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Calculus Applications of integration

Volume of solids of revolution

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Volume of Solids of Revolution

California Calculus • Standard 16.0 • Applications of Integration

Volume of Solids of Revolution is a topic in Applications of Integration in the California Calculus Standards. It is aligned to Standard 16.0, which requires students to use definite integrals in problems involving the volume of a solid.

A solid of revolution is formed by rotating a region about an axis. The disk method gives its volume as \(V=\pi\int_a^b [f(x)]^2\,dx\), with washer and shell methods for other cases.

California Calculus › Applications of Integration › Volume of Solids of Revolution  —  Standard 16.0

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Theory

Revolving a region about an axis sweeps out a solid of revolution. The disk method gives its volume as \(V=\pi\displaystyle\int_a^b [f(x)]^2\,dx\); the washer method handles a region between two curves, and the shell method revolves about the \(y\)-axis.

Slice the solid perpendicular to the axis of revolution. Each slice is a thin disk of radius \(f(x)\) and thickness \(dx\), with volume \(\pi[f(x)]^2\,dx\). Adding them gives the disk method:

\[V=\pi\int_a^b [f(x)]^2\,dx.\]

If the region lies between two curves, each slice is a washer with outer radius \(R\) and inner radius \(r\): \(V=\pi\int (R^2-r^2)\,dx\).

Revolving about the \(y\)-axis instead is often easier with the shell method, \(V=2\pi\int x\,f(x)\,dx\).

Key idea: a 2-D region plus its axis of revolution is enough to set up the integral — you do not need to draw the 3-D solid.
A region under a line revolved about the x-axis to form a cone Revolving the region under a straight line about the x-axis sweeps out a cone; each thin disk has radius equal to the curve height. x y region axis of revolution
A region revolved about the \(x\)-axis forms a cone (disk method).
An outer and inner radius forming a washer When a region between two curves is revolved, each slice is a washer with an outer radius R and an inner radius r. x y R r
A washer has outer radius \(R\) and inner radius \(r\).

The three volume formulas:

\[V_{\text{disk}}=\pi\int_a^b [f(x)]^2\,dx,\qquad V_{\text{washer}}=\pi\int_a^b (R^2-r^2)\,dx\]
disk volume is pi integral f squared; washer volume is pi integral R squared minus r squared
\[V_{\text{shell}}=2\pi\int_a^b x\,f(x)\,dx\quad(\text{about the } y\text{-axis})\]
shell volume is two pi integral x times f
Square the radius. The disk method uses \([f(x)]^2\), not \(f(x)\).

How to find a volume of revolution

  1. Sketch the region and identify the axis of revolution.
  2. Find the radius (the curve height) for a disk, or \(R\) and \(r\) for a washer.
  3. Integrate \(\pi\int [\text{radius}]^2\,dx\) over the interval.
Example 1 — Disk method (a cone)
Revolve the region under \(y=x\) on \([0,2]\) about the \(x\)-axis. Find the volume.
Solution

Disk method: radius \(x\), so \(V=\pi\displaystyle\int [f]^2\,dx\); antiderivative \(\dfrac{x^3}{3}\).

\(V\)\(=\)\(\pi\int_0^2 x^2\,dx\)
\(=\)\(\pi\left[\dfrac{x^3}{3}\right]_0^2\)
\(=\)\(\pi\cdot\dfrac{8}{3}=\dfrac{8\pi}{3}\)
volume equals eight pi over three cubic units
Example 2 — A square-root solid
Revolve the region under \(y=\sqrt{x}\) on \([0,4]\) about the \(x\)-axis.
Solution

Here \([f]^2=(\sqrt{x})^2=x\); antiderivative \(\dfrac{x^2}{2}\).

\(V\)\(=\)\(\pi\int_0^4 x\,dx\)
\(=\)\(\pi\left[\dfrac{x^2}{2}\right]_0^4\)
\(=\)\(\pi\cdot 8=8\pi\)
volume equals eight pi cubic units
Example 3 — Washer method
Revolve the region between \(y=\sqrt{x}\) and \(y=x^2\) about the \(x\)-axis.
Solution

Washer: \(V=\pi\displaystyle\int (R^2-r^2)\,dx\) with \(R=\sqrt{x}\), \(r=x^2\), so \(R^2-r^2=x-x^4\).

\(V\)\(=\)\(\pi\int_0^1 (x-x^4)\,dx\)
\(=\)\(\pi\left[\dfrac{x^2}{2}-\dfrac{x^5}{5}\right]_0^1\)
\(=\)\(\pi\left(\dfrac{1}{2}-\dfrac{1}{5}\right)=\dfrac{3\pi}{10}\)
volume equals three pi over ten cubic units
Example 4 — Set up the disk integral
Write the volume when \(y=f(x)\) on \([a,b]\) is revolved about the \(x\)-axis.
Solution

Each disk has area \(\pi[f(x)]^2\).

\(V\)\(=\)\(\pi\int_a^b [f(x)]^2\,dx\)
volume is pi times the integral of f squared

Common pitfalls

Square the radius. The disk integrand is \(\pi[f(x)]^2\); forgetting to square is the most common error.
Do not forget \(\pi\). Every disk and washer volume carries a factor of \(\pi\).
Washers subtract squares. Use \(R^2-r^2\), not \((R-r)^2\).

Frequently asked questions

What is the disk method?

Revolving a region about an axis, each slice is a disk of radius \(f(x)\); the volume is \(V=\pi\int_a^b [f(x)]^2\,dx\).

What is the washer method?

For a region between two curves, each slice is a washer; the volume is \(V=\pi\int (R^2-r^2)\,dx\) with outer radius \(R\) and inner radius \(r\).

Why is there a pi in the volume formula?

Because each slice is a circle of area \(\pi r^2\). Multiplying the circular area by the thickness and integrating brings the \(\pi\) along.

When do you use the shell method?

Usually when revolving about the \(y\)-axis: \(V=2\pi\int x\,f(x)\,dx\). It avoids solving the curve for \(x\).

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