# Calculus

## Applications of Integrals # Volume of a Solid of Revolution: Disks and Washers

If a region in the plane is revolved about a line in the same plane, the resulting object is known as a solid of revolution.

For example, a solid right circular cylinder can be generated by revolving a rectangle. Similarly, a solid spherical ball can be generated by revolving a semi-disk.

The line about which we rotate the shape is called the axis of revolution.

## The Disk Method

The disk method is used when we rotate a single curve y = f (x) around the x- (or y-) axis.

Suppose that y = f (x) is a continuous non-negative function on the interval [a, b].

The volume of the solid formed by revolving the region bounded by the curve $$y = f\left( x \right)$$ and the $$x-$$axis between $$x= a$$ and $$x = b$$ about the $$x-$$axis is given by

$V = \pi \int\limits_a^b {{{\left[ {f\left( x \right)} \right]}^2}dx} .$

The cross section perpendicular to the axis of revolution has the form of a disk of radius $$R = f\left( x \right).$$

Similarly, we can find the volume of the solid when the region is bounded by the curve $$x = f\left( y \right)$$ and the $$y-$$axis between $$y = c$$ and $$y = d,$$ and is rotated about the $$y-$$axis.

The resulting formula is

$V = \pi \int\limits_c^d {{{\left[ {f\left( y \right)} \right]}^2}dy} .$

## The Washer Method

We can extend the disk method to find the volume of a hollow solid of revolution.

Assuming that the functions $$f\left( x \right)$$ and $$g\left( x \right)$$ are continuous and non-negative on the interval $$\left[ {a,b} \right]$$ and $$g\left( x \right) \le f\left( x \right),$$ consider a region that is bounded by two curves $$y = f\left( x \right)$$ and $$y = g\left( x \right),$$ between $$x = a$$ and $$x = b.$$

The volume of the solid formed by revolving the region about the $$x-$$axis is

$V = \pi \int\limits_a^b {\left( {{{\left[ {f\left( x \right)} \right]}^2} - {{\left[ {g\left( x \right)} \right]}^2}} \right)dx} .$

At a point $$x$$ on the $$x-$$axis, a perpendicular cross section of the solid is washer-shape with the inner radius $$r = g\left( x \right)$$ and the outer radius $$R = f\left( x \right).$$

The volume of the solid generated by revolving about the $$y-$$axis a region between the curves $$x = f\left( y \right)$$ and $$x = g\left( y \right),$$ where $$g\left( y \right) \le f\left( y \right)$$ and $$c \le y \le d$$ is given by the formula

$V = \pi \int\limits_c^d {\left( {{{\left[ {f\left( y \right)} \right]}^2} - {{\left[ {g\left( y \right)} \right]}^2}} \right)dy} .$

## Volume of a Solid of Revolution for a Parametric Curve

If a bounding curve is defined in parametric form by the equations $$x = x\left( t \right),$$ $$y = y\left( t \right),$$ where the parameter $$t$$ varies from $$\alpha$$ to $$\beta,$$ then the volume of the solid generated by revolving the curve about the $$x-$$axis is given by

${V_x} = \pi \int\limits_\alpha ^\beta {{y^2}\left( t \right)\frac{{dx}}{{dt}}dt} .$

Respectively, when the curve is rotated about the $$y-$$axis, the volume of the solid of revolution is equal

${V_y} = \pi \int\limits_\alpha ^\beta {{x^2}\left( t \right)\frac{{dy}}{{dt}}dt} .$

## Volume of a Solid of Revolution for a Polar Curve

There are many curves that are given by a polar equation $$r = r\left( \theta \right).$$ To convert from polar coordinates $$\left( {r,\theta } \right)$$ to Cartesian coordinates $$\left( {x,y} \right),$$ we use the known formulas

$x = r\left( \theta \right)\cos \theta ,\;\; y = r\left( \theta \right)\sin \theta .$

So we come to the parametric form of the curve considered in the previous section.

It is important to keep in mind that the radius vector $$r$$ also depends on the parameter $$\theta.$$ Therefore, the derivatives $$\frac{{dx}}{{dt}}$$ and $$\frac{{dy}}{{dt}}$$ are written as

$\frac{{dx}}{{dt}} = \frac{{d\left( {r\left( \theta \right)\cos \theta } \right)}}{{dt}} = \frac{{d\left( {r\left( \theta \right)} \right)}}{{dt}}\cos \theta - r\left( \theta \right)\sin \theta ,$
$\frac{{dy}}{{dt}} = \frac{{d\left( {r\left( \theta \right)\sin \theta } \right)}}{{dt}} = \frac{{d\left( {r\left( \theta \right)} \right)}}{{dt}}\sin \theta + r\left( \theta \right)\cos \theta .$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Using the disk method, calculate the volume of the right circular cone of height $$H$$ and base radius $$R.$$

### Example 2

Find the volume of the solid obtained by rotating the sine function between $$x = 0$$ and $$x = \pi$$ about the $$x-$$axis.

