**Unformatted text preview: **liang (xl5432) – HW13 – berg – (55120)
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001 10.0 points 1 is the volume of the solid below the graph of
f having the rectangle
A = (x, y) : 0 ≤ x ≤ 7, 0 ≤ y ≤ 6 for its base. Thus the solid is the wedge The graph of the function
z z = f (x, y) = 7 − x 7 is the plane shown in y
z
(7, 6)
7 y x and so its volume is the area of triangular
face multiplied by the thickness of the wedge.
Consequently,
7 x I = 147 cu. units . Determine the value of the double integral
f (x, y) dxdy I =
A over the region
A = (x, y) : 0 ≤ x ≤ 7, 0 ≤ y ≤ 6 in the xy-plane by ﬁrst identifying it as the
volume of a solid below the graph of f . keywords: double integral, linear function,
volume under graph, volume, rectangular region, prism, triangle
002 Determine the value of the iterated integral
4 3 I = (1 + 2xy) dx dy .
0 1. I = 145 cu. units
2. I = 148 cu. units 10.0 points 1 1. I = 70 3. I = 147 cu. units correct 2. I = 74 4. I = 146 cu. units 3. I = 72 correct 5. I = 144 cu. units
Explanation:
The double integral
I = 5. I = 76
f (x, y) dxdy A 4. I = 78 Explanation: liang (xl5432) – HW13 – berg – (55120)
Integrating with respect to x and holding y
ﬁxed, we see that
3 (1 + 2xy) dx = 2 x+x y 1 x=3
x=1 In this case
3 I = (2 + 8y) dy = 2y + 4y 2 0 1
1
−
y 3+y I = 2
1 . 4 3 0 1 . Consequently,
. (3)(1 + 3)
(3 + 3) I = 2 ln Consequently, 004 I = 8 + 64 = 72 . = 2 ln (2) . 10.0 points Evaluate the iterated integral
4 4 I =
keywords: 1 003 dy = 2 ln(y) − ln(3 + y) Thus
4 2 1. I = 10.0 points 1 x y
dydx .
+
y x 15
ln (15)
2 Evaluate the iterated integral
3 3 I =
1 0 2. I = 15 ln (4) correct
2
dx dy .
(x + y)2 15
2 3. I = 4 ln 1. I = ln (2) 15
2 4. I = 15 ln
2. I = ln 6
5 5. I = 4 ln (15)
6
5 3. I = 1
ln
2 4. I = 1
ln (2)
2 5. I = 2 ln 6. I = Explanation:
Integrating with respect to y keeping x
ﬁxed, we see that 6
5 4
1 6. I = 2 ln (2) correct
Explanation:
Integrating the inner integral with respect
to x keeping y ﬁxed, we see that
3
0 15
ln (4)
2 x y
+
y x dy = = (ln(4)) x + 2
−
x+y 1
1
.
= 2 −
y 3+y 3
0 15
2 1
x 4 y2
2x 1 . Thus
4 2
dx =
(x + y)2 x ln(y) + I = (ln(4)) x +
1 = x2
2 15
2 1
x 15
ln(4) +
ln(x)
2 dx
4 .
1 liang (xl5432) – HW13 – berg – (55120) 3 when Consequently, A = I = 15 ln(4) . (x, y) : 0 ≤ x ≤ 2, 0 ≤ y ≤ 1 . 1. I = 7
3 2. I = 1
ln
6 7
3 3. I = 1
ln (3)
3
1
ln
4 5. I = 1
ln (3)
6 6. I = 10.0 points 1
ln
3 4. I = 005 1
ln (3)
4 Evaluate the iterated integral
ln(3) ln(4) I =
0 e2x−y dx dy . 0 1. I = 9
2. I = 8
3. I = 7
4. I = 5 correct Explanation:
Integrating with respect to x with y ﬁxed,
we see that
e2x−y dx = 0 ln(4) 1 2x−y
e
2 0 A = (x, y) : 0 ≤ x ≤ 2, 0 ≤ y ≤ 1 is a rectangle with sides parallel to the coordinate axes, the double integal can be interpreted as the iterated integral
1 Thus 0 2 xy 2
dx dy .
x2 + 3 2 42 − 1 −y
e .
