Volumes of solids of revolution
Find volumes of solids of revolution in Year 12 Specialist Mathematics for Queensland (QCAA). When a region is rotated a full turn about an axis it sweeps out a solid, and the disc method adds up thin circular slices to give its volume by integration.
You will learn to rotate a region about the \(x\)-axis using \(V=\pi\int[f(x)]^2\,dx\) and about the \(y\)-axis using \(V=\pi\int[g(y)]^2\,dy\), handle a region between two curves with the washer method, and leave every volume in exact \(\pi\) form — a key application of integral calculus in Unit 4.
Theory
A solid of revolution is formed by rotating a region about an axis. In Year 12 Specialist Mathematics (QCAA, Queensland) you find its volume by the disc method: each thin slice is a disc of radius equal to the function value, so about the \(x\)-axis \(V=\pi\displaystyle\int_a^b [f(x)]^2\,dx\) and about the \(y\)-axis \(V=\pi\displaystyle\int_c^d [g(y)]^2\,dy\). Answers are left in exact \(\pi\) form.
A solid of revolution is generated when a plane region is rotated a full turn about a straight line called the axis of revolution. If you slice the solid perpendicular to that axis, every cross-section is a disc (a circle).
Rotating the region under \(y=f(x)\) about the \(x\)-axis: a slice at position \(x\) is a disc of radius \(f(x)\) and thickness \(dx\), so its volume is \(\pi[f(x)]^2\,dx\). Adding the discs from \(x=a\) to \(x=b\) gives \(V=\pi\displaystyle\int_a^b [f(x)]^2\,dx\).
Rotating about the \(y\)-axis works the same way with the roles of \(x\) and \(y\) swapped. First make \(x\) the subject, \(x=g(y)\); a disc at height \(y\) has radius \(g(y)\), so \(V=\pi\displaystyle\int_c^d [g(y)]^2\,dy\).
When the region lies between two curves, each slice is a washer (a disc with a hole). With outer radius \(R\) and inner radius \(r\), the volume is \(V=\pi\displaystyle\int_a^b \big(R^2-r^2\big)\,dx\). Because \(\pi\) is a factor throughout, volumes are given in exact \(\pi\) form such as \(\dfrac{32\pi}{5}\).
Rotating the region under \(y=f(x)\) (between the curve, the \(x\)-axis and \(x=a,\,x=b\)) about the \(x\)-axis:
Rotating the region between \(x=g(y)\), the \(y\)-axis and \(y=c,\,y=d\) about the \(y\)-axis:
For a region between two curves (outer radius \(R\), inner radius \(r\)) rotated about the \(x\)-axis — the washer method:
How to find a volume of revolution
- Choose the axis and write the formula: about the \(x\)-axis use \(V=\pi\int_a^b [f(x)]^2\,dx\); about the \(y\)-axis first make \(x\) the subject, \(x=g(y)\), then use \(V=\pi\int_c^d [g(y)]^2\,dy\).
- Square the radius and simplify the integrand — for a region between two curves subtract the inner square from the outer square, \(R^2-r^2\).
- Integrate and substitute the limits, keeping \(\pi\) as a factor.
- Evaluate and state the exact volume in \(\pi\) form (a decimal only if asked).
Use \(V=\pi\displaystyle\int_a^b [f(x)]^2\,dx\) and square the radius \((x^2)^2=x^4\):
| \(V\) | \(=\) | \(\pi\int_0^1 \left(x^2\right)^2\,dx\) |
| \(=\) | \(\pi\int_0^1 x^4\,dx\) |
Integrate, then substitute the limits:
| \(=\) | \(\pi\left[\dfrac{x^5}{5}\right]_0^1\) | |
| \(=\) | \(\pi\left(\dfrac{1}{5}-0\right)\) | |
| \(=\) | \(\dfrac{\pi}{5}\) |
The volume is \(\dfrac{\pi}{5}\) units\(^3\).
