What is rotation about the y-axis?

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

What is wrong axis for the strip?

Rotation about the x-axis uses vertical strips (dx); rotation about the y-axis uses horizontal strips (dy).

What is limits in the wrong variable?

If you switch from dx to dy, the limits change to the y-range of the region.

What is sign error in the washer formula?

Outer minus inner, both squared. Reversing the order or forgetting to subtract gives nonsense.

§-Syllabus dot point

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 built up stage by stage, the washer method, the integral setup, and the handling of regions bounded by curves and lines, with worked examples.

Generated by Claude Opus 4.815 min answerUpdated 2026-06-20

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

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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 the y-axis. Extension 1 uses the disc method (and its washer extension for a region with a hole), not cylindrical shells. The single skill being tested is turning a picture of a spinning region into a definite integral of $\pi \times (\text{radius})^2$ , with the right radius, the right variable, and the right limits.

The idea worth internalising is that a solid of revolution is a stack of thin circular discs. Slice the solid perpendicular to the axis of rotation and every slice is a circle; the radius of that circle is the distance from the axis out to the boundary curve. Because a circle's area is $\pi r^2$ , the radius gets squared before you integrate, which is the step students most often drop. Once you see the solid as a stack of discs, the formula is just "add up the disc volumes", and adding up infinitely many infinitely thin discs is precisely what the integral does.

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 $dx$ . One disc has volume $\pi [f(x)]^2 \, dx$ (area of a circle times thickness), and summing them gives

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

Key fact

The volume of a solid of revolution is $V = \pi \displaystyle\int [\text{radius}]^2 \, d(\text{strip variable})$ , where the radius is the perpendicular distance from the axis of rotation to the boundary curve. Rotating about the x-axis pairs with vertical strips and $\int [f(x)]^2\,dx$ ; rotating about the y-axis pairs with horizontal strips and $\int [g(y)]^2\,dy$ (so you must rearrange to $x = g(y)$ first). The radius is squared before integrating, never after, and the factor of $\pi$ is never optional. For a region with a hole, use the washer form $\pi\int([\text{outer}]^2 - [\text{inner}]^2)$ , the difference of the squares.

The disc method, stage by stage

The disc method is best seen as a short film: a flat region spins about an axis, sweeps out a solid, and that solid is then sliced back into discs to add up. The four panels build exactly that picture for the region under $y = \sqrt{x}$ from $x = 0$ to $x = 4$ , rotated about the x-axis.

Stage 1, draw the region. Shade the region bounded by $y = \sqrt{x}$ , the x-axis and the line $x = 4$ . This flat region is what gets rotated; every later step refers back to it, so a clear sketch is not optional. Its right-hand edge has height $\sqrt{4} = 2$ .

Stage 2, take a representative disc. Picture a thin vertical strip of the region at a typical $x$ . When the region spins about the x-axis, that strip sweeps out a thin disc whose radius is the height of the curve, $r = f(x) = \sqrt{x}$ , and whose thickness is $dx$ . Its circular face is the ellipse straddling the axis.

Stage 3, sweep the whole region. A full turn about the x-axis sends every strip around, sweeping the region into a solid (here a horn-shaped paraboloid). Its outline is the curve above the axis and its mirror image below; the far end at $x = 4$ is a circular cap of radius $f(4) = 2$ .

Stage 4, sum the discs. Fill the solid with these thin discs. One disc has volume $\pi r^2\,dx = \pi[f(x)]^2\,dx$ , and adding them all from $a$ to $b$ turns the sum into an integral. For $f(x) = \sqrt{x}$ , $[f(x)]^2 = x$ , so

V = \pi \int_0^4 (\sqrt{x})^2 \, dx = \pi \int_0^4 x \, dx = \pi \left[\frac{x^2}{2}\right]_0^4 = 8\pi.

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 $dy$ , so

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, and convert the limits to $y$ -values. The axis of rotation decides the variable every time: rotate about the x-axis and integrate in $x$ ; rotate about the y-axis and integrate in $y$ . Forgetting to invert (and leaving the integral in $x$ ) is the single most common y-axis mistake.

The washer method (region with an 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, each cross section is an annulus, a washer, whose area is the big circle minus the hole, $\pi R^2 - \pi r^2$ :

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

This is the difference of the squares, $[f]^2 - [h]^2$ , and emphatically not $\pi\int(f - h)^2\,dx$ : squaring the gap would measure a disk whose radius is the gap, which is a different solid. Subtract the inner squared radius from the outer squared radius.

Setup recipe

Sketch the region (this is essential, do not skip).
Choose the axis of rotation.
Choose vertical strips (for x-axis rotation, integrate in $x$ ) or horizontal strips (for y-axis rotation, integrate in $y$ ).
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, and square it.
Evaluate $\pi \int [\text{radius}]^2 \, dx$ (or $dy$ ), 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, with $y$ -limits.
Squaring the difference instead of differencing the squares in a washer: use $[f]^2 - [h]^2$ .