2 1 2 ln(4)−y
e
− e−y =
=
2 I = 7
3 Explanation:
Since 5. I = 6 ln(4) correct xy 2
dx
x2 + 3 0 But to integrate
15
2 ln(3)
0 e−y dy = − 15 −y
e
2 15 − ln(3)
= −
e
−1 .
2 006 0 0 with respect to x with y ﬁxed we use the
substitution u = x2 + 3. For then Consequently,
15 1
−1
I = −
2 3 ln(3) du = 2x dx
= 5 . while Evaluate the double integral
I =
A xy 2
dx dy
x2 + 3 =⇒ u = 3, x = 2 10.0 points x = 0 =⇒ u = 7. In this case
2
0 xy 2
1
dx =
2+3
x
2 7
3 y2
y2
ln u
du =
u
2 7
3 . liang (xl5432) – HW13 – berg – (55120)
Thus
1
ln
2 7
3 1
= ln
6 7
3 I = 1 Thus
π/2 y 2 dy 2x cos(x + y) dx = π sin 0 y3 4 0
1
0 . + 2 cos Consequently, π
+y
2 π
+ y − 2 cos(y) .
2 In this case,
1
I = ln
6 7
3 I = . −π cos π
+y
2 + 2 sin
007 10.0 points 2x cos(x + y) dxdy 008 . 10.0 points Find the volume of the solid under the
graph of when A is the rectangle
π
,
2 0 I = −(4 − π) . A (x, y) : 0 ≤ x ≤ π/2 Consequently, Calculate the value of the double integral
I = π
+ y − 2 sin(y)
2 0 ≤ y ≤ π
2 f (x, y) = 5 + 3x2 + 4y . 1. I = −2π and above the rectangle
A = (x, y) : 1 ≤ x ≤ 2 , 0 ≤ y ≤ 2 . 2. I = (4 − π) 1. volume = 32 cu.units correct 3. I = 2π 2. volume = 39 cu.units
3. volume = 31 cu.units 4. I = π 4. volume = 37 cu.units
5. I = −2(4 − π) 5. volume = 30 cu.units 6. I = −(4 − π) correct
Explanation:
By treating I as an iterated integral, integrating ﬁrst with respect to x, we see that
2x cos(x + y) dx = 2x sin(x + y) Explanation:
The volume is given by the double integral
2 V =
0 sin(x + y) dx = 2 x sin(x + y) + cos(x + y) . (5 + 3x2 + 4y) dxdy 1 of f (x, y) over the rectangular region A. Integrating each term separately, we see that
2 −2 2 2 V = 2 2 5 dxdy +
0 1 0
2 3x2 dxdy 1 2 + 4y dxdy = 10 + 14 + 8.
0 1 liang (xl5432) – HW13 – berg – (55120)
Consequently,
volume = 32 cu.units .
009 √ 3y dx dy . I =
y 0 16
15 Explanation:
Treating I as an iterated integral, integrating ﬁrst with respect to x with y ﬁxed, we see
that 1
1. I =
correct
5
2. I = 19
15 5. I = − y 13
15 4. I = − Evaluate the iterated integral
1 2
2. I = − correct
3
3. I = − 10.0 points 4
5 1 y −1 y2 I = 2
3. I =
5 (5x − 2y) dx dy
1 = 4. I = 1
5. I = 5 −1 5 2
x − 2xy
2 y dy .
y2 Thus 3
5 1 Explanation:
As an interated integral, integrating ﬁrst
with respect to x, we see that
1 √ [ 3xy ]y y dy I =
0 1 = 3
0 I =
−1 = 5 2
(y − y 4 ) dy −
2
y3 y5
−
3
5 5
2 011
1 =
0 1
.
