Volumes of solids of revolution
Find volumes of solids of revolution in Year 12 VCE Specialist Mathematics. 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. It sits in the Calculus area of study of the VCE Mathematics Study Design (VCAA), within the Integral calculus topic of Unit 4.
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 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\(^3\).