How to find the volume of a solid of revolution generated by revolving a region bounded by the graph of a function around one of the axes using definite integrals? We will present examples based on the methods of disks and washers where the integration is parallel to the axis of rotation. A set of exercises with answers is presented at the end.

\LARGE {\text{Volume} = \color{red}{\int_{x_1}^{x_2} \pi [ f(x)^2 - h(x)^2 ] dx}}

\LARGE { \text{Volume} = \color{red}{\int_{y_1}^{y_2} \pi ( z(y) )^2 dy}}

\LARGE { \text{Volume} = \color{red}{\int_{y_1}^{y_2} \pi [ z(y)^2 - w(y)^2 ] dy}}

volume = (1/3) ? (radius)

= (1/3) ?(2)

= 8?/3

We now use definite integrals to find the volume defined above. If we let f(x) = x according to formula 1 above, the volume is given by the definite integral

\text{Volume} = \int_{x_1}^{x_2} \pi f(x)^2 dx = \int_{0}^{2} \pi x^2 dx = \left [\pi x^3 / 3 \right ]_0^2 = 8\pi/3

The first method works because y = x is a linear function and the volume generated is that of a right circular cone , however the second method work for shapes other than cones and will be used in the examples below.

The graph of y = √(r

\text{Volume} = \int_{x_1}^{x_2} \pi f(x)^2 dx = \int_{-r}^{r} \pi (\sqrt{r^2 - x^2})^2 dx = \int_{-r}^{r} \pi (r^2 - x^2) dx \\\\\\
= \pi\left [r^2 x - x^3/3 \right ]_{-r}^r = \pi\left [ (r^3 - r^3/3) - (-r^3 + r^3/3) \right ] = \dfrac{4}{3} \pi r^3

This is the very well known formula for the volume of the sphere. If you revolve a semi circle of radius r around the x axis, it will generate a sphere of radius r.

The shaded (red) region is bounded by the x axis, the line that passes through the points (0,0) and (1,1) and has the equation y = x, and the line that passes through the points (1,1) and (2,0) and has the equation y = -x + 2. Since the solid is generated by revolving through the y axis, we will use formula 4 given above to find the volume as follows. The limits of integration are y = 0 and y = 1

\text{Volume} = \int_{y_1}^{y_2} \pi [z(y)^2 - w(y)^2] dy

Solve y = - x + 2 for x

x = z(y) = - y + 2

Solve y = x for x

x = w(y) = y

Substitute in the formula for the volume to obtain

\text{Volume} = \int_{0}^{1} \pi [ (-y+2)^2 - y^2] dy \\
= \int_{0}^{1} \pi [ - 4 y + 4] dy \\ = \pi \left [-2y^2+4y \right ]_0^1 = 2\pi

We now use formula 2 above, washers with integration along the x axis. The limits of integration are x = 0 and x = 2. Let f(x) = 1 + x

\text{Volume} = \int_{x_1}^{x_2} \pi [ f(x)^2 - h(x)^2 ] dx = \int_{0}^{2} \pi [ (1+x^2)^2 - (\sqrt x)^2 ]dx\\
\\ = \pi \int_{0}^{2} [ 1+x^4+2x^2 - x ] dx = \pi \left [ x^5/5 + 2x^3/3 - x^2/2 + x \right]_0^2 = 176\pi / 15

The equation of the circle is given by

x

Solve the above equation for y to obtain two solutions each for one semicircle

y = R + √(r

The torus is generated by rotating the two halves semi circles the x axis hence the use of formula 2 given above to find the volume of the torus. Let f(x) = R + √(r

\text{Volume} = \int_{0}^{r} \pi [ f(x)^2 - h(x)^2 ] dx \\\\
= \pi \int_{0}^{r} [ (R + \sqrt(r^2 - x^2))^2 - (R - \sqrt(r^2 - x^2))^2 ]dx \\\\
= 4 R \pi \int_{0}^{r} [ \sqrt(r^2 - x^2) ]dx = 4 R \pi (1/2) \left [ x \sqrt{r^2-x^2} + r^2\arctan(\frac{x}{\sqrt{r^2-x^2}}) \right ]_0^r\\\\
= 4 R \pi ( \pi r^2 / 4) = {\pi}^2 R r^2

The total volume is twice the above, hence the volume of a torus is given by
\text{Volume} = 2 {\pi}^2 R r^2

(1) Find the volume of the solid generated when the region between the graphs of f(x) = x

(2) Find the volume generated when the finite region bounded by the curves y = x

(1) 26 Pi /5

(2) pi /10

Volume by Cylindrical Shells Method.

integrals and their applications in calculus.

Area under a curve.

Area between two curves.