Solution

Problems 5.3 Solutions

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?

Solution. We sweep out along the x-=axis. The section at x is a semicircle of radius y/2, so has area A(x) = (/2)(y/2)2 = (/8)(10 - x). Thus

V=

8

10

x2

(10 - x)dx = 10x -

0

8

2

10 0

=

25 4

.

2. A solid is formed over the region in the first quadrant bounded by the curve y = 4 - x so that

the section by any plane perpendicular to the x-axis is a square. What is the volume of this solid?

Solution. The section at x has area y2 = 4 - x, so

4

V = (4 - x)dx = 8 .

0

3. A solid is formed over the region in the first quadrant bounded by the curve y = 2x - x2 so that the section by any plane perpendicular to the x-axis is a semicircle. What is the volume of this solid?

Solution. As in problem 1,

dV = ( y )2 = (2x - x2)2dx = (4x2 - 4x3 + x4)dx .

22 8

8

Integrating dV from 0 to 2, we get

32 16 32 V = ( - + ) = 1.466 .

83 3 5

4. The region in the first quadrant bounded by y = x2 - 1, y = 0, x = 1, x = 4 is revolved around the x-axis. Find the volume of the resulting solid.

Solution. Here we find that at a typical x between 1 and 4, dV = r2dx = (x2 - 1)dx. Integrating, we get V = 18.

5. Find the volume of the solid obtained by rotating about the y-axis the region bounded by y = x2, x = 2 and the x-axis.

Solution. Here we will from 0 to 4. At a typical

use the washer method, y, dV = (R2 - r2)dy,

sweeping out and R = 2, r

=alonyg.

the y-axis, with y ranging Thus the volume is

4

V = (4 - y)dy = 8 .

0

6. The region in the first quadrant under the curve y2 = 2x - x2 is rotated about the y-axis. Find the volume of the resulting solid.

1

Solution. Here we sweep out along the x-axis from x = 0 to x = 2, using the shell method. At a typical x, dV = 2xydx = 2(2x2 - x3)dx, and

V = 2

2

(2x2

-

x3)dx

=

2(

2x3

0

3

-

x4 )

4

2 0

=

8 3

.

7. The region in the first quadrant bounded by y = x4 and x = 1 is revolved around the y-axis. Find the volume of the resulting solid.

Solution. Here let's use the shell method, sweeping out along the x-axis (compare this with problem 5). dV = 2xydx = 2x5dx. Integrating from 0 to 1, we get V = /53.

8. The region in the first quadrant bounded by y = x - x2 and y = x - x3 is revolved around the x-axis. Find the volume of the resulting solid.

Solution. Using the washer method, at a typical x between 0 and 1, dV = (R2 - r2)dx = [(x - x3)2 - (x - x2)2]dx. After some algebra, we obtain

V=

1

(2x3

-

3x4

+

x6)dx

=

1 (

-

3

+

1 )

=

.0428

.

0

257

9. The region bounded by the curves y = ex and y = e2x and the lines x = 0 and x = 1 is rotated around the x-axis for form a solid D. What is the volume of D?

Solution. We use the washer method: dV = (R2 - r2)dx = (e4x - e2x)dx. Thus

V ol(D) =

1

(e4x

-

e2x)dx

=

e2

(u - 1)du ,

0

21

making the substitution u = e2x. Thus

V ol(D) = (u - 1)2 e2 = (e2 - 1)2 . 22 1 4

10. The area between the curves y = x, y = x-1 and the lines x = 1 and x = 2 is rotated around the x-axis to form a solid D, and around the y-axis to form a solid E. What are the volumes of D and E?

Solution. To find the volume of D we use the washer method: dV = (R2-r2)dx = (x2-x-2)dx. Thus the volume is

V ol(D) =

2

(x2

1

-

x-2)dx

=

x2 (

2

+

1 )

x

2 1

=

((2

+

1 2

-

1 (

2

+

1))

=

.