How exam questions ask about volumes of revolution

The wording fixes both the method and the variable:

"The region bounded by $y = f(x)$ , the x-axis and $x = b$ is rotated about the x-axis." Disc method in $x$ : $V = \pi\int_a^b [f(x)]^2\,dx$ . Square the function, use the $x$ -limits.
"... rotated about the y-axis." Disc method in $y$ : rearrange to $x = g(y)$ first, then $V = \pi\int_c^d [g(y)]^2\,dy$ with $y$ -limits. Inverting and converting the limits is where the marks are won or lost.
"The region between $y = f(x)$ and $y = h(x)$ is rotated about the x-axis." Washer method: find the intersections for the limits, decide which curve is outer, then $V = \pi\int([f]^2 - [h]^2)\,dx$ .
"Find the volume of the solid formed", with a trig or inverse-trig boundary. Squaring may produce $\sin^2 x$ (use the half-angle identity) or $\frac{1}{1 + x^2}$ (an $\arctan$ integral). The squaring step links this dot point to the rest of the calculus topic.
"Give your answer in exact form" or no rounding instruction. Leave $\pi$ , surds and fractions in place, and label volumes "cubic units".
"Show that the volume is ..." Set up the integral, evaluate, and reach the stated value; the method marks are for the correct disc or washer set-up even before the arithmetic.

Worked example

Rotate about the x-axis (disc method)

To find the volume when $y = x$ on $[0, 1]$ is rotated about the x-axis, each thin vertical strip at position $x$ sweeps out a disc of radius $f(x) = x$ and thickness $dx$ . The disc formula gives $V = \pi \int_0^1 [f(x)]^2 \, dx$ .

V = \pi \int_0^1 x^2 \, dx = \pi \left[\frac{x^3}{3}\right]_0^1 = \frac{\pi}{3}.

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

Final answer: $\dfrac{\pi}{3}$ .

Rotate about the y-axis (disc method, inverting the function)

To find the volume when $y = x^2$ on the region $0 \le y \le 4$ is rotated about the y-axis, horizontal strips are needed. From $y = x^2$ , we get $x = \sqrt{y}$ , so $x^2 = y$ . The disc radius at height $y$ is $g(y) = \sqrt{y}$ , so the integrand is $[g(y)]^2 = y$ .

Evaluate the integral:

V = \pi \int_0^4 y \, dy = \pi \left[\frac{y^2}{2}\right]_0^4 = \pi \cdot 8 = 8\pi.

Final answer: $8\pi$ .

Rotate a region with a hole (washer method)

For the region between $y = x^2$ and $y = 2x$ rotated about the x-axis, first find where the curves meet: $x^2 = 2x$ gives $x = 0$ or $x = 2$ . On $[0, 2]$ , test $x = 1$ : $2(1) = 2 > 1 = 1^2$ , so $y = 2x$ is the outer curve. The washer method subtracts the inner volume from the outer.

Set up and evaluate the washer integral:

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

Final answer: $\dfrac{64\pi}{15}$ .

Rotate a sine boundary (half-angle identity)

For $y = \sin x$ on $[0, \pi]$ rotated about the x-axis, the integrand is $\sin^2 x$ . Since $\sin^2 x$ has no standard antiderivative in this form, use the half-angle identity $\sin^2 x = \frac{1 - \cos 2x}{2}$ to convert it to an integrable expression.

Integrate and evaluate:

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

Final answer: $\dfrac{\pi^2}{2}$ .

Rotate an arctan-type boundary

For $y = \frac{1}{\sqrt{x^2 + 1}}$ on $[0, 1]$ rotated about the x-axis, squaring gives $[f(x)]^2 = \frac{1}{x^2 + 1}$ . This is the standard arctan integrand.

Evaluate the integral:

V = \pi \int_0^1 \frac{1}{x^2 + 1} \, dx = \pi \left[\arctan x\right]_0^1 = \pi\left(\frac{\pi}{4} - 0\right) = \frac{\pi^2}{4}.

Final answer: $\dfrac{\pi^2}{4}$ .

Exam technique

Sketch the region first: the picture tells you the radius, the limits and whether there is a hole. Write the disc or washer formula with the radius squared before you substitute, because markers award a method mark for the correct set-up even if the arithmetic slips. Match the variable to the axis: x-axis pairs with $\int [f(x)]^2\,dx$ ; y-axis pairs with $\int [g(y)]^2\,dy$ after you invert to $x = g(y)$ and convert the limits to $y$ -values. For a washer use the difference of the squares $[f]^2 - [h]^2$ , never $(f - h)^2$ . If squaring produces $\sin^2 x$ use the half-angle identity, and if it produces $\frac{1}{1 + x^2}$ recognise the $\arctan$ integral. Keep the answer exact (carry $\pi$ , surds and fractions) and label it "cubic units".

Exam-style practice questions

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

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.