David Gamble Zhu MAT 267 ONLINE B Spring 2018
Assignment Section 12.2 due 04/08/2018 at 11:59pm MST
1. (1 point) Evaluate the iterated integral I=Z1
0Z1+x
1−x
(6x2+
6y)dydx
Solution:
SOLUTION
I=Z1
0Z1+x
1−x
(6x2+6y)dydx
=Z1
06x2y+3y2y=1+x
y=1−xdx
=Z1
0h6x2(1+x) + 3(1+x)2−6x2(1−x)−3(1−x)2idx
=Z1
012x3+12xdx
=3x4+6x21
0
=9
Correct Answers:
•9
2. (1 point)
Suppose Ris the shaded region in the figure, and
f(x,y)is a continuous function on R. Find the limits
of integration for the following iterated integrals.
(a) ZZ
R
f(x,y)dA =ZB
AZD
C
f(x,y)dydx
A =
B =
C =
D =
(b) ZZ
R
f(x,y)dA =ZF
EZH
G
f(x,y)dx dy
E =
F =
G =
H =
Solution:
SOLUTION
The region Rconsists of the points on or inside the triangle
with vertices (−2,−2),(4,−2),(−2,2).
(a) The region is bounded below by the line y=−2 and above
by the line through the points (4,−2),(−2,2), which has equa-
tion y+2=−2
3(x−4).
The bounds for xare −2≤x≤4. Thus
ZZR
f(x,y)dydx =Z4
−2Z−2
3(x−4)−2
−2
f(x,y)dydx
(b) The region is bounded on the left by the line x=−2 and on
the right by the line through the points (4,−2),(−2,2), which
has equation x=−3
2(y−2)−2.
The bounds for yare −2≤x≤2. Thus
ZZR
f(x,y)dydx =Z2
−2Z−3
2(y−2)−2
−2
f(x,y)dydx
Correct Answers:
•-2
•4
•-2
•2-0.666667*(x+2)
•-2
•2
•-2
•-[2+1.5*(y-2)]
3. (1 point) Find the volume of the solid bounded by the
planes x=0,y=0,z=0, and x+y+z=8.
Solution:
SOLUTION
1
The region D, intersection of the solid with the xy-plane, is
shown below.
The region is bounded below by y=0 and above by y=8−x,
with 0 ≤x≤8. Thus the volume is given by
V=Z8
0Z8−x
0
(8−x−y)dydx
=Z8
0(8−x)y−y2
2y=8−x
y=0
dx
=Z8
0(8−x)2−(8−x)2
2dx
=Z8
0
(8−x)2
2dx
=1
2Z8
0
(8−x)2dx
Using the substitution u=8−x du =−dx, yields
V=−1
2Z0
8
u2dx
=1
2u3
38
0
=83
6
Correct Answers:
•85.3333
4. (1 point) Consider the integral Z64
0Z8√x
0
f(x,y)dydx.
Sketch the region of integration and change the order of inte-
gration.
Zb
aZg2(y)
g1(y)
f(x,y)dxdy
a=b=
g1(y) = g2(y) =
Solution:
SOLUTION
The region of integration is shown below.
Because the region is
D={(x,y)|0≤x≤64,0≤y≤8√x}
=(x,y)|0≤y≤64,y2
64 ≤x≤64
we have
Z64
0Z8√x
0
f(x,y)dydx =Z64
0Z64
y2/64
f(x,y)dxdy
Thus a=0,b=64,g1(y) = y2
64 and g2(y) = 64.
Correct Answers:
•0
•64
•yˆ2/64
•64
5. (1 point) Consider the integral Z4
0Z√16−y
0
f(x,y)dxdy. If
we change the order of integration we obtain the sum of two
integrals:
Zb
aZg2(x)
g1(x)
f(x,y)dydx +Zd
cZg4(x)
g3(x)
f(x,y)dydx
a=b=
g1(x) = g2(x) =
c=d=
g3(x) = g4(x) =
Solution:
SOLUTION
The region of integration is shown below.
