Volume of a solid of revolution

What you will be able to do

• find the volume of a solid made by turning a region about the $x$-axis or the $y$-axis;
• use discs for a solid with no hole and washers for a solid with a hole;
• check a result against the volume of a cone or a sphere.

Where this sits. The Indian syllabus (NCERT Class 12, Chapter 8, and JEE Main) asks for areas only. It does not ask for volumes of solids of revolution. A-Level Further Maths, IB Analysis and Approaches Higher Level and AP Calculus (AB and BC) do ask for them, with discs and washers. Exact lists differ between specifications, so check your own syllabus.

Slicing a solid

Cut a loaf of bread into thin slices. The volume of the loaf is the sum of the volumes of the slices. A solid of revolution is cut in the same way.

Take the region under the curve $y = f(x)$ from $x = a$ to $x = b$, and turn it once around the $x$-axis. A slice at position $x$, cut across the axis, is a thin disc. Its radius is $y = f(x)$ and its thickness is $dx$. The face of the disc is a circle, so its volume is

Adding all the discs from $x = a$ to $x = b$ gives

Two checks you already know

A cone. Turn the line $y = \dfrac{r}{h}x$ from $x = 0$ to $x = h$. The radius at the end is $r$ and the height is $h$. Then $V = \pi\displaystyle\int_0^h \dfrac{r^2}{h^2}x^2\,dx = \dfrac{1}{3}\pi r^2 h$. This is the cone formula.

A sphere. Turn the half circle $y = \sqrt{R^2 - x^2}$ from $x = -R$ to $x = R$. Then $V = \pi\displaystyle\int_{-R}^{R} (R^2 - x^2)\,dx = \pi\left(2R^3 - \dfrac{2R^3}{3}\right) = \dfrac{4}{3}\pi R^3$. This is the sphere formula.

About the y-axis

Sometimes the region is turned about the $y$-axis. The slices are now horizontal discs. At height $y$ the radius is $x$, so the disc has volume $\pi x^2\,dy$. Write $x^2$ in terms of $y$ first, then integrate in $y$.

For example, turn $y = x^2$ about the $y$-axis from $y = 0$ to $y = 4$. Here $x^2 = y$, so $V = \pi\displaystyle\int_0^4 y\,dy = 8\pi$. A cylinder of the same radius $2$ and height $4$ has volume $16\pi$, and the bowl holds exactly half of it.

Washers

Now take a region between two curves, an upper curve $y = R(x)$ and a lower curve $y = r(x)$, with $R \ge r \ge 0$. Turned about the $x$-axis, a slice is a disc with a hole in it, called a washer. Its area is the area of the big circle minus the area of the hole.

A worked example

Find the volume made by turning the region between $y = x$ and $y = x^2$ about the $x$-axis.

1. Where they meet. $x = x^2$ gives $x = 0$ and $x = 1$.
2. Which is outer. For $0 < x < 1$, $x$ is above $x^2$. So $R = x$ and $r = x^2$.
3. Integrate. $V = \pi\displaystyle\int_0^1 (x^2 - x^4)\,dx = \pi\left(\dfrac{1}{3} - \dfrac{1}{5}\right) = \dfrac{2\pi}{15}$.

Practise with new numbers

Try it in Python

The program cuts the cone made by $y = \dfrac{x}{2}$, for $0 \le x \le 6$, into $n$ thin discs and adds their volumes. The exact volume is $\dfrac{1}{3}\pi r^2 h$ with $r = 3$ and $h = 6$.

Common mistakes

• Forgetting to square the radius. The face of a disc has area $\pi y^2$, not $\pi y$.
• Forgetting $\pi$. The integral of $y^2$ is only the part under the $\pi$.
• Writing $(R - r)^2$ for a washer. The area of a washer is $\pi R^2 - \pi r^2$, which is $\pi(R^2 - r^2)$. Square each radius first, then subtract.
• Integrating in $x$ when the region is turned about the $y$-axis. Write $x^2$ in terms of $y$ and use $dy$ with limits in $y$.
• Using $x$ limits for a $dy$ integral. The limits belong to the variable you integrate.

Remember these

• About the $x$-axis, discs: $V = \pi\displaystyle\int_a^b y^2\,dx$.
• About the $y$-axis, discs: $V = \pi\displaystyle\int_c^d x^2\,dy$.
• Washers: $V = \pi\displaystyle\int (R^2 - r^2)\,dx$, outer radius $R$ and inner radius $r$.
• A cone, $\dfrac{1}{3}\pi r^2 h$, and a sphere, $\dfrac{4}{3}\pi R^3$, are two quick checks.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. The region under $y = 3x$ from $x = 0$ to $x = 2$ is turned about the $x$-axis. Find the volume.

2. The region between the $y$-axis and the curve $x = y^2$, for $0 \le y \le 2$, is turned about the $y$-axis. Find the volume.

3. A pipe has outer radius $3$ cm, inner radius $1$ cm and length $4$ cm. Show that washers give its volume as $32\pi$ cm$^3$.

4. Why is the area of a washer $\pi(R^2 - r^2)$ and not $\pi(R - r)^2$?

5. The half circle $y = \sqrt{4 - x^2}$ is turned about the $x$-axis. Find the volume and say what solid it is.

References

• OpenStax, Calculus Volume 1, Chapter 6, section 6.2 (determining volumes by slicing: discs and washers).
• NCERT, Mathematics Part II for Class XII, Chapter 8, for the area part of the unit that comes before this lesson.