What is wrong variable for the axis?

Rotating about the y-axis means integrating π x^2\,dy in terms of y, not π y^2\,dx.

What are mismatched limits?

Use x-limits when integrating in x and y-limits when integrating in y.

Write the disc-method volume for rotating y = f(x), a x b, about the x-axis. [1 mark]

Find the volume when y = sqrt(x), 0 x 4, is rotated about the x-axis. [3 marks]

§-Syllabus dot point

VICSpecialist MathematicsSyllabus dot point

How does integration give the volume of a solid formed by rotating a region about an axis, and how do we choose between disc, washer and shell setups?

The use of definite integrals to find the volume of a solid of revolution generated by rotating a region about the $x$ -axis or $y$ -axis, using the disc and washer (annulus) methods, and the setting up of the appropriate integral

A focused answer to the VCE Specialist Mathematics Unit 4 key-knowledge point on volumes of revolution. The disc and washer methods for rotation about the x-axis and y-axis, choosing the integration variable, and setting up the integral, with a verified worked example.

Generated by Claude Opus 4.87 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The disc method
The washer (annulus) method
Rotation about the $y$ -axis
Setting up from the wording
Examples in context
Try this

What this dot point is asking

VCAA wants you to compute the volume of a solid formed when a region is rotated about the $x$ -axis or $y$ -axis, by setting up and evaluating a definite integral. You need the disc method for a region against the axis and the washer (annulus) method when there is a gap, and you must choose the integration variable to match the axis of rotation.

The disc method

Slice the solid into thin discs perpendicular to the axis of rotation. Rotating the region under $y = f(x)$ about the $x$ -axis, a slice at position $x$ is a disc of radius $f(x)$ and thickness $dx$ , with volume $\pi [f(x)]^2\,dx$ . Summing over the interval gives

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

The radius is the distance from the axis to the curve, here simply $y = f(x)$ . The square is essential: the disc area is $\pi r^2$ .

The washer (annulus) method

When the region lies between two curves $y_{\text{outer}}$ and $y_{\text{inner}}$ (both measured from the axis), each slice is a washer: a disc with a hole. Its area is the outer disc minus the inner disc, so

V = \int_a^b \pi\left([y_{\text{outer}}]^2 - [y_{\text{inner}}]^2\right) dx.

Subtract the squares of the radii, not the radii themselves; $[y_o]^2 - [y_i]^2 \ne (y_o - y_i)^2$ .

Rotation about the $y$ -axis

To rotate about the $y$ -axis, slice horizontally. Each disc has radius $x$ (the distance from the $y$ -axis to the curve) and thickness $dy$ , so you express $x$ as a function of $y$ and integrate with $y$ -limits:

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

The rule is: match the integration variable to the axis. Rotation about the $x$ -axis integrates in $x$ with radius $y$ ; rotation about the $y$ -axis integrates in $y$ with radius $x$ .

Worked example

A solid from a parabola

The region bounded by $y = x^2$ , the $x$ -axis, and the line $x = 2$ is rotated about the $x$ -axis. Find the volume.

Step 1: Set up the disc integral

Rotating about the $x$ -axis, each thin vertical slice of the region sweeps out a disc. The disc at position $x$ has radius equal to the function value $y = x^2$ and thickness $dx$ , so its volume is $\pi (x^2)^2\,dx$ . Integrating over $x$ from $0$ to $2$ accumulates all the discs.

V = \int_0^2 \pi (x^2)^2\,dx = \pi\int_0^2 x^4\,dx.

Step 2: Integrate and evaluate

Apply the power rule to $x^4$ , then substitute the upper limit $x = 2$ and subtract the value at $x = 0$ . The lower limit contributes zero, simplifying the arithmetic.

\pi\int_0^2 x^4\,dx = \pi\left[\frac{x^5}{5}\right]_0^2 = \pi\left(\frac{2^5}{5} - 0\right) = \pi\cdot\frac{32}{5} = \frac{32\pi}{5}.

Checking the squaring: the radius is $x^2$ , so the squared radius is $(x^2)^2 = x^4$ , and $\int x^4\,dx = \frac{x^5}{5}$ ; evaluating at $2$ gives $\frac{32}{5}$ , confirming the arithmetic.

Final answer: $V = \dfrac{32\pi}{5} \approx 20.1$ cubic units.