The point Ahas coordinates (√12,4)and the curve
connecting the point (4,0)to Ahas equation y=16 −x2.
Thus
Z4
0Z√16−y
0
f(x,y)dxdy =Z√12
0Z4
0
f(x,y)dydx+Z4
√12 Z16−x2
0
f(x,y)dydx
Correct Answers:
•0
•3.4641
•0
•4
•3.4641
•4
•0
•16-xˆ2
6. (1 point) Consider the integral Z16
1Z4lnx
0
f(x,y)dydx.
Sketch the region of integration and change the order of inte-
gration.
Zb
aZg2(y)
g1(y)
f(x,y)dxdy
a=b=
g1(y) = g2(y) =
Solution:
SOLUTION
The region of integration is shown below.
2
The point Ahas coordinates (16,4ln(16)).
Because the region is
D={(x,y)|1≤x≤16,0≤y≤4lnx}
=n(x,y)|0≤y≤4ln(16),ey/4≤x≤16o
we have
Z16
1Z4lnx
0
f(x,y)dydx =Z4ln(16)
0Z16
ey/4f(x,y)dxdy
Thus, a=0,b=4ln(16),g1(y) = ey/4and g2(y) = 16.
Correct Answers:
•0
•11.0904
•exp(y/4)
•16
7. (1 point) Evaluate the integral by reversing the order of
integration.
Z1
0Z4
4y
ex2dxdy =
Solution:
SOLUTION
The region of integration is shown below.
The region is bounded below by y=0 and above by y=x
4, with
0≤x≤4.
Thus
Z1
0Z4
4y
ex2dxdy =R4
0R
x
4
0ex2dydx
=R4
0hyex2iy=x
4
y=0dx
=R4
0
x
4ex2dx
Using the substitution u=x2,du =2xdx, yields
=1
8R16
0eudu
=1
8e16 −1
Correct Answers:
•1.11076E+06
8. (1 point)
Consider the following integral. Sketch its region of
integration in the xy-plane.
Z7
0Z49
y2ysinx2dx dy
(a) Which graph shows the region of integration in
the xy-plane? [?/A/B/C/D]
(b) Write the integral with the order of integration re-
versed:
Z7
0Z49
y2ysinx2dx dy =ZB
AZD
C
ysinx2dydx
with limits of integration
A =
B =
C =
D =
(c) Evaluate the integral.
A B
C D
(Click on a graph to enlarge it)
Solution:
SOLUTION
(a) The region is bounded on the left by the function x=y2
and on the right by the vertical line x=49. The bounds for y
are 0 ≤y≤7. Thus the region corresponds to graph A.
(b) The region is bounded below by y=0 and above by y=√x,
3
while 0 ≤x≤49. Thus
Z7
0Z49
y2ysinx2dx dy =Z49
0Z√x
0
ysin(x2)dydx
(c)
R49
0R√x
0ysin(x2)dydx =R49
0hy2
2i√x
0sin(x2)dx
=1
2R49
0xsin(x2)dx [substitution: u=x2,du =2xdx]
=1
4R2401
0sin(u)du
=1
4[−cos(u)]2401
0
=1−cos(2401)
4
Correct Answers:
•A
•0
•49
•0
•sqrt(x)
•[1-cos(7ˆ4)]/4
9. (1 point)
Consider the following integral. Sketch its region of
integration in the xy-plane.
Z0
−2Z0
−√4−x22xy dy dx
(a) Which graph shows the region of integration in
the xy-plane? [?/A/B/C/D]
(b) Evaluate the integral.
A B
C D
(Click on a graph to enlarge it)
Solution:
SOLUTION
(a) The region is bounded below by y=−√4−x2, which rep-
resents the lower part of a circle centered at the origin and of
radius 2, and above by y=0. Since −2≤x≤0, the region rep-
resents the quarter of a disk in the third quadrant and it matches
graph D.
(b)
Z0
−2Z0
−√4−x22xy dy dx =2Z0
−2
xy2
20
−√4−x2
dx
=−1Z0
−2
x(4−x2)dx
=−14x2
2−x4
40
−2
=−124
4−24
2
=4
Correct Answers:
•D
•4
10. (1 point) Set up a double integral in rectangular coordi-
nates for calculating the volume of the solid under the graph of
the function f(x,y) = 23 −x2−y2and above the plane z=7.
