Austin Cholley Zhu MAT 267 ONLINE A Fall 2021
Assignment Section 12.5 due 09/19/2021 at 11:59pm MST
Problem 1. (1 point)
Evaluate ZZZB
zex+ydV where Bis the box determined by
0≤x≤3, 0 ≤y≤5, and 0 ≤z≤3.
The value is .
Solution:
SOLUTION
ZZZB
zex+ydV =Z3
0Z5
0Z3
0
zexeydz dy dx
=Z3
0
exdx Z5
0
eydy Z3
0
zdz
= [ex]3
0[ey]5
0z2
23
0
=e3−1e5−19
2
Answer(s) submitted:
•12660.56680
(correct)
Correct Answers:
•12660.5668095718
Problem 2. (1 point)
Evaluate the triple integral
ZZZE
xy dV where Eis the solid tetrahedon with vertices
(0,0,0),(3,0,0),(0,2,0),(0,0,3).
Solution:
SOLUTION
The plane thorough the points (3,0,0),(0,2,0),(0,0,3)has equa-
tion x
3+y
2+z
3=1. Thus
E=n(x,y,z)|0≤x≤3,0≤y≤21−x
3,0≤z≤31−x
3−y
2o
and
ZZZE
xy dV =Z3
0Z2(1−x
3)
0Z3(1−x
3−y
2)
0
xy dz dy dx
=Z3
0Z2(1−x
3)
0
xy 31−x
3−y
2dy dx
=Z3
0Z2(1−x
3)
03x1−x
3y−3xy2
2dy dx
=−Z3
0h3
2x1−x
3y2−x
2y3i2(1−x
3)
0
dx
=Z3
03
2x1−x
321−x
32−x
221−x
33dx
=Z3
0
6x1−x
33−4x1−x
33
dx
=Z3
0
2x1−x
33
dx
[ Using the substitution u=1−x
3,du =−1
3dx ]
=−18Z1
0
(1−u)u3du
=18Z1
0
(u3−u4)du
=0.9
Answer(s) submitted:
•.4
(incorrect)
Correct Answers:
•0.9
1
Problem 3. (1 point)
Evaluate the triple integral
ZZZE
x2eydV where Eis bounded by the parabolic cylinder z=
25 −y2and the planes z=0,x=5,and x=−5.
Solution:
SOLUTION
A picture of the solid is shown below.
We have
E=(x,y,z)| −5≤x≤5,−5≤y≤5,0≤z≤25 −y2.
Thus
ZZZE
x2eydV =Z5
−5Z5
−5Z25−y2
0
x2eydz dy dx
=Z5
−5Z5
−5
x2ey(25 −y2)dy dx
=Z5
−5
x2dx Z5
−5
ey(25 −y2)dy
=x3
35
−5Z5
−5
ey(25 −y2)dy
=253
3Z5
−5
ey(25 −y2)dy
Using integration by parts with u=25 −y2
,dv =eydy,du =
−2ydy,v=ey
,yields Rey(25 −y2)dy =ey(25 −y2) + 2Ryeydy.
Using integration by parts again, with u=y,dv =eydy,du =
dy,v=ey
,yields
Rey(25 −y2)dy =ey(25 −y2) + 2[yey−Reydy] = ey(25 −y2) +
2yey−2ey. So
ZZZE
x2eydV =253
3ey(25 −y2) + 2yey−2ey5
−5
=253
3h2(5)e5−2e5+2(5)e−5+2e−5i
=253
38e5+12e−5
Answer(s) submitted:
•98948.8440
(correct)
Correct Answers:
•98948.8440153835
Problem 4. (1 point)
Evaluate the triple integral
ZZZE
x dV where Eis the solid bounded by the paraboloid x=
5y2+5z2and x=5.
Solution:
SOLUTION
A picture of the solid with its projection on the yz-plane is given
below.
The projection of Eonto the yz- plane is the disk 5 =5y2+5z2or
y2+z2=1.
Using polar coordinates y=rcosθand z=rsinθ, we get
ZZZE
x dV =ZZDZ5
5y2+5z2x dxdA =1
2ZZD(5)2−(5y2+5z2)2dA
=25
2Z2π
0Z1
01−r4r drdθ=25
2[θ]2π
0Z1
0r−r5dr
=25πr2
2−r6
61
0
=25
3π
Answer(s) submitted:
•((125pi )/3)
(incorrect)
Correct Answers:
•26.1799
2
Problem 5. (1 point)
Evaluate the triple integral
ZZZE
zdV where Eis the solid bounded by the cylinder y2+z2=
144 and the planes x=0,y=3xand z=0 in the first octant.