Setting up from the wording

The first decision is the axis of rotation, because it fixes the integration variable and the radius. Rotation about the $x$ -axis integrates $\pi y^2$ with respect to $x$ , using $x$ -limits; rotation about the $y$ -axis integrates $\pi x^2$ with respect to $y$ , using $y$ -limits. The second decision is disc versus washer: if the region touches the axis along its whole width, use a single disc; if there is a gap between the region and the axis (typically because two curves bound it), use a washer and subtract the squared inner radius. Sketching the region and a representative slice before writing the integral prevents almost every set-up error, since the slice shows the radius and the thickness directly.

Many VCAA items ask only for the integral (a "write down" mark) and then for its evaluation, so practise stopping cleanly at the correct integrand before computing. Keeping the constant $\pi$ outside the integral and squaring the radius first keeps the working tidy.

Examples in context

Example 1. Rotating $y = r$ (a horizontal line) for $0 \le x \le h$ about the $x$ -axis gives a cylinder, $V = \int_0^h \pi r^2\,dx = \pi r^2 h$ .

Example 2. Rotating the region between $y = x$ and $y = x^2$ on $[0, 1]$ about the $x$ -axis uses the washer $\pi(x^2 - x^4)$ , since $y = x$ is outer.

Try this

Q1. Write the disc-method volume for rotating $y = f(x)$ , $a \le x \le b$ , about the $x$ -axis. [1 mark]

Cue. $\int_a^b \pi [f(x)]^2\,dx$ .

Q2. Find the volume when $y = \sqrt{x}$ , $0 \le x \le 4$ , is rotated about the $x$ -axis. [3 marks]

Cue. $\int_0^4 \pi x\,dx = \pi\left[\frac{x^2}{2}\right]_0^4 = 8\pi$ .

Q3. State the washer integrand for outer radius $R(x)$ and inner radius $r(x)$ . [1 mark]

Cue. $\pi(R^2 - r^2)$ .

Exam-style practice questions

Practice questions written in the style of VCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

VCAA 2022 Exam 14 marksThe curve

y^2 = x - 1

, where

2 \le x \le 5

, is rotated about the

x

-axis to form a solid of revolution. (a) Write down the definite integral, in terms of

x

, for the volume. (b) Evaluate the volume exactly.

Show worked answer →

(a) For rotation about the $x$ -axis, the disc method gives $V = \displaystyle\int_a^b \pi y^2\,dx$ . Since $y^2 = x - 1$ and $2 \le x \le 5$ :

$V = \displaystyle\int_2^5 \pi(x - 1)\,dx$ .

(b) Antidifferentiate: $\displaystyle\int (x - 1)\,dx = \dfrac{(x - 1)^2}{2}$ . Evaluate:

$V = \pi\left[\dfrac{(x - 1)^2}{2}\right]_2^5 = \pi\left(\dfrac{16}{2} - \dfrac{1}{2}\right) = \pi\left(8 - \dfrac{1}{2}\right) = \dfrac{15\pi}{2}$ .

Markers reward the disc integral with $y^2 = x - 1$ substituted, and the exact value $\frac{15\pi}{2}$ .

VCAA 2023 Exam 25 marksThe region bounded by

y = x

and

y = x^2

for

0 \le x \le 1

is rotated about the

x

-axis. (a) Explain why the washer method is required and identify the outer and inner radii. (b) Calculate the exact volume of the solid generated.

Show worked answer →

(a) On $[0, 1]$ , $y = x$ lies above $y = x^2$ , so the region has a gap from the axis and each cross-section is a washer. The outer radius is $y_{\text{outer}} = x$ and the inner radius is $y_{\text{inner}} = x^2$ .

(b) $V = \displaystyle\int_0^1 \pi\left(x^2 - (x^2)^2\right)dx = \pi\int_0^1 \left(x^2 - x^4\right)dx$ .

$= \pi\left[\dfrac{x^3}{3} - \dfrac{x^5}{5}\right]_0^1 = \pi\left(\dfrac{1}{3} - \dfrac{1}{5}\right) = \pi\cdot\dfrac{2}{15} = \dfrac{2\pi}{15}$ .

Markers reward identifying the washer with correct radii, subtracting the squares, and the exact volume $\frac{2\pi}{15}$ .