# Volume of a Solid of Revolution

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.

## Formulas to calculate the volume generated by revolving graphs of functions around one of the axes

### Formula 1 - Disk around the x axis

If f is a function such that f(x) ≥ 0 for all x in the interval [x_{1}, x

_{2}], the volume of the solid generated by revolving, around the x axis, the region bounded by the graph of f, the x axis (y = 0) and the vertical lines x = x

_{1}and x =

_{2}is given by the integral

### Formula 2 - Washer around the x axis

If f and h are functions such that f(x) ≥ h(x) for all x in the interval [x_{1}, x

_{2}], the volume of the solid generated by revolving, around the x axis, the region bounded by the graphs of f and h, between x = x

_{1}and x = x

_{2}is given by the integral

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

### Formula 3 - Disk around the y axis

If z is a function of y such that x = z(y) and z(y)≥ 0 for all y in the interval [y1 , y2], the volume of the solid generated by revolving, around the y axis, the region bounded by the graph of z, the y axis (x = 0) and the horizontal lines y = y1 and y = y2 is given by the integral
\LARGE { \text{Volume} = \color{red}{\int_{y_1}^{y_2} \pi ( z(y) )^2 dy}}

### Formula 4 - Washer around the y axis

If z and w are functions of y such that z(y) ≥ w(y) for all y in the interval [ y1 , y2 ], the volume of the solid generated by revolving the region bounded by the graphs of z and w, between y = y1 and y = y2, around the y axis , is given by the integral
\LARGE { \text{Volume} = \color{red}{\int_{y_1}^{y_2} \pi [ z(y)^2 - w(y)^2 ] dy}}

## Examples to Find Volume of a Solid of Revolution Using Definite Integrals

### Example 1

Find the volume of the solid generated by revolving the region bounded by the graph of y = x, y = 0, x = 0 and x = 2.(see figure below).### Solution to Example 1

We present two methods__Method 1__This problem may be solved using the formula for the volume of a right circular cone.

volume = (1/3) π (radius)

^{ 2}height

= (1/3) π(2)

^{ 2}2

= 8π/3

__Method 2__

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.

### Example 2

Find the volume of the solid generated by revolving the semicircle y = √ (r^{ 2}- x

^{ 2}) around the x axis, radius r > 0.

### Solution to Example 2

The graph of y = √(r

^{ 2}- x

^{ 2}) is shown above and y ≥ 0 from x = -r to x = r. The volume is given by formula 1 as follows

\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.

### Example 3

Find the volume of the solid generated by revolving the shaded (red) region about the y axis.### Solution to Example 3

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

### Example 4

Find the volume of the solid generated by the rotation of curves y = 1 + x^{2}and y = √x, around the x axis and limited to the left by x = 0 and to the right by x = 2.

### Solution to Example 4

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

^{2}and h(x) = √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

### Example 5

Find the volume of the torus generated when the circle with center at (0,R) and radius r is rotated around the x axis.### Solution to Example 5

The equation of the circle is given by

x

^{ 2}+ (y - R)

^{ 2}= r

^{ 2}

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

y = R + √(r

^{ 2}- x

^{ 2}) , upper semi circle , and y = R - √(r

^{ 2}- x

^{ 2}) , lower semi circle

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

^{ 2}- x

^{ 2}) and h(x) = R - √(r

^{ 2}- x

^{ 2}). Because of the symmetry of the circle and therefore the torus with respect to the y axis, we integrate from x = 0 to x = r then double the answer to find the total volume.

\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

### Exercises

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

^{ 2}+ 2 and h(x) = x is revolved about the x axis and over the interval [0,1].

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

^{3}, y = x

^{2}is revolved about the y axis.(hint: you need to find the points of intersections of the two curves)

### Answers to Above Exercises

(1) 26 Pi /5

(2) pi /10

### More Links and References

Volume by Cylindrical Shells Method .

integrals and their applications in calculus.

Area under a curve .

Area between two curves .