Learn9709Paper 1 · Pure Mathematics 1IntegrationVolumes of revolution

Integration · Subtopic 5 of 5

Volumes of Revolution

Spinning a region around an axis makes a solid. Slicing it into thin discs gives $V=\pi\displaystyle\int y^2\,dx$ about the $x$ -axis and $V=\pi\displaystyle\int x^2\,dy$ about the $y$ -axis, plus regions between two graphs.

Cambridge 9709 Paper 1Edexcel YMA01 WMA14≈ 45 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

Explain the volume formula by slicing the solid into thin discs.
Find volumes of revolution about the $x$ -axis using $V=\pi\displaystyle\int_a^b y^2\,dx$ .
Find volumes about the $y$ -axis using $V=\pi\displaystyle\int_c^d x^2\,dy$ .
Find the volume when a region between two graphs is rotated.

Pause and think

Rotate the line $y=x$ from $x=0$ to $x=3$ through $360^\circ$ about the $x$ -axis. What solid do you get, and what is its volume?

Section 2 of 14

Core idea

Start hereA vase from a curve

Put a curve on a potter's wheel and spin it: it sweeps out a 3D shape, a solid of revolution. How much water does it hold? Slice it like a loaf: each slice is a thin disc, a tiny cylinder whose volume we know. Add them up with an integral.

The big idea

Rotating $y=f(x)$ for $a\le x\le b$ through $360^\circ$ about the $x$ -axis gives volume $V=\pi\displaystyle\int_a^by^2\,dx$ . About the $y$ -axis: $V=\pi\displaystyle\int_c^dx^2\,dy$ .

1. The disc method

Spinning y = √x: the disc at x

x = 0.20

The bold line is the edge-on view of one disc. Its area is

\pi y^2=\pi x

; adding them from

0

4

V=\pi\displaystyle\int_0^4x\,dx=8\pi

V=\pi\int y^2\,dx

$\text{disc: radius }y,\ \text{thickness }\delta x$
A thin slice perpendicular to the axis.
$\delta V\approx\pi y^2\,\delta x$
Cylinder volume $\pi r^2h$ .
$V=\lim_{\delta x\to0}\sum\pi y^2\,\delta x=\pi\int_a^by^2\,dx$
Add all discs; thinner slices give the exact value.

Worked example $y=\sqrt x$ , $0\le x\le4$ , about the $x$ -axis

Find the volume.

Goal: Exact volume.
Strategy: Square $y$ , integrate.

$y^2=x$
Square first.
$V=\pi\int_0^4x\,dx=\pi\left[\tfrac{x^2}2\right]_0^4=8\pi$
Integrate.

Answer

8\pi\approx25.1

✓ CheckIt fits inside a cylinder of radius

2

, length

4

(

16\pi

): exactly half ✓ (a paraboloid is half its cylinder).

Worked exampleCheck: the cone formula

Rotate $y=\frac rhx$ , $0\le x\le h$ , about the $x$ -axis.

Goal: Show $V=\frac13\pi r^2h$ .
Strategy: Disc method with letters.

$V=\pi\int_0^h\frac{r^2}{h^2}x^2\,dx$
Square $y$ .
$=\pi\frac{r^2}{h^2}\cdot\frac{h^3}3=\tfrac13\pi r^2h$
Integrate; simplify.

Answer

\frac13\pi r^2h

✓.

✓ CheckMatches the known cone formula.

2. Washers (hollow solids)

Worked exampleBetween $y=x$ and $y=x^2$ , about the $x$ -axis

Find the volume.

Goal: Volume of a hollow solid.
Strategy: Outer volume minus inner volume.

$x=x^2\Rightarrow x=0,1;\ \ y=x\text{ is outside}$
Limits; on $(0,1)$ , $x>x^2$ .
$V=\pi\int_0^1(x^2-x^4)\,dx$
$y_{\text{out}}^2-y_{\text{in}}^2$ .
$=\pi\left(\tfrac13-\tfrac15\right)=\tfrac{2\pi}{15}$
Integrate.

Answer

\frac{2\pi}{15}

✓ CheckPositive and less than the outer cone's

\frac\pi3

✓.

Common misconceptionSquaring the difference Use

\pi\int(y_{\text{out}}^2-y_{\text{in}}^2)dx

, not

\pi\int(y_{\text{out}}-y_{\text{in}})^2dx

Common misconceptionForgetting π Every disc is a circle: the

\pi

is essential.

Pause and think

$y=x^2$ from $y=0$ to $y=4$ rotated about the $y$ -axis?

Summary

$V=\pi\int y^2dx$ or $\pi\int x^2dy$ .
Washers: subtract squares.

Solid of revolution: Shape swept by rotating a region.

Rotating y = √x about the x-axis

Each thin slice becomes a disc of radius

y

, so

V=\pi\displaystyle\int_0^4 y^2\,dx=\pi\int_0^4 x\,dx=8\pi

HistoryArchimedes' favourite result. Archimedes proved that a sphere has exactly

\tfrac23

the volume of the cylinder that just contains it. He was so proud of this that he asked for a sphere inside a cylinder to be carved on his tomb. The Roman writer Cicero found the overgrown tomb in Syracuse in 75 BC by spotting that carving.

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