Unformatted text preview: nanni (arn437) – Assignment 18 – guntel – (54940)
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001 and above the rectangle
A= (x, y ) : 1 ≤ x ≤ 2 , 0 ≤ y ≤ 2 . 1. volume = 62 cu.units correct 10.0 points 2. volume = 69 cu.units Find the value of the integral 3. volume = 61 cu.units 2 I= 1 f (x, y ) dx
0 4. volume = 67 cu.units when
5. volume = 60 cu.units f (x, y ) = 4x + x2 y . Explanation:
The volume is given by the double integral 8
1. I = 8 − y 2
3 2 (2 + 9x2 + 8y ) dxdy V= 2. I = 8 + y 0 3. I = 4y + 4y 2 2 V= 2 2 9x2 dxdy 2 dxdy +
0 1 0
2 8
5. I = 8 − y
3 1 2 + 8y dxdy = 4 + 42 + 16.
0 1 Consequently, 6. I = 4y + y 2 volume = 62 cu.units . Explanation:
Since x varies but y does not,
2 4 x + x2 y d x =
0 1 of f (x, y ) over the rectangular region A. Integrating each term separately, we see that 2 8
4. I = 8 + y correct
3 I= 2 1
2 x2 + x3 y
3 003 2
0 . 10.0 points Evaluate the integral Consequently, xexy dxdy I=
A 8
I = 8+ y .
3 over the rectangle
A = { (x, y ) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 3 }. keywords:
1. I = 13
e − 4 correct
3 Find the volume of the solid under the
graph of 2. I = 13
e −4
6 f (x, y ) = 2 + 9x2 + 8y 3. I = 13
e −3
3 002 10.0 points nanni (arn437) – Assignment 18 – guntel – (54940) 2 13
8 4. I = 13
e −3
6 2. I = ln 5. I = 13
e −2
6 3. I = 13
1
ln
2
4 6. I = 13
e −2
3 4. I = 1
4
ln
2
13 5. I = ln xexy dxdy I= 13
correct
4 6. I = ln Explanation:
Since the integral with respect to y in 4
13 A can be evaluated easily using substitution
(or directly making the substitution in one’s
head), while the integral with respect to x requires integration by parts, this suggests that
we should represent the double integral as the
repeated integral
1 I= xe
0 3 dy dx. 0 Now after integration the inner integral becomes
3
exy
= e3x − 1 .
0 0 1
−1 3 (e 3xy 2
dy dx ,
4 + x2 xy 3
4 + x2 1
−1 − 1) dx = 0 e3x
3 1 −x 0 , 0 2x
dx = ln(4 + x2 )
4 + x2 Consequently, and so
I=
004 13
e −4
3 . I = ln 10.0 points Determine the value of the double integral
I=
A 3xy 2
dA
4 + x2 over the rectangle
A= 1. I = (x, y ) : 0 ≤ x ≤ 3, −1 ≤ y ≤ 1
13
1
ln
2
8 = 2x
.
4 + x2 But 1 I= −1 3xy 2
dy =
4 + x2 Thus
3x 1 I= integrating ﬁrst with respect to y . Now after
integration with respect to y with x ﬁxed, we
see that 3
xy Explanation:
The double integral over the rectangle A
can be represented as the iterated integral . 13
4 . 3
0 . ...
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This note was uploaded on 11/02/2011 for the course M 408N taught by Professor Gualdini during the Fall '10 term at University of Texas.
 Fall '10
 Gualdini
 Calculus

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