chester (crc2876) – HW13 – meth – (91845)
1
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001
10.0 points
Evaluate the double integral
I
=
i i
A
3
dxdy
with
A
=
b
(
x, y
) : 2
≤
x
≤
7
,
2
≤
y
≤
4
B
by frst identiFying it as the volume oF a solid.
1.
I
= 36
2.
I
= 34
3.
I
= 28
4.
I
= 32
5.
I
= 30
correct
Explanation:
The value oF
I
is the volume oF the solid
below the graph oF
z
=
f
(
x, y
) = 3 and above
the region
A
=
b
(
x, y
) : 2
≤
x
≤
7
,
2
≤
y
≤
4
B
.
Since
A
is a rectangle, this solid is a box with
base
A
and height 3. Its volume, thereFore, is
given by
length
×
width
×
height
= (7

2)
×
(4

2)
×
3
.
Consequently,
I
= 30
.
keywords: volume, double integral, rectangu
lar region, rectangular solid
002
10.0 points
The graph oF the Function
z
=
f
(
x, y
) = 6

x
is the plane shown in
z
6
x
y
Determine the value oF the double integral
I
=
i i
A
f
(
x, y
)
dxdy
over the region
A
=
b
(
x, y
) : 0
≤
x
≤
6
,
0
≤
y
≤
4
B
in the
xy
plane by frst identiFying it as the
volume oF a solid below the graph oF
f
.
1.
I
= 71 cu. units
2.
I
= 72 cu. units
correct
3.
I
= 70 cu. units
4.
I
= 68 cu. units
5.
I
= 69 cu. units
Explanation:
The double integral
I
=
i i
A
f
(
x, y
)
dxdy
is the volume oF the solid below the graph oF
f
having the rectangle
A
=
b
(
x, y
) : 0
≤
x
≤
6
,
0
≤
y
≤
4
B
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2
for its base. Thus the solid is the wedge
z
6
6
x
y
(6
,
4)
and so its volume is the area of triangular
face multiplied by the thickness of the wedge.
Consequently,
I
= 72 cu. units
.
keywords: double integral, linear function,
volume under graph, volume, rectangular re
gion, prism, triangle
003
10.0 points
Find the value of the integral
I
=
i
2
0
f
(
x, y
)
dx
when
f
(
x, y
) =
x

4
x
2
y .
1.
I
= 2 +
32
3
y
2
2.
I
= 2

4
y
3.
I
=
y

4
y
2
4.
I
=
y

16
y
2
5.
I
= 2 +
32
3
y
6.
I
= 2

32
3
y
correct
Explanation:
Since
x
varies but
y
does not,
I
=
i
2
0
(
x

4
x
2
y
)
dx
=
b
1
2
x
2

4
3
x
3
y
B
2
0
.
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 Summer '10
 KUSHNER
 Multivariable Calculus

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