For E we use the shell method:dV = 2xhdx = 2x(x-1 - x-2)dx = 2(1 - x-1)dx. Thus

2

V ol(E) = 2

(1

-

x-1)dx

=

2(x

-

ln x)

2 1

=

2(2

-

ln

2

-

1)

=

2(1

-

ln

2)

.

1

2

11. An ellipsoid is formed by rotating the curve 4x2 + y2 = 1 around the x-axis. What is its volume? What is the volume of the ellipsoid obtained by rotating this curve about the y-axis?

Solution. By symmetry, the volume is twice the volume of the solid obtained by rotating the region in the first quadrant bounded by the curve y = 1 - 4x2.

a) For rotation about the x-axis we use the disc method. We have dV = y2dx = (1 - 4x2)dx. To find the range of x we solve the equation 1 - 4x2 = 0; this gives x = 1/2. Thus

V olume = 2

1/2 0

(1

-

4x2)dx

=

2[x

-

4 x3 3

1/2 0

=

2

14 -

23

1 8

2 =.

3

b). To rotate about the y-axis we use the shell method: dV = 2xydx = 2x 1 - 4x2dx. Thus

1/2

V olume = 2

2x 1 - 4x2dx .

0

Let u = 1 - 4x2, du = -8xdx. Then when x = 0, u = 1 and when x = 1/2, u = 0. Thus

V olume = -

2

0 u1/2du = - 2 u3/2

1

23

0 1

=

3

.

12. A pencil sharpener is made by drilling a cone out of a sphere; the cone has as its axis a diameter of the sphere, and its vertex is on the surface of the sphere. If the ratio of the height to base radius in the cone is 4 to 1, and the sphere has a 1 inch radius, what is the volume of the pencil sharpener?

Solution. We can model this as a solid of revolution as follows. The sphere is obtained by rotating the curve y = 1 - x2 around the x-axis. Let's put the vertex of the cone at (-1,0) and its axis along the x-axis. The given information tells us that the cone is generated by rotating (around the x-axis) the line of slope 1/4 and x-intercept -1. The equation of this line is y = (x + 1)/4 (see the figure). Now, we calculate the volume using the method of washers:

dV = (R2 - r2)dx = (1 - x2) - (1 + x)2 dx 16

The range of integration is from -1 to the x-coordinate of the point of intersection of the two curves. To find that, we solve

1 - x2 = (1 + x) which simplifies to 17x2 + 2x - 15 = 0 , 4

which has the solutions x = -1, 15/17. Thus the volume is

15/17

(1 - x2) - (1 + x)2

dx =

1024 = 3.71

-1

16

867

cubic inches. Should a problem like this occur on an examination it suffices to end the argument with the definite integral to be computed. In cases like this, the actual computation is arithmetically tedious, and is best done on the computer (I used MAPLE).

3

13. Consider the region in the first quadrant bounded by y = sin x, x = 0, x = . Find the volume

of the solid obtained by rotating this region about the x-axis. Hint: If you use the disc method you'll need to know: cos(2x) = 1 - 2 sin2 x.

Solution. The disc method gives us: dV = y2dx = sin2 xdx. Thus the volume is

sin2 xdx = 1

(1 - cos(2x))dx ,

0

20

by the hint. This gives us

V olume

=

1 2

x-

sin(2x) 2

0

=

2

.

14. Now use the shell method to solve problem 13. Hint: The derivative of sin x - x cos x is x sin x.

Solution. For the shell method: dV = 2xydx. The volume is

x sin xdx =

sin x

- x cos x

0

=

,

0

by the hint.

15. A solid is formed over the region in the first quadrant bounded by the curve y = 2x - x2 so that the section by any plane perpendicular to the x-axis is a semicircle. What is the volume of this solid?

Solution. As in problem 1,

dV = ( y )2 = (2x - x2)2dx = (4x2 - 4x3 + x4)dx .

22 8

8

Integrating dV from 0 to 2, we get

32 16 32 V = ( - + ) = 1.466 .

83 3 5

4

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download