Volumes of Revolution (AI HL)

Spin a 2D region a full turn about an axis and it sweeps out a solid - this page is about finding that solid's volume with a single integral. The idea is a small step up from area: square the boundary function first, then integrate. It's part of the broader Integration topic.

11 questions on this sub-topic.

Practise volumes of revolution → Try exam-style questions

The formula (and which way round to use it)

Covered under IB syllabus reference AHL5.12. The formula itself is in the booklet - the skill being tested is choosing the correct axis version and setting up \(y^2\) (or \(x^2\)) correctly before you integrate.

Volume of revolution

\[V=\pi\int_a^b y^2\,dx\]

Rotating the region under \(y=f(x)\) a full turn about the \(x\)-axis sweeps out this volume; swap \(x\) and \(y\) to rotate about the \(y\)-axis instead.

✓ In the formula booklet

Choosing \(dx\) or \(dy\)

\[x\text{-axis} \Rightarrow \pi\!\int y^2\,dx \qquad y\text{-axis} \Rightarrow \pi\!\int x^2\,dy\]

The variable you integrate with respect to always matches the axis being rotated about - mixing this up is the single most common setup error.

Need help evaluating \(\int y^2\,dx\) on your calculator? See the parent Integration page.

Worked examples

1
Medium
No calc
[3 marks]

The region bounded by \(y = x^2\), the \(x\)-axis and the lines \(x = 0\) and \(x = 2\) is rotated \(360^\circ\) about the \(x\)-axis.

(a)  Write down an integral for the volume.

(b)  Calculate the exact volume.

Worked solution

(a)   \(V = \pi \displaystyle\int_0^2 x^4\, dx\) A1

(b)   \(V = \pi \left[\dfrac{x^5}{5}\right]_0^2\) M1
\(= \pi \cdot \dfrac{32}{5} = \dfrac{32\pi}{5}\) A1

A1 Integral M1 Integrate A1 Exact answer
2
Hard
GDC
[5 marks]

The region between \(y = \sin x\) and the \(x\)-axis from \(x=0\) to \(x=\pi\) is rotated \(2\pi\) about the \(x\)-axis.

(a) Find the volume, giving your answer to 3 significant figures.

(b) Give the exact value of \(V\) in terms of \(\pi.\)

Worked solution

(a) \(V = \pi\int_0^{\pi} \sin^2 x\,dx.\) M1 A1
Evaluating, \(\int_0^{\pi}\sin^2 x\,dx = \tfrac{\pi}{2},\) so \(V = \tfrac{\pi^2}{2}\) M1
\(\approx 4.93.\) A1

(b) \(V=\dfrac{\pi^2}{2}\) exactly (from the unrounded integral). A1

M1 Stating the correct Formula A1 \(\sin^2 x\) M1 Evaluate A1 Correct answer of \(4.93\) A1 Exact value

Common mistakes

Ready to practise properly?

10 volumes-of-revolution questions, marked instantly like the real exam.

Quick answers

What is the formula for a volume of revolution?

\(V=\pi\int_a^b y^2\,dx\) for a rotation about the \(x\)-axis, or swap \(x\) and \(y\) to rotate about the \(y\)-axis instead.

Do I integrate with respect to x or y?

Match the variable to the axis of rotation: integrate with respect to \(x\) for a rotation about the \(x\)-axis, and with respect to \(y\) for a rotation about the \(y\)-axis.

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