Daniel Mirandoli Ruedemann MAT 267 ONLINE B Summer 2019
Assignment Section 12.6 due 08/12/2019 at 11:59pm MST
1. (1 point) What are the rectangular coordinates of the point
whose cylindrical coordinates are
(r=2,θ=4
3π,z=−5)?
x=
y=
z=
Solution:
SOLUTION
x=rcosθ=2cos4
3π
y=rsinθ=2sin4
3π
z=−5
Correct Answers:
•-1
•-1.73205
•-5
2. (1 point)
What are the rectangular coordinates of the point whose
cylindrical coordinates
are (r=0,θ=1.8,z=7)?
x=
y=
z=
Solution:
SOLUTION
x=rcosθ=0cos(1.8)
y=rsinθ=0sin(1.8)
z=7
Correct Answers:
•0
•0
•7
3. (1 point) What are the cylindrical coordinates of the point
whose rectangular coordinates are (x=4,y=5,z=4)?
r=
θ=
z=
Solution:
SOLUTION
r2=x2+y2=42+52=41,so r=√41
The point (4,5)is in the first quadrant of the xy-plane, so
θ=arctany
x=arctan5
4
z=4
Thus, one set of cylindrical coordinates is
√41,arctan5
4,4
Correct Answers:
•6.40312
•0.896055
•4
4. (1 point) What are the cylindrical coordinates of the point
whose rectangular coordinates are (x=−2,y=4,z=−4)?
r=
θ=
z=
Solution:
SOLUTION
r2=x2+y2= (−2)2+ (4)2=20,so r=√20
The point (−2,4)is in the second quadrant of the xy-plane, so
θ=arctany
x+π=arctan(−2) + π
z=−4
Thus, one set of cylindrical coordinates is
√20,arctan(−2) + π,−4
Correct Answers:
•4.47213595499958
•2.0344439357957
•-4
5. (1 point) Express the point given in Cartesian coordinates
in cylindrical coordinates (r,θ,z).
A) 3√2
2,3√2
2,8=( , , )
B)−3√2
2,3√2
2,8=(, , )
C)3√2
2,−3√2
2,8=(, , )
D)−3√2
2,−3√2
2,8=(, , )
Solution:
SOLUTION
(A) r2=3√2
22+3√2
22=32so r=3.
The point 3√2
2,3√2
2is in the first quadrant of the xy-
plane, so θ=arctan3√2
2/3√2
2=arctan(1) = π/4
z=8.
Thus one set of cylindrical coordinates is (3,π/4,8).
(B) r2=−3√2
22+3√2
22=32so r=3.
The point −3√2
2,3√2
2 is in the second quadrant
of the xy-plane, so θ=arctan3√2
2/−3√2
2+π=
arctan(−1) + π=−π/4+π=3π/4
1
z=8.
Thus one set of cylindrical coordinates is (3,3π/4,8).
(C) r2=3√2
22+−3√2
22=32so r=3.
The point 3√2
2,−3√2
2 is in the fourth quadrant of
the xy-plane, so θ=arctan−3√2
2/3√2
2+2π=
arctan(−1) + 2π=−π/4+2π=7π/4
z=8.
Thus one set of cylindrical coordinates is (3,7π/4,8).
(D) r2=−3√2
22+−3√2
22=32so r=3.
The point −3√2
2,−3√2
2 is in the third quadrant of
the xy-plane, so θ=arctan−3√2
2/−3√2
2+π=
arctan(1) + π=π/4+π=5π/4
z=8.
Thus one set of cylindrical coordinates is (3,5π/4,8).
Correct Answers:
•3
•0.785398
•8
•3
•2.35619
•8
•3
•5.49779
•8
•3
•3.92699
•8
6. (1 point) Use cylindrical coordinates to evaluate the triple
integral RRREpx2+y2dV , where Eis the solid bounded by the
circular paraboloid z=9−4x2+y2and the xy-plane.
Solution:
SOLUTION
The paraboloid z=9−4x2+y2=9−4r2intersects the
xy-plane in the circle x2+y2=9
4or r2=9
4⇒r=3
2, so in
cylindrical coordinates, Eis given by
E=(r,θ,z)|0≤θ≤2π,0≤r≤3
2,0≤z≤9−4r2.
