•0
•8
3
10. (1 point)
Suppose the solid Win the figure consists of the
points below the xy-plane that are between concentric
spheres centered at the origin of radii 4 and 10. Find the
limits of integration for an iterated integral of the form
ZZZ
W
f dV =ZB
AZD
CZF
E
f(ρ,φ,θ)ρ2sin(φ)dρdφdθ.
A =
B =
C =
D =
E =
F =
If necessary, enter ρas rho, φas phi, and θas theta.
(Drag to rotate)
Solution:
SOLUTION
ZZZ
W
f dV =Z2π
0Zπ
π/2Z10
4
f(ρ,φ,θ)ρ2sin(φ)dρdφdθ..
Correct Answers:
•0
•6.28319
•1.5708
•3.14159
•4
•10
11. (1 point) FInd the volume of the solid that lies within the
sphere x2+y2+z2=9, above the xy plane, and outside the cone
z=8px2+y2.
Solution:
SOLUTION
In spherical coordinates, the cone z=8px2+y2has equa-
tion ρcosφ=8ρsinφ⇒cosφ=8sinφ⇒tanφ=1
8.
The cone opens upwards so φ=arctan1
8.
Since the region of integration is outside the cone and above the
xy-plane, it must be arctan 1
8≤φ≤π
2.
Thus, in spherical coordinates, the solid is described by:
E=(ρ,θ,φ)|0≤ρ≤3,0≤θ≤2π,arctan1
8≤φ≤π
2
V=ZZE
dV =Z2π
0Z3
0Zπ
2
arctan(1
8)ρ2sinφdφdρdθ
=Z2π
0
dθZ3
0
ρ2dρZπ
2
arctan(1
8)sinφdφ
= [θ]2π
0ρ3
33
0
[−cosφ]
π
2
arctan(1
8)
=2π33
3cosarctan1
8
=2π33(8)
3√65
Correct Answers:
•56.1119919804601
12. (1 point)
Find an equation for the paraboloid z=x2+y2in spherical
coordinates. (Enter rho, phi and theta for ρ,φand θ, respec-
tively.)
equation:
Solution:
SOLUTION
The paraboloid has equation rho =cos(phi)
sin2(phi).
Correct Answers:
•rho = [cos(phi)]/([sin(phi)]ˆ2)
13. (1 point)
The region Wis the cone shown below.
The angle at the vertex is π/3, and the top is flat and at a height
of 5√3.
Write the limits of integration for RWdV in the following co-
ordinates (do not reduce the domain of integration by taking
advantage of symmetry):
(a) Cartesian:
With a=,b=,
c=,d=,
e=, and f=,
Volume = Rb
aRd
cRf
ed d d
(b) Cylindrical:
With a=,b=,
c=,d=,
e=, and f=,
Volume = Rb
aRd
cRf
ed d d
(c) Spherical:
With a=,b=,
4
c=,d=,
e=, and f=,
Volume = Rb
aRd
cRf
ed d d
Solution:
SOLUTION
(a) Since the cone has an angle of π/3 at its vertex, it has
equation
z=√3px2+y2.
The top of the cone, the plane with equation z=5√3, intersects
the cone in the circle x2+y2=25. Thus, in cartesian coordi-
nates we have
ZW
dV =Z5
−5Z√25−x2
−√25−x2Z5√3
√3√x2+y21dzdydx.
(b) In cylindrical coordinates the cone has equation z=√3r,
so the integral becomes
ZW
dV =Z2π
0Z5
0Z5√3
√3r
r dz dr dθ.
(c) In spherical coordinates, the cone has equation φ=π/6
and the plane z=5√3 is ρcosφ=5√3. Thus we have
ZW
dV =Z2π
0Zπ/6
0Z5√3/cosφ
0
ρ2sinφdρdφdθ.
Correct Answers:
•-5
•5
•-[sqrt(5ˆ2-xˆ2)]
•sqrt(5ˆ2-xˆ2)
•sqrt(3)*sqrt(xˆ2+yˆ2)
•8.66025
•1
•z
•y
•x
•0
•2*pi
•0
•5
•sqrt(3)*r
•8.66025
•r
•z
•r
•theta
•0
•2*pi
•0
•pi/6
•0
•8.66025/[cos(phi)]
•rhoˆ2*sin(phi)
•rho
•phi
•theta
14. (1 point) Evaluate the integral.
Z5
0Z√25−x2
−√25−x2Z√25−x2−z2
−√25−x2−z2
1
(x2+y2+z2)1/2dydzdx =
Solution:
SOLUTION
z=−p25 −x2−y2and z=p25 −x2−y2represent, respec-
tively, the lower and upper part of the sphere x2+y2+z2=25.
In spherical coordinates, the sphere has equation ρ=5.
y=−√25 −x2and y=√25 −x2represent, respectively, the
lower and upper part of the circle x2+y2=25, which is the
intersection of the sphere ρ=5 with the xy-plane.
Since 0 ≤x≤5, the region of integration consists of the hemi-
sphere ρ=5 with x>0.
Thus, in spherical coordintes, the integral becomes
Zπ/2
−π/2Zπ
0Z5
0
1
(ρ2)1/2ρ2sinφdρdφdθ=Zπ/2
−π/2Zπ
0Z5
0
ρsinφdρdφdθ
= [θ]π/2
−π/2ρ2
25
0
[−cosφ]π
0
=π52
2(2) = 25π
Correct Answers:
•25*pi
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