Volume of Revolution: Method

What a volume of revolution is for

When a flat region is rotated a full turn about a straight line, it sweeps out a three-dimensional solid, a solid of revolution. Finding the volume of that solid is a standard application of integration in the Form 5 Integration chapter of Add Math.

If the region sits under a curve $y=f(x)$ and is rotated about the x-axis, every thin strip of width $dx$ becomes a thin disc of radius $y$ , and adding up all the discs gives the volume. This method turns an area calculation into a volume calculation, and it appears whenever a question asks for the volume generated when a shaded region is rotated $360^{\circ}$ (a full turn) about an axis.

When to reach for it

Reach for a volume of revolution whenever a question shows a region bounded by a curve and an axis and then rotates it, look for phrases such as 'the region is rotated through $360^{\circ}$ about the x-axis' or 'the solid generated when the shaded region is revolved about the y-axis'. The word 'volume' together with 'rotated' or 'revolved' is the clearest signal.

If the rotation is about the x-axis you integrate with respect to $x$ and need $y$ written in terms of $x$ ; if it is about the y-axis you integrate with respect to $y$ and need $x$ written in terms of $y$ . Matching the axis of rotation to the variable of integration is the first decision to make.

The method, step by step

Rotation about the x-axis

V=\pi\int_{a}^{b} y^{2}\,dx

Rotation about the y-axis

V=\pi\int_{a}^{b} x^{2}\,dy

1
Identify the axis of rotation
This decides which formula to use and which variable you integrate with respect to.
2
Rearrange the curve
Get $y$ in terms of $x$ for the x-axis, or $x$ in terms of $y$ for the y-axis.
3
Read off the limits
Find the lower and upper limits $a$ and $b$ that bound the region along the axis of rotation.
4
Square the expression
Form $y^{2}$ (or $x^{2}$ ), expanding any bracket in full before you integrate.
5
Set up and integrate
Write $\pi\int_{a}^{b}$ and integrate the squared expression term by term.
6
Substitute the limits
Put in the upper limit, then the lower limit, and subtract.
7
Multiply by pi
State the volume, giving your answer in cubic units or in terms of $\pi$ .

Worked example

Q1[4 marks]

The region is bounded by the line $y=2x+1$ , the x-axis, and the lines $x=0$ and $x=1$ . Find the volume of the solid generated when this region is rotated through $360^{\circ}$ about the x-axis.

Give your answer in terms of $\pi$ .

Show worked solution

The rotation is about the x-axis, so use $V=\pi\int_{a}^{b} y^{2}\,dx$ with $a=0$ and $b=1$ .

Here $y=2x+1$ , so square it: $y^{2}=(2x+1)^{2}=4x^{2}+4x+1$ . Expand the bracket in full, do not square term by term.

Set up the integral:

V=\pi\int_{0}^{1}\left(4x^{2}+4x+1\right)dx

Integrate term by term:

V=\pi\left[\frac{4x^{3}}{3}+2x^{2}+x\right]_{0}^{1}

Substitute the upper limit $x=1$ : $\frac{4}{3}+2+1=\frac{4}{3}+3=\frac{13}{3}$ . The lower limit $x=0$ gives $0$ .

Therefore $V=\pi\times\frac{13}{3}=\frac{13\pi}{3}$ cubic units.

Common mistakes to avoid

Forgetting to square the curve, integrating $y$ instead of $y^{2}$ , which gives an area, not a volume.
Squaring a bracket term by term, e.g. writing $(2x+1)^{2}=4x^{2}+1$ and dropping the middle term $4x$ .
Leaving out the $\pi$ in the final answer.
Integrating with respect to the wrong variable, using $dx$ when the rotation is about the y-axis.
Using the wrong limits: they must be the $x$ -values (or $y$ -values) that bound the region along the axis of rotation.

How one-to-one teaching helps

The step most students slip on is squaring the curve: they either integrate $y$ instead of $y^{2}$ , or they square a bracket term by term and lose the middle term. In a one-to-one lesson our teachers watch that exact line as you write it, so the habit of expanding $(2x+1)^{2}$ in full becomes automatic and the $\pi$ is never dropped.

Because your working is shown step by step, every method mark stays visible and easy to credit. Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English.

To see how we teach volumes of revolution, message us on WhatsApp to arrange a one-hour paid class from RM50/hr at the teacher's rate.

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Frequently asked questions

Which formula do I use for a volume of revolution?

About the x-axis, use $V=\pi\int_{a}^{b} y^{2}\,dx$ ; about the y-axis, use $V=\pi\int_{a}^{b} x^{2}\,dy$ . Choose the one that matches the axis of rotation, and make sure the curve is written in terms of the correct variable before you square it.

Do I need to memorise the volume of revolution formula?

Yes, this is a formula you are expected to know and write from memory. Practise pairing the right variable of integration with the axis of rotation so the setup becomes automatic.

Why is there a

\pi

in the formula?

Each thin slice of the solid is a disc whose area is $\pi r^{2}$ , with the radius equal to $y$ (or $x$ ). Integrating $\pi y^{2}$ adds up all the discs, so the $\pi$ comes from the area of a circle and must appear in your answer.

What if the region is rotated about the y-axis instead?

Then integrate with respect to $y$ : rearrange the curve to get $x$ in terms of $y$ , square it, and use $V=\pi\int_{a}^{b} x^{2}\,dy$ with the $y$ -limits that bound the region.

Source:SRC-DSKP-EN

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

How to find a volume of revolution