Solution:
SOLUTION
A picture of the solid Eis given below.
ZZZE
zdV =Z4
0Z12
3xZ√144−y2
0
zdzdydx =Z4
0Z12
3x
1
2144 −y2dy dx
=1
2Z4
0144y−y3
3y=12
y=3x
dx =1
2Z4
01728 −576 −432x+9x3dx
=1
21152x−432 x2
2+9x4
44
0
=1
2(4608 −3456 +576) = 864
Answer(s) submitted:
•864
(correct)
Correct Answers:
•864
Problem 6. (1 point)
Use a triple integral to find the volume of the solid bounded by the
parabolic cylinder y=6x2and the planes z=0,z=4 and y=8.
Solution:
SOLUTION
V=Zq4
3
−q4
3Z8
6x2Z4
0
dz dy dx =Zq4
3
−q4
3Z8
6x24dy dx
=4Zq4
3
−q4
3
[y]y=8
y=6x2dx =4Zq4
3
−q4
38−6x2dx
=48x−6x3
3q4
3
−q4
3
=128
3r8
6
Answer(s) submitted:
•49.26722
(correct)
Correct Answers:
•49.2672
3
Problem 7. (1 point)
Find the volume of the solid enclosed by the paraboloids z=
16x2+y2and z=18 −16x2+y2.
Solution:
SOLUTION
The paraboloids intersect when 16x2+y2=18 −
16x2+y2⇒x2+y2=9
16 , thus the intersection is the circle
x2+y2=9
16 ,z=9.
The projection of Eonto the xy-plane is the disk x2+y2≤9
16 , so
E=(x,y,z)|x2+y2≤9
16 ,16x2+y2≤z≤18 −16x2+y2
Let D=(x,y)|x2+y2≤9
16 .
Then using polar coordinates x=rcosθand y=rsinθ,we have
V=ZZZE
dV =ZZDZ18−16(x2+y2)
16(x2+y2)dz dA =ZZD18 −32x2+y2dA
=Z2π
0Z3
4
018 −32r2r dr dθ=Z2π
0
dθZ3
4
018r−32r3dr
= [θ]2π
09r2−8r4
3
4
0=81
16 π
Answer(s) submitted:
•((81pi )/(16))
(correct)
Correct Answers:
•15.9043
4
Problem 8. (1 point)
Express the integral ZZZE
f(x,y,z)dV as an iterated integral in six
different ways, where E is the solid bounded by z=0,x=0,z=
y−6xand y=12.
1. Zb
aZg2(x)
g1(x)Zh2(x,y)
h1(x,y)
f(x,y,z)dzdydx
a=b=
g1(x) = g2(x) =
h1(x,y) = h2(x,y) =
2. Zb
aZg2(y)
g1(y)Zh2(x,y)
h1(x,y)
f(x,y,z)dzdxdy
a=b=
g1(y) = g2(y) =
h1(x,y) = h2(x,y) =
3. Zb
aZg2(z)
g1(z)Zh2(y,z)
h1(y,z)
f(x,y,z)dxdydz
a=b=
g1(z) = g2(z) =
h1(y,z) = h2(y,z) =
4. Zb
aZg2(y)
g1(y)Zh2(y,z)
h1(y,z)
f(x,y,z)dxdzdy
a=b=
g1(y) = g2(y) =
h1(y,z) = h2(y,z) =
5. Zb
aZg2(x)
g1(x)Zh2(x,z)
h1(x,z)
f(x,y,z)dydzdx
a=b=
g1(x) = g2(x) =
h1(x,z) = h2(x,z) =
6. Zb
aZg2(z)
g1(z)Zh2(x,z)
h1(x,z)
f(x,y,z)dydxdz
a=b=
g1(z) = g2(z) =
h1(x,z) = h2(x,z) =
Solution:
SOLUTION
1. Z2
0Z12
6xZy−6x
0
f(x,y,z)dz dy dx
2. Z12
0Zy
6
0Zy−6x
0
f(x,y,z)dz dx dy
3. Z12
0Z12
zZy−z
6
0
f(x,y,z)dx dy dz
4. Z12
0Zy
0Zy−z
6
0
f(x,y,z)dx dz dy
5. Z2
0Z12−6x
0Z12
6x+z
f(x,y,z)dy dz dx
6. Z12
0Z2−z
6
0Z12
6x+z
f(x,y,z)dy dx dz
Answer(s) submitted:
•0
•2
•6x
•12
•0
•y-6x
•0
•12
•0
•(y/6)
•0
•y-6x
•0
•12
•z
•12
5
•0
•(((y-z))/6)
•0
•12
•0
•y
•0
•(((y-z))/6)
•0
•2
•0
•12-6x
•6x+z
•12
•0
•12
•0
•2-(z/6)
•6x+z
•12
(correct)
Correct Answers:
•0
•2
•6*x
•2*6
•0
•y-6*x
•0
•12
•0
•y/6
•0
•y-6*x
•0
•12
•z
•2*6
•0
•(y-z)/6
•0
•12
•0
•y
•0
•(y-z)/6
•0
•2
•0
•2*6-6*x
•6*x+z
•2*6
•0
•12
•0
•2-z/6
•6*x+z
•2*6
6
Problem 9. (1 point)
Express the integral ZZZE
f(x,y,z)dV as an iterated integral in six
different ways, where E is the solid bounded by z=0,z=8yand
x2=49 −y.