5 10.0 points Evaluate the double integral
y −1 y2 I = 1. I = − 7
15 2(y 2 − y 3 ) dy y3 y4
−
3
4 (5x − 2y) dx dy . 2
.
3 10.0 points Evaluate the iterated integral
3π/2 1 −2 I = − Consequently, 010 −1 Consequently, y 3/2 − y 2 dy . 2 5/2 1 3
y − y
I = 3
5
3 1 cos(θ) I =
0 1. I = e − 2
2. I = 1
−2
e 3. I = 2(e − 1) 0 2 esin(θ) dr dθ . 1 .
−1 liang (xl5432) – HW13 – berg – (55120)
1
−1
e 4. I = 2 6 The integral can be written as the repeated
integral correct √ 1 y I = 5. I = 2e 0 (8x − y) dx y dy. Now 6. I = 0 √ Explanation:
After simple integration
cos(θ) sin(θ) 2e y 2re
sin(θ) = 2 cos(θ) e . 1 In this case, 0 I = sin(θ) 2 cos(θ) e sin(θ) dθ = 2e 0 y y But then
I = 3π/2 4x − xy √ = 4y − y 3/2 − 3y 2 . 0 0 2 (8x − y) dx = cos(θ) sin(θ) dr = y 3π/2
0 (4y − y 3/2 − 3y 2) dx
= . 2
2y 2 − y 5/2 − y 3
5 1
0 . Consequently, Consequently,
I = 2 1
−1
e .
013 012 10.0 points 3
.
5 I = 10.0 points Evaluate the double integral Find the value of the double integral
I =
I =
A (8x − y) dxdy D 4y
dxdy
+1 x2 when D is the region when A is the region
(x, y) : y ≤ x ≤ √ y, 0≤y≤1 . (x, y) : 0 ≤ x ≤ 1,
in the xy-plane. 1. I = 4
5 1. I = 2 ln(2) 2. I = 3
correct
5 2. I = ln(2) correct 3. I = 9
10 3. I = 2 4. I = 7
10 4. I = 4 ln(2) 5. I = 1
Explanation: 5. I = 1
6. I = 4 0≤y≤ √ x liang (xl5432) – HW13 – berg – (55120)
Explanation:
As an iterated integral, integrating ﬁrst
with respect to y, we see that
√ 1 x I =
0 0 4y
dy
2+1
x x dx . Explanation:
After integration with respect to y we see
that
1 4y
dy
x2 + 1 0 I =
√
0 x x
= 2 2
.
x +1 In this case,
x
dx =
2+1
x I = 2
0 4x sin(y)
0 y2
= 2 2
x +1 1 7. I = 4 (1 − cos(1))
8. I = 2 (cos(1) − 1) Now
√ 7 ln(x2 + 1) 1
0 0 dx 4x sin(x2 ) dx = 0 −2 cos(x2 ) 1
0 , using substitution in the second integral.
Consequently,
. Consequently,
I = ln(2) 1 = x2 . I = 2 (1 − cos(1)) .
015 10.0 points The graph of
f (x, y) = 6xy keywords:
014 10.0 points over the bounded region A in the ﬁrst quadrant enclosed by Evaluate the double integral
I = 4x cos(y) dxdy
D y = 9 − x2 and the x, y-axes is the surface when D is the bounded region enclosed by the
graphs of
y = 0, y = x2 , x = 1. 1. I = 2 (1 − sin(1))
2. I = 4 (1 − sin(1))
3. I = 4 (sin(1) − 1)
4. I = 4 (cos(1) − 1)
5. I = 2 (sin(1) − 1)
6. I = 2 (1 − cos(1)) correct Find the volume of the solid under this graph
over the region A.
1. Volume = 81
cu. units
2 liang (xl5432) – HW13 – berg – (55120)
243
cu. units
8 4. Volume = 3. I = 243
cu. units correct
4 3. Volume = 2. I = 243
cu. units
2 2. Volume = 4. I = 1
0
1 2y
2
1
2 f (x, y) dx dy which in turn can be written as the repeated
integral
6xy dy dx.
