# π Calculus I Problem Set 11 Answers Answer

```Calculus I
Problem Set 11 Answers
p
1. A solid is formed over the region in the first quadrant bounded by the curve y = 10 x so that the section
by any plane perpendicular to the x-axis is a semicircle. What is the volume of this solid?
Answer. We sweep out along the x-=axis. The section at x is a semicircle of radius y=2, so has area A(x) =
(π =2)(y=2)2 = (π =8)(10 x). Thus
V
=
3.5
3
2.5
2
1.5
1
0.5
0
p
=
π
8
Z
10
(10
x)dx =
0
π
10x
8
x2 10 25π
=
2 0
4
y=
:
p
10
x
dx
2
0
4
6
8
10
p
2. A solid is formed over the region in the first quadrant bounded by the curve y = 4
by any plane perpendicular to the x-axis is a square. What is the volume of this solid?
Answer. The section at x has area y 2 = 4
x so that the section
x, so
Z
V
4
=
(4
x)dx = 8 :
0
y=
2
1.5
=
p
=
p
4
x
dx
1
=
0.5
0
p
0
0.5
1
1.5 2
2.5
3
3.5
p
4
3. A solid is formed over the region in the first quadrant bounded by the curve y = 2x x 2 so that the section
by any plane perpendicular to the x-axis is a semicircle. What is the volume of this solid?
Answer. As in problem 1,
π y 2 π
π
2 2
2
( ) = (2x x ) dx = (4x
4x3 + x4 )dx :
2 2
8
8
Integrating dV from 0 to 2, we get
π 32
32
2π
V= (
16 + ) =
:
8 3
5
15
dV
=
y = 2x
1
p
=
=
0.8
0.6
p
=
0.4
0.2
=
0
0
0.5
1
1.5
2
x2
p
=
p
dx
4. The region in the first quadrant bounded by y =
Find the volume of the resulting solid.
p2
x
1; y = 0; x = 1; x = 4 is revolved around the x-axis.
Answer. Here we find that at a typical x between 1 and 4, dV
V = 18π .
=
=
π (x 2
y = 2x
p
p
= π r 2 dx =
1)dx. Integrating, we get
x2
4
3
=
2
=
1
0
=
0
0.5 1
1.5 2
2.5 3
3.5 4
p
=
=
p
dx
5. Find the volume of the solid obtained by rotating about the y-axis the region bounded by y = x 2 ; x = 2 and
the x-axis.
Answer. Here we will use the washer method, sweeping out along the y-axis, with y ranging from 0 to 4. At
p
a typical y, dV = (π R 2 π r2 )dy, and R = 2; r = y. Thus the volume is
V
=
=π
Z
4
y)dy = 8π
(4
0
:
p
=
p
7
6
5
=
=
=
=
4
3
2
1
0
0
p
=
dy
p
y = x2
=
0.5
1
1.5
2
=
2.5
6. The region in the first quadrant under the curve y = 2x
the resulting solid.
x 2 is rotated about the y-axis. Find the volume of
Answer. Here we sweep out along the x-axis from x = 0 to x = 2, using the shell method. At a typical x,
dV = 2π xydx = 2π (2x 2 x3 )dx, and
V
=
= 2π
Z
2
(2x
x3 )dx = 2π (
2
0
2x3
3
x4 2 8π
) =
4 0
3
:
p
=
p
=
=
=
p
1
=
0.8
=
0.6
=
p
dx
p
2x
=
0.4
0.2
0
p
=
0
0.5
1
1.5
2
x2
7. The region in the first quadrant bounded by y = x 4 and x = 1 is revolved around the y-axis. Find the volume
of the resulting solid.
p
Here let’s use the shell method, sweeping out along the x-axis (compare this with problem 5).
=
dV = 2π xydx = 2π x 5 dx. Integrating from 0 to 1, we get Vp= π =53.
=
p
=
p
=
=
=
=
2.5
p
=
=
2
y = x4
=
1.5
p
1
=
=
x=1
0.5
0
0
0.2
0.4
0.6
0.8
dx
1
x 2 and y = x
8. The region in the first quadrant bounded by y = x
the volume of the resulting solid.
Answer. Using the washer method, at a typical x between 0 and 1, dV
p
x2=)2 ]dx. After some algebra, we obtain
=
p
Z
V
1
=
(2x
0
=
3
p
= = π(
3x4 + x6 )dx
=
p
= (π R 2
π r2 )dx = π [(x
3 1
+ ) = :0428π
5 7
x 3 )2
:
=
=
=
p
0.3
y=x
=
0.4
=
=
=
=
p
=
1
2
x3 is revolved around the x-axis. Find
y=x
0.2
x2
=
=
0.1
0
0
0.2
0.4
0.6
0.8
1
dx
9. The average value of a function y = f (x) defined over an interval [a; b] is defined to be
yave =
Z
1
b
a
b
f (x)dx :
a
x3
(x
Find the average of y = sin x over the interval [0; π ].
yave =
1
π
Z
π
sin xdx
0
=
1
π
( cos x)j0
π
=
2
:
π
10. Let g(x) = x2 + x3 for x in the interval 0 x 10. Find the average, or mean, value of g on the interval.
Find the average slope of the graph of y = g(x) on the interval.
Answer. The average value of the function is
1
10
Z
10
2
3
(x + x )dx =
0
1 x3
10 3
+
x4 10 100 1000
=
+
4 0
3
4
= 283:33
Since the slope is y0 = g0 (x)m the average slope is
1
10
Z
10
g0 (x)dx =
0
1
(g(10)
10
g(0)) = 110 :
:
```