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Unformatted text preview: 3 }. when
A= (x, y ) : 0 ≤ x ≤ 1, 0≤y≤1 . 1. I = 7 5
9
ln 3 +
2
2 2. I = 7 5
9
ln 3 −
2
2 3. I = 7 9
ln 3 − 5
2 11
4. I =
π
6 4. I = 7 2
5
ln 3 −
9
2 5
5. I = π
3 5. I = 7 3 ln 3 − 1. I = 3
π
2 4
2. I = π correct
3
3. I = 2π (x, y ) : 0 ≤ x ≤ 1, 0≤y≤1 is a rectangle with sides parallel to the coordinate axes, the double integral can be represented as the iterated integral
1 1 I=
0 0 5 + x2
dx dy .
1 + y2 Now
1
0 5
2 Explanation:
Since the area of the rectangle A is 6, the
average value of f over A is given by Explanation:
Since
A= correct 1
1
5 + x2
dx =
5 x + x3
2
2
1+y
1+y
3 I= 0 7(x + y ) ln x dxdy.
A The integral with respect to y can be carried
out immediately, whereas the integral with respect to x would require integration by parts,
so this suggests that I should be written as a
repeated integral 1 . 1
6 I= 7
6 3 3 (x + y ) ln x dy dx,
1 0 rogers (grr459) – HW13 – rusin – (55220)
integrating ﬁrst with respect to y . After integration the inner integral becomes Consequently, 3
1
9
xy + y 2 ln x = 3x ln x + ln x.
2
2
0
Now is the time to integrate by parts, for then
3 3x ln x +
1 9
ln x dx
2 volume = 7 cu. units . keywords: double integral, volume of solid,
integral over rectangle, volume under plane 3
9
9
32
x ln x − x2 + x ln x − x
2
4
2
2 = 4 3
1 008 . 10.0 points Evaluate the iterated integral Thus
9
5
ln 3 −
2
2 I=7 . 1 2 I= y
0 007 16 − x2 dx dy . 10.0 points Find the volume of the solid lying under the
plane
z = 7 − x − 2y
and above the rectangle
A= 0 (x, y ) : 1 ≤ x ≤ 2 , 0≤y≤2 . 1. volume = 5 cu. units
2. volume = 7 cu. units correct
3. volume = 4 cu. units
4. volume = 8 cu. units
5. volume = 6 cu. units
Explanation:
The volume of the solid is given by the
double integral π1
+
32
√
π
3
2. I = 8
+
3
2 1. I = 2 π1
−
32
√
π
3
4. I = 2
correct
+
3
2
√
π
3
5. I = 2
−
3
2
√
π
3
6. I = 8
−
3
2
3. I = 8 Explanation:
To integrate
2 V=
A y (7 − x − 2y ) dxdy . 0 As a repeated integral this becomes
2
1 0 (7 − x − 2y ) dy dx 2 =
1 7y − xy − y 2 1 (10 − 2x) dx = dx = 4 cos u du , 2
0 dx 2 = with respect to x while keeping y ﬁxed we use
the substitution x = 4 sin u. For then 2 V= 16 − x2 dx 10x − x2 and
x=0 =⇒ u=0 x=2 =⇒ u= 2
1 . π
.
6 rogers (grr459) – HW13 – rusin – (55220) 5 Consequently, Thus
π /6 2 16 − x2 dx = 4y y
0 4 cos2 u du 31
1
.
−=
42
4 I= 0 π /6 = 8y (1 + cos 2u) du
010 0 since Find the value of the integral cos 2u = 2 cos2 u − 1 . Consequently, I= 1 π /6 1
sin 2u
dy
2
0
0
√
3 121
π
,
=8
+
y
0
6
4
2 I=8 y u+ and so √
3
π
+
I=2
3
2
009 A (x, y ) : 0 ≤ y ≤ x − 4, 4 ≤ x ≤ 6 .
22
3 2. I = 23
3 Evaluate the integral
x2 I=
0 0 1. I = (3x − 5y ) dydx . {4(x − 4) − 3y } dxdy when A is the region 1. I = . 10.0 points 1 10.0 points 3. I = 7
4. I = 8 3
20 20
correct
3 5. I = 7
2. I =
20 Explanation:
The integral can be written as the repeated
integral 3. I = 1
20 4. I = 1
correct
4 x −4 6 I= 9
5. I =
20 4 0 {4(x − 4) − 3y } dy dx. Now Explanation:
After integration with respect to y we see
that
1 5
3xy − y 2
2 I=
0 1 =
0 = x2
0 {4(x − 4) − 3y } dy 0 dx = 5
3 x3 − x4 d x
2
34 15
x− x
4
2 x −4 0 . x −4
0 . Consequently
6 1 3
4(x − 4)y − y 2
2 I=
4 5
(x − 4)2 dx =
2 5
(x − 4)3
6 6
4 , rogers (grr459) – HW13 – rusin – (55220) 6
y and so
I= 011 20
3 4 . 10.0 points Reverse the order of integration in the integral
x
2x 2 2 f (x, y ) dy dx , I=
0 0 but make no attempt to evaluate either integral.
2 4 1. I = f (x, y ) dx dy
y 0 Integration is taken ﬁrst with respect to y for
ﬁxed x along the dashed vertical line.
To change the order of integration, now ﬁx
y and let x vary along the solid horizontal line
in
y y 4 2. I = f (x, y ) dx dy
0 4 2
4 2 3. I = f (x, y ) dx dy correct
y/2 0 2y 2 4. I = f (x, y ) dx dy
0 0
y /2 4 5. I = x f (x, y ) dx dy
0
2 4 6. I = f (x, y ) dx dy
0 2 0 2y Explanation:
The region of integration is the set of all
points
(x, y ) : 0 ≤ y ≤ 2x , 0 ≤ x ≤ 2
in the plane bounded by the xaxis and the
graphs of
y = 2x , x = 2. This is the shaded region in Since the equation of the slant line is x = y/2,
integration in x is along the line from (y/2, y )
to (2, y ) for ﬁxed y , and then from y = 0 to
y = 4.
Consequently, after changing the order of
integration,
4 2 I= f (x, y ) dx dy
0 . y/2 keywords: double integral, reverse order integration, linear function,...
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This note was uploaded on 09/22/2013 for the course M 408 D taught by Professor Textbookanswers during the Fall '07 term at University of Texas at Austin.
 Fall '07
 TextbookAnswers

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