Daniel Mirandoli Ruedemann MAT 267 ONLINE B Summer 2019
Assignment Section 12.5 due 08/12/2019 at 11:59pm MST
1. (1 point) Evaluate ZZZB
zex+ydV where Bis the box deter-
mined by
0≤x≤2, 0 ≤y≤5, and 0 ≤z≤1.
The value is .
Solution:
SOLUTION
ZZZB
zex+ydV =Z2
0Z5
0Z1
0
zexeydz dy dx
=Z2
0
exdx Z5
0
eydy Z1
0
zdz
= [ex]2
0[ey]5
0z2
21
0
=e2−1e5−11
2
Correct Answers:
•470.915471613476
2. (1 point) Evaluate the triple integral
ZZZE
xy dV where Eis the solid tetrahedon with vertices
(0,0,0),(10,0,0),(0,7,0),(0,0,1).
Solution:
SOLUTION
The plane thorough the points (10,0,0),(0,7,0),(0,0,1)has
equation x
10 +y
7+z=1. Thus
E=n(x,y,z)|0≤x≤10,0≤y≤71−x
10 ,0≤z≤1−x
10 −y
7o
and
ZZZE
xy dV =Z10
0Z7(1−x
10 )
0Z1−x
10 −y
7
0
xy dz dy dx
=Z10
0Z7(1−x
10 )
0
xy 1−x
10 −y
7dy dx
=Z10
0Z7(1−x
10 )
01x1−x
10 y−xy2
7dy dx
=−Z10
0h1
2x1−x
10 y2−1
3
x
7y3i7(1−x
10 )
0
dx
=Z10
01
2x1−x
10 71−x
10 2−1
3
x
771−x
10 3dx
=Z10
0
49
2x1−x
10 3−49
3x1−x
10 3
dx
=Z10
0
49
6x1−x
10 3
dx
[ Using the substitution u=1−x
10 ,du =−1
10 dx ]
=−2450
3Z1
0
(1−u)u3du
=2450
3Z1
0
(u3−u4)du
=245
6
Correct Answers:
•40.8333
3. (1 point) Evaluate the triple integral
ZZZE
x8eydV where Eis bounded by the parabolic cylinder
z=49 −y2and the planes z=0,x=7,and x=−7.
Solution:
SOLUTION
A picture of the solid is shown below.
We have
E=(x,y,z)| −7≤x≤7,−7≤y≤7,0≤z≤49 −y2.
Thus
ZZZE
x8eydV =Z7
−7Z7
−7Z49−y2
0
x8eydz dy dx
=Z7
−7Z7
−7
x8ey(49 −y2)dy dx
=Z7
−7
x8dx Z7
−7
ey(49 −y2)dy
=x9
97
−7Z7
−7
ey(49 −y2)dy
=279
9Z7
−7
ey(49 −y2)dy
Using integration by parts with u=49 −y2
,dv =eydy,du =
−2ydy,v=ey
,yields Rey(49−y2)dy =ey(49−y2)+2Ryeydy.
Using integration by parts again, with u=y,dv =eydy,du =
dy,v=ey
,yields
Rey(49 −y2)dy =ey(49 −y2)+2[yey−Reydy] = ey(49 −y2)+
1
2yey−2ey. So
ZZZE
x8eydV =279
9ey(49 −y2) + 2yey−2ey7
−7
=279
92(7)e7−2e7+2(7)e−7+2e−7
=279
912e7+16e−7
Correct Answers:
•118008406832.069
4. (1 point) Evaluate the triple integral
ZZZE
x dV where Eis the solid bounded by the paraboloid
x=6y2+6z2and x=6.
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 6 =6y2+6z2
or y2+z2=1.
Using polar coordinates y=rcosθand z=rsin θ, we get
ZZZE
x dV =ZZDZ6
6y2+6z2x dxdA =1
2ZZD(6)2−(6y2+6z2)2dA
=18Z2π
0Z1
01−r4r drdθ=18 [θ]2π
0Z1
0r−r5dr
=36πr2
2−r6
61
0
=12π
Correct Answers:
•37.6991
5. (1 point) Evaluate the triple integral
ZZZE
zdV where Eis the solid bounded by the cylinder
y2+z2=225 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 =Z5
0Z15
3xZ√225−y2
0
zdzdydx =Z5
0Z15
3x
1
2225 −y2dy dx
=1
2Z5
0225y−y3
3y=15
y=3x
dx =1
2Z5
03375 −1125 −675x+9x3dx
=1
22250x−675 x2
2+9x4
45
0
=1
211250 −16875
2+5625
4=16875
8
Correct Answers:
•2109.38
6. (1 point) Use a triple integral to find the volume of the
solid bounded by the parabolic cylinder y=9x2and the planes
z=0,z=8 and y=11.
