Volumes of Revolution (AA HL)

Q: What is the formula for a volume of revolution about the x-axis?

V = pi times the integral from a to b of y^2 dx, for the solid formed when the region under a curve is rotated fully about the x-axis.

Q: How do I find a volume of revolution about the y-axis?

Use the same formula with x and y swapped: V = pi times the integral of x^2 dy, rewriting the curve's equation to give x in terms of y first.

Spin the region under a curve fully around an axis and you get a solid - a volume of revolution. Integrating \(y^2\) (or \(x^2\)) along the axis of rotation, scaled by \(\pi\), gives its exact volume. This page covers the formula for rotating about either axis, worked examples, and the mistakes that lose marks. It's part of the broader Integration topic.

19 questions on this sub-topic.

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The formula

Covered under IB syllabus reference AHL5.17: the area of the region enclosed by a curve and the \(y\)-axis in a given interval, and volumes of revolution about the \(x\)-axis or \(y\)-axis. The core formula is in the formula booklet - the skill is picking the right variable to integrate with respect to.

Volume about the x-axis

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

Rotating the region under a curve fully around the \(x\)-axis between \(x=a\) and \(x=b\).

✓ In the formula booklet

Volume about the y-axis

\[V = \pi\int_c^d x^2\,dy\]

Swap \(x\) and \(y\) roles: rewrite the curve as \(x\) in terms of \(y\), then integrate with respect to \(y\) between the two \(y\)-limits.

Same formula, x and y swapped

Need the full syllabus wording and formula-booklet reference table? See Integration.

Worked examples

Hard

GDC

[5 marks]

The region bounded by \(y = x^2\), \(y = 4\) and the \(y\)-axis (first quadrant) is rotated about the \(y\)-axis.

Find the volume.

Worked solution

\(V = \pi\int_0^4 x^2\,dy\) with \(x^2\) M1 \(= y.\) A1
\(V = \pi\int_0^4 y\,dy\) M1 \(= \pi\left[\tfrac{y^2}{2}\right]_0^4\) A1 \(= 8\pi \approx 25.1.\) A1

M1 Formula (about \(y\)) A1 \(x^2=y\) M1 Set up A1 Antiderivative A1 Limits, \(8\pi\)

Medium

GDC

[5 marks]

Find the area between \(y=4-x^2\) and the x-axis.

Worked solution

Curve meets axis at \(x\) M1
\(= \pm 2.\) A1
\(\int_{-2}^{2}(4 - x^2)\,dx\) M1
\(= [4x - \tfrac{x^3}{3}]_{-2}^{2}\) A1
\(= \tfrac{32}{3} \approx 10.7.\) A1

M1 Roots A1 \(\pm2\) M1 Integral A1 Antiderivative A1 Correct answer of \(\approx10.7\)

Easy

Calculator

[3 marks]

Find the area enclosed between the curve \(y = x^2\), the \(x\)-axis and the lines \(x = 1\) and \(x = 3\).

Worked solution

\(A = \int_1^3 x^2\,dx.\) M1
\([\tfrac{x^3}{3}]_1^3 = \tfrac{27}{3} - \tfrac13 = \tfrac{26}{3}\) A1 \(\approx 8.67.\) A1

M1 Set up integral A1 Antiderivative A1 Correct answer of \(\approx8.67\)

Common mistakes

Forgetting to square \(y\) (or \(x\)) before integrating. The formula is \(\pi\int y^2\,dx\), not \(\pi\int y\,dx\) - skipping the square is one of the most common slips on this topic.
Dropping the \(\pi\). Every volume-of-revolution answer needs the constant \(\pi\) out front - it's easy to lose when the working gets long.
Not rewriting the curve before rotating about the y-axis. A rotation about the \(y\)-axis needs \(x^2\) as a function of \(y\), so an equation given as \(y=f(x)\) must be rearranged into \(x=g(y)\) first.

Full topicIntegration RelatedIntegration by Parts FormulasAA HL formula booklet explained PrintAA HL formula sheet

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Quick answers

What is the formula for a volume of revolution about the x-axis?

\(V = \pi\int_a^b y^2\,dx\), for the solid formed when the region under a curve is rotated fully about the \(x\)-axis.

How do I find a volume of revolution about the y-axis?

Use the same formula with \(x\) and \(y\) swapped: \(V = \pi\int_c^d x^2\,dy\), rewriting the curve's equation to give \(x\) in terms of \(y\) first.

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