How to Use Solids of Revolution to Find Volumes

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Encyclopedia>Functions>Integrals>How to Use Solids of Revolution to Find Volumes

Gabriel’s horn

The figure you see above is called “Gabriel’s Horn”, and is known for having a finite volume while having an infinite surface area. That’s pretty cool! If you cut it vertically crosswise, you get very thin rings. If you integrate over all these rings, you get the volume. The computation of the volume of Gabriel’s Horn is written in Example 1.

A solid of revolution is a way to find the volume of three-dimensional figures with the help of two-dimensional figures. A solid of revolution can be thought of as a collection of very thin slices. These discs can be looked at as two-dimensional, and when they’re put together, they make up a three-dimensional body. You can also think of them as thin slices of bread that, when put together, make up a loaf of bread. The volume is defined as

V = \int_{a}^{b} A (x) d x

where $A (x)$ is the area of a slice. You can see from the figure that the thin slices are circular. The radius $r$ of a solid of revolution is given by $r = f (x)$ . That means the area of each slice is given by

A (x) = π r^{2} = π {(f (x))}^{2}

Then, taking the final step, the total volume is given by:

Formula

Volume of a Solid of Revolution

V = \int_{a}^{b} π {(f (x))}^{2} d x

Example 1

Find the volume of the solid of revolution about the $x$ -axis of $f (x) = \frac{1}{x}$ , with $x$ going from 1 to 10

Insert what you know into the formula for the volume and calculate:

\begin{array}{l} V & = \int_{1}^{10} π {(\frac{1}{x})}^{2} d x \\ = π \int_{1}^{10} \frac{1}{x^{2}} d x \\ = π \int_{1}^{10} x^{- 2} d x \\ = - π x^{- 1} |_{1}^{10} \\ = - π \cdot 1 0^{- 1} - (- π \cdot 1^{- 1}) \\ = - \frac{π}{10} + π \\ = \frac{9 π}{10} \end{array}

The volume of the solid is $\frac{9 π}{10}$ .

Example 2

Find the volume of the solid of revolution about the $x$ -axis for $f (x) = \cos x$ , with $x$ going from 0 to $π$

Example of volume of the solid of revolution

Because you get

\cos^{2} x

when you insert the function into the formula, you have to find the indefinite integral first. You can insert the limits after you’ve done that. Here’s the integration:

\begin{array}{l} \int π \cos^{2} x d x \\ = π \int \cos^{2} x d x \\ \overset{*}{=} π (\sin x \cos x + \int \sin^{2} x d x) \\ = π (\sin x \cos x + \int 1 - \cos^{2} x d x) \\ = π (\sin x \cos x + x - \int \cos^{2} x d x) \\ = π \sin x \cos x + π x - π \int \cos^{2} x d x \end{array}

\begin{array}{l} \int π \cos^{2} x d x & = π \int \cos^{2} x d x \\ \overset{*}{=} π (\sin x \cos x + \int \sin^{2} x d x) \\ = π (\sin x \cos x + \int 1 - \cos^{2} x d x) \\ = π (\sin x \cos x + x - \int \cos^{2} x d x) \\ = π \sin x \cos x + π x - π \int \cos^{2} x d x \end{array}

\begin{aligned} u & = \cos x & v^{'} & = \cos x \\ u^{'} & = - \sin x & v & = \sin x \end{aligned}

This gives you the equation below that you have to solve for $\int π \cos^{2} x d x$ :

\begin{array}{l} \int π \cos^{2} x d x & = π \sin x \cos x + π x \\ - π \int \cos^{2} x d x \\ 2 \cdot \int π \cos^{2} x d x & = π \sin x \cos x + π x | \div 2 \\ \int π \cos^{2} x d x & = \frac{π}{2} \sin x \cos x + \frac{π x}{2} \end{array}

\begin{array}{l} \int π \cos^{2} x d x & = π \sin x \cos x + π x - π \int \cos^{2} x d x \\ 2 \cdot \int π \cos^{2} x d x & = π \sin x \cos x + π x | \div 2 \\ \int π \cos^{2} x d x & = \frac{π}{2} \sin x \cos x + \frac{π x}{2} \end{array}

Now you can insert the limits:

\begin{array}{l} V & = \int_{0}^{π} π \cos^{2} x d x \\ = \frac{π}{2} \sin x \cos x + \frac{π x}{2} |_{0}^{π} \\ = \frac{π}{2} \sin π \cos π + \frac{π^{2}}{2} \\ - (\frac{1}{2} \sin 0 \cos 0 + \frac{0}{2}) \\ = \frac{π^{2}}{2} \end{array}

\begin{array}{l} V & = \int_{0}^{π} π \cos^{2} x d x \\ = \frac{π}{2} \sin x \cos x + \frac{π x}{2} |_{0}^{π} \\ = \frac{π}{2} \sin π \cos π + \frac{π^{2}}{2} - (\frac{1}{2} \sin 0 \cos 0 + \frac{0}{2}) \\ = \frac{π^{2}}{2} \end{array}

You have found the volume of the solid of revolution to be

\frac{π^{2}}{2}

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