### Example 3

Calculate the volume of the solid obtained by rotating the region bounded by the parabola $$y = {x^2}$$ and the square root function $$y = \sqrt x$$ around the $$x-$$axis.

### Example 4

Find the volume of the solid obtained by rotating the region bounded by two parabolas $y = {x^2} + 1, y = 3 - {x^2}$ about the $$x-$$axis.

### Example 1.

Using the disk method, calculate the volume of the right circular cone of height $$H$$ and base radius $$R.$$

Solution.

The slant height of the cone is defined by the equation:

$x = R - \frac{R}{H}y.$

Hence, the volume of the cone is given by

$V = \pi \int\limits_0^H {{{\left[ {x\left( y \right)} \right]}^2}dy} = \pi \int\limits_0^H {{{\left[ {R - \frac{R}{H}y} \right]}^2}dy} = \pi {R^2}\int\limits_0^H {\left( {1 - \frac{{2y}}{H} + \frac{{{y^2}}}{{{H^2}}}} \right)dy} = \pi {R^2}\left. {\left( {H - \frac{{{y^2}}}{H} + \frac{{{y^3}}}{{3{H^2}}}} \right)} \right|_0^H = \pi {R^2}\left( {\cancel{H} -\cancel{H} + \frac{H}{3}} \right) = \frac{{\pi {R^2}H}}{3}.$

### Example 2.

Find the volume of the solid obtained by rotating the sine function between $$x = 0$$ and $$x = \pi$$ about the $$x-$$axis.

Solution.

By the disk method,

$V = \pi \int\limits_0^\pi {{{\left[ {\sin x} \right]}^2}dx} = \frac{\pi }{2}\int\limits_0^\pi {\left( {1 - \cos 2x} \right)dx} = \frac{\pi }{2}\left. {\left( {x - \frac{{\sin 2x}}{2}} \right)} \right|_0^\pi = \frac{\pi }{2}\left[ {\left( {\pi - 0} \right) - \left( {0 - 0} \right)} \right] = \frac{{{\pi ^2}}}{2}.$

### Example 3.

Calculate the volume of the solid obtained by rotating the region bounded by the parabola $$y = {x^2}$$ and the square root function $$y = \sqrt x$$ around the $$x-$$axis.

Solution.

Both curves intersect at the points $$x = 0$$ and $$x = 1.$$ Using the washer method, we have

$V = \pi \int\limits_0^1 {\left( {{{\left[ {\sqrt x } \right]}^2} - {{\left[ {{x^2}} \right]}^2}} \right)dx} = \pi \int\limits_0^1 {\left( {x - {x^4}} \right)dx} = \pi \left. {\left( {\frac{{{x^2}}}{2} - \frac{{{x^5}}}{5}} \right)} \right|_0^1 = \pi \left( {\frac{1}{2} - \frac{1}{5}} \right) = \frac{{3\pi }}{{10}}.$

### Example 4.

Find the volume of the solid obtained by rotating the region bounded by two parabolas $y = {x^2} + 1, y = 3 - {x^2}$ about the $$x-$$axis.

Solution.

First we determine the boundaries $$a$$ and $$b:$$

${x^2} + 1 = 3 - {x^2},\;\; \Rightarrow 2{x^2} = 2,\;\; \Rightarrow {x^2} = 1,\;\;\Rightarrow x_{1,2} = \pm 1.$

Hence the limits of integration are $$a = -1,$$ $$b = 1.$$ We sketch the bounding region and the solid of revolution:

$V = \pi \int\limits_a^b {\left( {{{\left[ {f\left( x \right)} \right]}^2} - {{\left[ {g\left( x \right)} \right]}^2}} \right)dx} = \pi \int\limits_{ - 1}^1 {\left( {{{\left( {3 - {x^2}} \right)}^2} - {{\left( {{x^2} + 1} \right)}^2}} \right)dx} = \pi \int\limits_{ - 1}^1 {\left( {{{\left[ {3 - {x^2}} \right]}^2} - {{\left[ {{x^2} + 1} \right]}^2}} \right)dx} = \pi \int\limits_{ - 1}^1 {\left( {8 - 8{x^2}} \right)dx} = 8\pi \int\limits_{ - 1}^1 {\left( {1 - {x^2}} \right)dx} = 8\pi \left. {\left( {x - \frac{{{x^3}}}{3}} \right)} \right|_{ - 1}^1 = 8\pi \left[ {\left( {1 - \frac{1}{3}} \right) - \left( { - 1 + \frac{1}{3}} \right)} \right] = 8\pi \cdot \frac{4}{3} = \frac{{32\pi }}{3}.$