M408L hwk9 solutions - Pham Quoc Homework 9 Due 3:00 am...

Pham, Quoc – Homework 9 – Due: Mar 27 2007, 3:00 am – Inst: Eric Katerman 1
This print-out should have 15 questions.
Multiple-choice questions may continue on
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before answering. 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

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
.