Instructions: Please enter the integrand in the first answer box.
Depending on the order of integration you choose, enter dx and
dy in either order into the second and third answer boxes with
only one dx or dy in each box. Then, enter the limits of integra-
tion.
ZB
AZD
C
A =
B =
C =
D =
Solution:
SOLUTION
The function f(x,y) = 23 −x2−y2intersects the plane z=7
when x2+y2=16.
Thus the region of integration is
D=n(x,y)| −4≤x≤4,−√16 −x2≤y≤√16 −x2o
=n(x,y)| −4≤y≤4,−p16 −y2≤x≤p16 −y2o.
The volume is then
V=R4
−4R√16−x2
−√16−x223 −x2−y2−7dydx or
V=R4
−4R√16−y2
−√16−y223 −x2−y2−7dx dy.
Correct Answers:
•16-xˆ2-yˆ2; dx; dy; -4; 4; -[sqrt(16-yˆ2)]; sqrt(16-yˆ2)
4
11. (1 point)
Suppose Ris the shaded region in the figure, and
f(x,y)is a continuous function on R. Find the limits
of integration for the following iterated integral.
(a) ZZ
R
f(x,y)dA =ZB
AZD
C
f(x,y)dydx
A =
B =
C =
D =
Solution:
SOLUTION
The region is bounded below by the line through the points
(−4,−2),(2,1). This line has equation y=1
2(x+4)−2.
The upper bound is the line y=2, while −4≤x≤2.
Thus
ZZR
f(x,y)dA =Z2
−4Z2
1
2(x+4)−2
f(x,y)dydx
Correct Answers:
•-4
•2
•0.5*(x+4)-2
•2
12. (1 point)
Consider the following integral. Sketch its region of
integration in the xy-plane.
Z3
0Ze3
ey
x
ln(x)dx dy
(a) Which graph shows the region of integration in
the xy-plane? [?/A/B/C/D]
(b) Write the integral with the order of integration re-
versed:
Z3
0Ze3
ey
x
ln(x)dx dy =ZB
AZD
C
x
ln(x)dydx
with limits of integration
A =
B =
C =
D =
(c) Evaluate the integral.
A B
C D
(Click on a graph to enlarge it)
Solution:
SOLUTION
The region is bounded on the left by the function x=eyor,
equivalently, y=ln x, and on the right by the vertical line x=e3.
The limits for yare 0 ≤y≤3. Thus
R=(x,y)|0≤y≤3,ey≤x≤e3
=(x,y)|1≤x≤e3,0≤y≤lnx
5
(a) The graph of this region is shown in figure B.
(b) Z3
0Ze3
ey
x
ln(x)dx dy =Ze3
1Zlnx
0
x
ln(x)dydx
(c) Ze3
1Zlnx
0
x
ln(x)dydx =Ze3
1
x dx =x2
2e3
1
=e6−1
2
Correct Answers:
•B
•1
•eˆ3
•0
•ln(x)
•[eˆ(2*3)-1]/2
13. (1 point) Find the volume of the region under the graph
of f(x,y) = 4x+y+1 and above the region y2≤x, 0 ≤x≤16.
volume =
Solution:
SOLUTION
The region of integration is shown below.
Thus,
Volume =Z4
−4Z16
y2(4x+y+1)dx dy =Z4
−4
(2(x2)+(y+1)x)
x=16
x=y2
dy
=Z4
−4
2(256 −y4)+(y+1)(16 −y2)dy.
Expanding the second binomial product, we have
Volume =Z4
−4
2(256 −y4)+(16 +16y−y2−y3)dy
=2(256y−y5
5) + 16(y+y2
2)−y3
3−y4
4
4
−4
=2(2048 −2048
5) + 16(8)−128
3=50432
15 .
Correct Answers:
•4*4*4ˆ5/5+4*4ˆ3/3
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