Square the radius \((2x)^2=4x^2\) and set up the integral:
| \(V\) | \(=\) | \(\pi\int_0^3 (2x)^2\,dx\) |
| \(=\) | \(\pi\int_0^3 4x^2\,dx\) |
Integrate and substitute the limits:
| \(=\) | \(\pi\left[\dfrac{4x^3}{3}\right]_0^3\) | |
| \(=\) | \(\pi\left(\dfrac{4\times 27}{3}-0\right)\) | |
| \(=\) | \(36\pi\) |
Compare with \(V=k\pi\):
| \(k\) | \(=\) | \(36\) |
\(k=36\), so the volume is \(36\pi\) units\(^3\).
About the \(y\)-axis, make \(x^2\) the subject so the radius squared is in terms of \(y\):
| \(y\) | \(=\) | \(x^2\) |
| \(x^2\) | \(=\) | \(y\) |
Apply \(V=\pi\displaystyle\int_c^d [g(y)]^2\,dy\) with \([g(y)]^2=x^2=y\):
| \(V\) | \(=\) | \(\pi\int_0^3 x^2\,dy\) |
| \(=\) | \(\pi\int_0^3 y\,dy\) |
Integrate and substitute the limits:
| \(=\) | \(\pi\left[\dfrac{y^2}{2}\right]_0^3\) | |
| \(=\) | \(\pi\left(\dfrac{9}{2}-0\right)\) | |
| \(=\) | \(\dfrac{9\pi}{2}\) |
The volume is \(\dfrac{9\pi}{2}\) units\(^3\).
On \([0,1]\) the line \(y=x\) is above \(y=x^2\), so the outer radius is \(x\) and the inner is \(x^2\):
| \(V\) | \(=\) | \(\pi\int_0^1 \left[(x)^2-\left(x^2\right)^2\right]\,dx\) |
| \(=\) | \(\pi\int_0^1 \left(x^2-x^4\right)\,dx\) |
Integrate each term, then substitute the limits:
| \(=\) | \(\pi\left[\dfrac{x^3}{3}-\dfrac{x^5}{5}\right]_0^1\) | |
| \(=\) | \(\pi\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\) | |
| \(=\) | \(\pi\times\dfrac{5-3}{15}\) | |
| \(=\) | \(\dfrac{2\pi}{15}\) |
The volume is \(\dfrac{2\pi}{15}\) units\(^3\).
Common pitfalls
Frequently asked questions
What is the formula for the volume of a solid of revolution?
About the \(x\)-axis, \(V=\pi\displaystyle\int_a^b [f(x)]^2\,dx\); about the \(y\)-axis, \(V=\pi\displaystyle\int_c^d [g(y)]^2\,dy\). Each slice is a disc whose radius is the distance from the axis to the curve.
Why do you square the function in the volume formula?
Each cross-section is a disc of radius \(r=f(x)\) and area \(\pi r^2=\pi[f(x)]^2\). Multiplying that area by the thickness and adding up the discs gives the integral of \(\pi[f(x)]^2\).
How do you find a volume when rotating about the y-axis?
Make \(x\) the subject, \(x=g(y)\), so the disc radius is in terms of \(y\). Then integrate with respect to \(y\) between the \(y\)-limits: \(V=\pi\displaystyle\int_c^d [g(y)]^2\,dy\).
What is the washer method?
When the region lies between two curves, each slice is a disc with a hole (a washer). With outer radius \(R\) and inner radius \(r\), \(V=\pi\displaystyle\int_a^b (R^2-r^2)\,dx\) — square each radius, then subtract.
Should the answer be left in terms of pi?
Yes. Since \(\pi\) is a factor of every volume of revolution, leave the exact answer in \(\pi\) form, such as \(8\pi\) or \(\dfrac{32\pi}{5}\). Give a decimal only when the question asks for one.
What are the units of a volume of revolution?
Volume is measured in cubic units, written units\(^3\). The number multiplies \(\pi\), so a typical answer looks like \(\dfrac{25\pi}{2}\) units\(^3\).