Solution:
SOLUTION
V=Zq11
9
−q11
9Z11
9x2Z8
0
dz dy dx =Zq11
9
−q11
9Z11
9x28dy dx
=8Zq11
9
−q11
9
[y]y=11
y=9x2dx =8Zq11
9
−q11
911 −9x2dx
=811x−9x3
3q11
9
−q11
9
=352
3r11
9
Correct Answers:
•129.717
2
7. (1 point) Find the volume of the solid enclosed by the
paraboloids z=25x2+y2and z=18 −25 x2+y2.
Solution:
SOLUTION
The paraboloids intersect when 25x2+y2=18 −
25x2+y2⇒x2+y2=9
25 , thus the intersection is the cir-
cle x2+y2=9
25 ,z=9.
The projection of Eonto the xy-plane is the disk x2+y2≤9
25 ,
so
E=(x,y,z)|x2+y2≤9
25 ,25x2+y2≤z≤18 −25 x2+y2
Let D=(x,y)|x2+y2≤9
25 .
Then using polar coordinates x=rcosθand y=rsin θ,we have
V=ZZZE
dV =ZZDZ18−25(x2+y2)
25(x2+y2)dz dA =ZZD18 −50x2+y2dA
=Z2π
0Z3
5
018 −50r2r dr dθ=Z2π
0
dθZ3
5
018r−50r3dr
= [θ]2π
09r2−25
2r4
3
5
0=81
25 π
Correct Answers:
•10.1788
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−3xand 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. Z4
0Z12
3xZy−3x
0
f(x,y,z)dz dy dx
2. Z12
0Zy
3
0Zy−3x
0
f(x,y,z)dz dx dy
3. Z12
0Z12
zZy−z
3
0
f(x,y,z)dx dy dz
4. Z12
0Zy
0Zy−z
3
0
f(x,y,z)dx dz dy
3
5. Z4
0Z12−3x
0Z12
3x+z
f(x,y,z)dy dz dx
6. Z12
0Z4−z
3
0Z12
3x+z
f(x,y,z)dy dx dz
Correct Answers:
•0
•4
•3*x
•4*3
•0
•y-3*x
•0
•12
•0
•y/3
•0
•y-3*x
•0
•12
•z
•4*3
•0
•(y-z)/3
•0
•12
•0
•y
•0
•(y-z)/3
•0
•4
•0
•4*3-3*x
•3*x+z
•4*3
•0
•12
•0
•4-z/3
•3*x+z
•4*3
9. (1 point) Express the integral ZZZE
f(x,y,z)dV as an iter-
ated integral in six different ways, where E is the solid bounded
by z=0,z=7yand x2=64 −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. Z8
−8Z64−x2
0Z7y
0
f(x,y,z)dz dy dx
2. Z64
0Z√64−y
−√64−yZ7y
0
f(x,y,z)dz dx dy
3. Z448
0Z64
z
7Z√64−y
−√64−y
f(x,y,z)dx dy dz
4
4. Z64
0Z7y
0Z√64−y
−√64−y
f(x,y,z)dx dz dy
5. Z8
−8Z7(64−x2)
0Z64−x2
z
7
f(x,y,z)dy dz dx
6. Z448
0Z√64−z
7
−√64−z
7Z64−x2
z
7
f(x,y,z)dy dx dz
Correct Answers:
•-8
•8
•0
•64 - x**2
•0
•7*y
•0
•64
•-sqrt(64-y)
•sqrt(64-y)
•0
•7*y
•0
•448
•z/7
•64
•-sqrt(64-y)
•sqrt(64-y)
•0
•64
•0
•7*y
•-sqrt(64-y)
•sqrt(64-y)
•-8
•8
•0
•7*(64-xˆ2)
•z/7
•64 - xˆ2
•0
•448
•-sqrt(64-z/7)
•sqrt(64-z/7)
•z/7
•64-xˆ2
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