NSWMaths Extension 1Syllabus dot point

How do we compute the volume of a solid generated by rotating a region around an axis?

Calculate volumes of revolution about the x-axis and y-axis using the disc method

A focused answer to the HSC Maths Extension 1 dot point on volumes of revolution. The disc method for rotation about the x-axis and y-axis, the integral setup, and the handling of regions bounded by curves and lines, with worked examples.

Generated by Claude OpusReviewed by Better Tuition Academy8 min answerUpdated 2026-05-20

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What this dot point is asking

NESA wants you to set up and evaluate the integral for the volume of a solid of revolution generated by rotating a planar region around the x-axis or y-axis. Extension 1 uses the disc (or washer) method, not shells.

The answer

The disc method (rotation about the x-axis)

Consider a region bounded by $y = f(x)$ , the x-axis, and the vertical lines $x = a$ and $x = b$ (with $f(x) \ge 0$ on $[a, b]$ ).

When rotated about the x-axis, a thin vertical strip at position $x$ of width $dx$ sweeps out a thin disc of radius $f(x)$ and thickness $dx$ . Volume of one disc: $\pi [f(x)]^2 \, dx$ .

Total volume:

V = \pi \int_a^b [f(x)]^2 \, dx.

Rotation about the y-axis

For a region with $x = g(y)$ on $[c, d]$ (with $g(y) \ge 0$ ), rotating about the y-axis gives discs of radius $g(y)$ and thickness $dy$ . Volume:

V = \pi \int_c^d [g(y)]^2 \, dy.

If the curve is given as $y = f(x)$ and you rotate about the y-axis, invert to find $x = g(y)$ first.

The washer method (rotation about an axis, with inner curve)

If the region is bounded above by $y = f(x)$ and below by $y = h(x)$ (with $0 \le h(x) \le f(x)$ ) and rotated about the x-axis:

V = \pi \int_a^b \bigl([f(x)]^2 - [h(x)]^2\bigr) \, dx.

This subtracts the inner volume from the outer.

Setup recipe

Sketch the region (this is essential, do not skip).
Choose the axis of rotation.
Choose vertical strips (for x-axis rotation, integrate $dx$ ) or horizontal strips (for y-axis rotation, integrate $dy$ ).
Identify the limits of integration: the x-range or y-range of the region.
Write the radius (or outer minus inner) as a function of the integration variable.
Set up and evaluate $\pi \int (\text{radius})^2 \, dV$ or the washer version.

Common pitfalls in setup

Confusing the bound with the function: $y = x^2$ rotated about the x-axis between $0$ and $2$ has radius $f(x) = x^2$ , so the integrand is $\pi (x^2)^2 = \pi x^4$ , not $\pi x^2$ .
Wrong integration variable: rotating about the y-axis requires $dy$ as the strip width.
Sign issues: $[f(x)]^2$ is always non-negative, but writing $f(x)^2$ is unambiguous, while $f(x^2)$ would be the squared input.

Worked example

Simple disc, x-axis

Find the volume when $y = x$ between $x = 0$ and $x = 1$ is rotated about the x-axis.

$V = \pi \int_0^1 x^2 \, dx = \pi \cdot \frac{1}{3} = \frac{\pi}{3}$ .

(This is a cone of radius 1 and height 1, so the formula $\frac{1}{3} \pi r^2 h = \frac{\pi}{3}$ confirms.)

Disc, y-axis

Find the volume when $y = x^2$ between $y = 0$ and $y = 4$ is rotated about the y-axis.

$x = \sqrt{y}$ , so $x^2 = y$ .

$V = \pi \int_0^4 y \, dy = \pi \cdot \frac{16}{2} = 8 \pi$ .

Washer

Find the volume when the region between $y = x^2$ and $y = 2 x$ is rotated about the x-axis.

Intersections: $x^2 = 2 x$ , so $x = 0$ or $x = 2$ .

On $[0, 2]$ , $2 x \ge x^2$ , so $y = 2 x$ is outer and $y = x^2$ is inner.

$V = \pi \int_0^2 [(2 x)^2 - (x^2)^2] \, dx = \pi \int_0^2 (4 x^2 - x^4) \, dx = \pi \left[ \frac{4 x^3}{3} - \frac{x^5}{5} \right]_0^2$

$= \pi \left( \frac{32}{3} - \frac{32}{5} \right) = \pi \cdot \frac{160 - 96}{15} = \frac{64 \pi}{15}$ .

Trig integrand

Find the volume when $y = \sin x$ between $x = 0$ and $x = \pi$ is rotated about the x-axis.

$V = \pi \int_0^\pi \sin^2 x \, dx$ .

Use $\sin^2 x = \frac{1 - \cos 2 x}{2}$ :

$V = \pi \int_0^\pi \frac{1 - \cos 2 x}{2} \, dx = \frac{\pi}{2} \left[ x - \frac{\sin 2 x}{2} \right]_0^\pi = \frac{\pi}{2} (\pi - 0) = \frac{\pi^2}{2}$ .

Inverse-trig integrand

Find the volume when $y = \frac{1}{\sqrt{x^2 + 1}}$ between $x = 0$ and $x = 1$ is rotated about the x-axis.

$V = \pi \int_0^1 \frac{1}{x^2 + 1} \, dx = \pi \left[ \arctan x \right]_0^1 = \pi \cdot \frac{\pi}{4} = \frac{\pi^2}{4}$ .

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2022 HSC Q144 marksThe region bounded by

y = x^2

, the x-axis and the line

x = 2

is rotated about the x-axis. Find the volume of the solid formed.

Show worked answer →

Disc method about the x-axis: $V = \pi \int_a^b y^2 \, dx$ .

Here $y = x^2$ , so $y^2 = x^4$ . Limits: $x = 0$ to $x = 2$ .

$V = \pi \int_0^2 x^4 \, dx = \pi \left[ \frac{x^5}{5} \right]_0^2 = \pi \cdot \frac{32}{5} = \frac{32 \pi}{5}$ .

Markers reward writing the disc formula, the squared integrand, the correct limits, and the exact answer.

2021 HSC Q224 marksThe region bounded by

y = \sqrt{x}

, the y-axis and the line

y = 2

is rotated about the y-axis. Find the volume of the solid formed.

Show worked answer →

Disc method about the y-axis: $V = \pi \int_a^b x^2 \, dy$ .

Express $x$ in terms of $y$ : $x = y^2$ , so $x^2 = y^4$ . Limits: $y = 0$ to $y = 2$ .

$V = \pi \int_0^2 y^4 \, dy = \pi \left[ \frac{y^5}{5} \right]_0^2 = \frac{32 \pi}{5}$ .

Markers reward inverting the function to $x$ in terms of $y$ , the disc formula, and the integration step.