1. Zb
aZg2(x)
g1(x)Zh2(x,y)
h1(x,y)
f(x,y,z)dzdydx
a=b=
g1(x) = g2(x) =
h1(x,y) = h2(x,y) =
2. Zb
aZg2(y)
g1(y)Zh2(x,y)
h1(x,y)
f(x,y,z)dzdxdy
a=b=
g1(y) = g2(y) =
h1(x,y) = h2(x,y) =
3. Zb
aZg2(z)
g1(z)Zh2(y,z)
h1(y,z)
f(x,y,z)dxdydz
a=b=
g1(z) = g2(z) =
h1(y,z) = h2(y,z) =
4. Zb
aZg2(y)
g1(y)Zh2(y,z)
h1(y,z)
f(x,y,z)dxdzdy
a=b=
g1(y) = g2(y) =
h1(y,z) = h2(y,z) =
5. Zb
aZg2(x)
g1(x)Zh2(x,z)
h1(x,z)
f(x,y,z)dydzdx
a=b=
g1(x) = g2(x) =
h1(x,z) = h2(x,z) =
6. Zb
aZg2(z)
g1(z)Zh2(x,z)
h1(x,z)
f(x,y,z)dydxdz
a=b=
g1(z) = g2(z) =
h1(x,z) = h2(x,z) =
Solution:
SOLUTION
1. Z7
−7Z49−x2
0Z8y
0
f(x,y,z)dz dy dx
2. Z49
0Z√49−y
−√49−yZ8y
0
f(x,y,z)dz dx dy
3. Z392
0Z49
z
8Z√49−y
−√49−y
f(x,y,z)dx dy dz
4. Z49
0Z8y
0Z√49−y
−√49−y
f(x,y,z)dx dz dy
5. Z7
−7Z8(49−x2)
0Z49−x2
z
8
f(x,y,z)dy dz dx
6. Z392
0Z√49−z
8
−√49−z
8Z49−x2
z
8
f(x,y,z)dy dx dz
Answer(s) submitted:
•-7
•7
•0
•49-xˆ(2)
•0
•8y
•0
•49
•-sqrt(49-y)
•sqrt(49-y)
•0
•8y
•0
•392
7
•(z/8)
•49
•-sqrt(49-y)
•sqrt(49-y)
•0
•49
•0
•8y
•-sqrt(49-y)
•sqrt(49-y)
•-7
•7
•0
•8(49-xˆ(2))
•(z/8)
•49-xˆ(2)
•0
•392
•-sqrt(49-(z/8))
•sqrt(49-(z/8))
•(z/8)
•49-xˆ(2)
(correct)
Correct Answers:
•-7
•7
•0
•49 - x**2
•0
•8*y
•0
•49
•-sqrt(49-y)
•sqrt(49-y)
•0
•8*y
•0
•392
•z/8
•49
•-sqrt(49-y)
•sqrt(49-y)
•0
•49
•0
•8*y
•-sqrt(49-y)
•sqrt(49-y)
•-7
•7
•0
•8*(49-xˆ2)
•z/8
•49 - xˆ2
•0
•392
•-sqrt(49-z/8)
•sqrt(49-z/8)
•z/8
•49-xˆ2
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