Now the inner integral is equal to
3xy √ 9−x2 0 1 2 6. I = f (x, y) dx dy
0 y (x, y) : x/2 ≤ y ≤ 1 , 0 ≤ x ≤ 2
in the plane bounded by the y-axis and the
graphs of
x
y = 1.
y = ,
2
This is the shaded region in 9−x2 0 2 0 Explanation:
The region of integration is the set of all
points A 0 y/2 5. I = f (x, y) dxdy, √ y f (x, y) dx dy
0 Explanation:
The volume of the solid under the graph of
f is given by the double integral 3 2 f (x, y) dx dy
0 0 V = 2y f (x, y) dx dy correct
0 243
cu. units
16 5. Volume = 8 = 3x(9 − x2 ). y Thus 1
3 V =3
0 3
x(9 − x2 ) dx = − (9 − x2 )2
4 3
0 .
x Consequently,
2 243
cu. units .
Volume =
4
016 10.0 points Reverse the order of integration in the integral
2 y 1 I = f (x, y) dy dx ,
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 x/2 1 but make no attempt to evaluate either integral.
2 1 1. I = f (x, y) dx dy
0 y/2 x
2 liang (xl5432) – HW13 – berg – (55120)
Since the equation of the slant line is x = 2y,
integration in x is along the line from (0, y)
to (2y, y) for ﬁxed y, and then from y = 0 to
y = 1.
Consequently, after changing the order of
integration,
1 f (x, y) dx dy √ 3
2 . 0 3 keywords: double integral, reverse order integration, linear function,
017 1 2y I =
0 10.0 points Reverse the order of integration in the integral
I = f (x, y) dx dy
3/2 018 3/2 10.0 points Reverse the order of integration in the double integral f (x, y) dx dy I = f (x, y) dx dy ,
0 9y 2
4y 2 1 f (x, y) dx dy cor3/2 y2 4 2 0 √ 0 but make no attempt to evaluate either integral. 3 √ 16 rect f (x, y) dy dx
0 f (x, y) dx dy x2 2. I = f (x, y) dy dx
0 0 3 0 4
16 4x2 3/2 3. I = 16
2y 2 1 4. I = √ f (x, y) dx dy
3/2 1 5. I = x 1. I =
√ √ 4 3. I = f (x, y) dy dx
0 3 x2
16 4. I = 3
2y 2 Explanation:
The region of integration is similar to the
one in the ﬁgure 4 f (x, y) dy dx correct √ 0 f (x, y) dx dy
3/2 . 3 f (x, y) dy dx, 1. I = 2. I = √ x/2 but make no attempt to evaluate either integral.
√ 4y 2 1 1
√ 3 4 This shaded region is enclosed by the graphs
of 4y 2 = x, y = 1 , and x = 3. To change
the order of integration, ﬁrst ﬁx y. Then,
as the graph shows, x varies from 3 to 4y 2 .
To cover the region√ integration, therefore,
of
y must vary from 3/2 to 1. Hence, after
changing the order of integration,
I = 4 9 x
√ 16 5. I = x f (x, y) dy dx
0 0
4 6. I =
0 16
√ f (x, y) dy dx
x liang (xl5432) – HW13 – berg – (55120)
Explanation:
The region of integration is the set of all
points
(x, y) : y 2 ≤ x ≤ 16 , 0 ≤ y ≤ 4
in the plane bounded by the y-axis and the
graphs of
√
y = x,
y = 4.
This is the shaded region in
y
4 x
16
(not drawn to scale). Integration is taken ﬁrst
with respect to x for ﬁxed y along the solid
horizontal line.
To change the order of integration, ﬁrst ﬁx
x in the interval [0, 16] and let y vary along
the vertical dashed line. Then y varies from
√
0 to x. Hence, after changing the order of
integration,
16 I =
0 4
√ f (x, y) dy dx . x keywords: double integral, reverse order integration, parabola, 10 ...

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