Pham, Quoc – Homework 9 – Due: Mar 27 2007, 3:00 am – Inst: Eric Katerman
1
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The due time is Central
time.
001
(part 1 of 1) 10 points
Evaluate the double integral
I
=
Z Z
A
2
dxdy
with
A
=
n
(
x, y
) : 4
≤
x
≤
6
,
3
≤
y
≤
6
o
by first identifying it as the volume of a solid.
1.
I
= 12
correct
2.
I
= 10
3.
I
= 8
4.
I
= 14
5.
I
= 6
Explanation:
The value of
I
is the volume of the solid
below the graph of
z
=
f
(
x, y
) = 2 and above
the region
A
=
n
(
x, y
) : 4
≤
x
≤
6
,
3
≤
y
≤
6
o
.
Since
A
is a rectangle, this solid is a box with
base
A
and height 2. Its volume, therefore, is
given by
length
×
width
×
height
= (6

4)
×
(6

3)
×
2
.
Consequently,
I
= 12
.
keywords: volume, double integral, rectangu
lar region, rectangular solid
002
(part 1 of 1) 10 points
Evaluate the iterated integral
I
=
Z
3
1
n
Z
4
0
2
(
x
+
y
)
2
dx
o
dy .
1.
I
= ln
3
2
2.
I
= 2 ln
15
7
correct
3.
I
= ln
15
7
4.
I
=
1
2
ln
3
2
5.
I
= 2 ln
3
2
6.
I
=
1
2
ln
15
7
Explanation:
Integrating the inner integral with respect
to
x
keeping
y
fixed, we see that
Z
4
0
2
(
x
+
y
)
2
dx
=
h

2
x
+
y
i
4
0
= 2
n
1
y

1
4 +
y
o
.
In this case
I
= 2
Z
3
1
n
1
y

1
4 +
y
o
dy
= 2
h
ln
y

ln(4 +
y
)
i
3
1
.
Consequently,
I
= 2 ln
‡
(3)(1 + 4)
(4 + 3)
·
= 2 ln
15
7
.
keywords: iterated integral, rational function,
log integral
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Pham, Quoc – Homework 9 – Due: Mar 27 2007, 3:00 am – Inst: Eric Katerman
2
003
(part 1 of 1) 10 points
Evaluate the iterated integral
I
=
Z
ln 4
0
ˆ
Z
ln 3
0
e
2
x

y
dx
!
dy .
1.
I
= 6
2.
I
= 3
correct
3.
I
= 4
4.
I
= 2
5.
I
= 5
Explanation:
Integrating with respect to
x
with
y
fixed,
we see that
Z
ln 3
0
e
2
x

y
dx
=
1
2
h
e
2
x

y
i
ln 3
0
=
1
2
‡
e
2 ln 3

y

e

y
·
=
‡
3
2

1
2
·
e

y
.
Thus
I
= 4
Z
ln 4
0
e

y
dy
=

4
h
e

y
i
ln 4
0
=

4
‡
e

ln 4

1
·
.
Consequently,
I
=

4
‡
1
4

1
·
= 3
.
keywords:
004
(part 1 of 1) 10 points
Determine the value of the double integral
I
=
Z Z
A
3
xy
2
9 +
x
2
dA
over the rectangle
A
=
n
(
x, y
) : 0
≤
x
≤
4
,

2
≤
y
≤
2
o
,
integrating first with respect to
y
.
1.
I
= 8 ln
‡
25
9
·
correct
2.
I
= 8 ln
‡
9
25
·
3.
I
= 8 ln
‡
25
18
·
4.
I
= 4 ln
‡
25
18
·
5.
I
= 4 ln
‡
9
25
·
6.
I
= 4 ln
‡
25
9
·
Explanation:
The double integral over the rectangle
A
can be represented as the iterated integral
I
=
Z
4
0
Z
2

2
3
xy
2
9 +
x
2
dy
¶
dx ,
integrating first with respect to
y
. Now after
integration with respect to
y
with
x
fixed, we
see that
Z
2

2
3
xy
2
9 +
x
2
dy
=
h
xy
3
9 +
x
2
i
2

2
=
16
x
9 +
x
2
.
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 Spring '08
 RAdin
 2 g, Pham, Eric Katerman

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