Thus
ZZZEpx2+y2dV =Z2π
0Z3
2
0Z9−4r2
0
√r2r dz dr dθ
=Z2π
0
dθZ3
2
0Z9−4r2
0
r2dz dr
=2πZ3
2
0
r2(9−4r2)dr
=2πZ3
2
0
(9r2−4r4)dr
=2π9r3
3−4r5
5
3
2
0
=81
10 π
Correct Answers:
•25.4469
7. (1 point) Find the volume of the solid enclosed by the
paraboloids z=9x2+y2and z=32 −9x2+y2.
Solution:
SOLUTION
The two paraboloids intersect when
9x2+y2=32 −9x2+y2or x2+y2=16
9. So
V=ZZx2+y2≤16
932 −9x2+y2−9x2+y2dA
=Z2π
0Z4
3
0
(32 −18r2)r dr dθ
=Z2π
0
dθZ4
3
0
(32r−18r3)dr
= [θ]2π
016r2−9
2r4
4
3
0
=2π16(4
3)2−9
2(4
3)4
=2π128
9=256
9π
Correct Answers:
•89.3609
8. (1 point) Find the volume of the ellipsoid x2+y2+9z2=
36.
Solution:
SOLUTION
In cylindrical coordinates, the ellipsoid has equation r2+
9z2=36 and it intersects the xy-plane in the circle r2=36 or
r=6, so the solid is given by
E=((r,θ,z)|0≤r≤6,0≤θ≤2π,−√36 −r2
√9≤z≤√36 −r2
√9)
Thus the volume is given by
ZZZ dV =Z2π
0Z6
0Z√36−r2
√9
−√36−r2
√9
r dz dr dθ=4
√9
πZ6
0
rp36 −r2dr
Using the substitution u=36 −r2,du =−2rdr, yields
ZZZ dV =−2
√9
πZ0
36
√udu =2
√9
π2
3u3/236
0
=4(6)3
3√9
π
Correct Answers:
•301.59289474462
9. (1 point)
Find an equation for the plane y=4 in cylindrical coordi-
nates. (Type theta for θin your answer.)
equation:
Solution:
SOLUTION
The plane has equation r=4
sin(theta).
Correct Answers:
•r = 4/[sin(theta)]
2
10. (1 point) Suppose f(x,y,z) = x2+y2+z2and Wis the
solid cylinder with height 9 and base radius 2 that is centered
about the z-axis with its base at z=−2. Enter θas theta.
(a) As an iterated integral,
ZZZ
W
f dV =ZB
AZD
CZF
E
dz dr dθ
with limits of integration
A =
B =
C =
D =
E =
F =
(b) Evaluate the integral.
Solution:
SOLUTION
W={(r,θ,z)|0≤r≤2,0≤θ≤2π,−2≤z≤7}.
f(r,θ,z) = r2+z2.
(a)
ZZZW
f dV =Z2π
0Z2
0Z7
−2
(r2+z2)r dz dr dθ
(b)
Z2π
0Z2
0Z7
−2
(r2+z2)r dz dr dθ= [θ]2π
0Z2
0r3z+rz3
37
−2
dr
=2πZ2
07r3+343
3r+2r3+8
3rdr
=2π9r4
4+117 r2
22
0
=540π
Correct Answers:
•(rˆ2+zˆ2)*r
•0
•2*pi
•0
•2
•-2
•7
•pi*3240/6
11. (1 point)
Suppose the solid Win the figure is one-quarter of a
circular cylinder of height 4 and radius 8 centered about
the z-axis in the first octant. Find the limits of integration
for an iterated integral of the form
ZZZ
W
f dV =ZB
AZD
CZF
E
f(r,θ,z)dzrdr dθ.
A =
B =
C =
D =
E =
F =
If necessary, enter θas theta.
(Drag to rotate)
Solution:
SOLUTION
ZZZ
W
f dV =Zπ/2
0Z8
0Z4
0
f(r,θ,z)dzrdr dθ.
Correct Answers:
•0
•pi/2
•0
•8
•0
•4
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