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Volumes of revolution (WACE Year 12 Mathematics Specialist Unit 4)

Syllabus dot point

“Find the volume of a solid generated by rotating a region about the x-axis or y-axis using the disc method”

7 min read

Quick answer

Rotating y=f(x)y = f(x) about the xx-axis from aa to bb sweeps out discs of radius f(x)f(x), so the volume is π∫ab[f(x)]2 dx\pi\int_a^b [f(x)]^2\,dx. Rotating about the yy-axis instead integrates π∫cd[x]2 dy\pi\int_c^d [x]^2\,dy with xx written as a function of yy.

Jump to a section
  1. What this dot point is asking
  2. What a solid of revolution is
  3. The disc method
  4. Rotating about the y-axis
  5. Rotating the region between two curves
  6. Setting limits
  7. Matching the integral to the axis
  8. Rotation about a line other than an axis
  9. Exam-style questions

What this dot point is asking

SCSA wants you to set up and evaluate volumes of revolution by the disc method, choosing the axis, the variable of integration and the limits correctly.

What a solid of revolution is

A solid of revolution is generated by rotating a plane region a full turn about a fixed straight line, the axis of revolution. Because of the rotation, every cross-section perpendicular to the axis is a circle (or, between two curves, an annulus), so the solid is rotationally symmetric. Familiar shapes arise this way: rotating a rectangle gives a cylinder, rotating a right triangle gives a cone, and rotating a semicircle gives a sphere. The disc method computes the volume of any such solid by summing the areas of these circular cross-sections along the axis, which is why the technique reduces to a single definite integral of πr2\pi r^2.

The disc method

When a region under y=f(x)y = f(x) is rotated about the xx-axis, each thin slice becomes a disc of radius f(x)f(x) and thickness dxdx, with area π[f(x)]2\pi[f(x)]^2. Summing gives

Volume about the x-axis

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

The radius is the distance from the axis to the curve, which here is just f(x)f(x). The squared radius is essential: forgetting to square is the classic error.

The reasoning behind the formula is a slicing argument. Cut the solid into thin slabs perpendicular to the axis of rotation. Each slab is approximately a cylinder (a disc) of radius f(x)f(x) and thickness Δx\Delta x, so its volume is approximately π[f(x)]2 Δx\pi[f(x)]^2\,\Delta x. Adding the slabs and letting Δx→0\Delta x \to 0 turns the sum into the definite integral π∫ab[f(x)]2 dx\pi\int_a^b [f(x)]^2\,dx. Seeing the volume as an accumulation of disc areas explains why the integrand is the cross-sectional area πr2\pi r^2 and why the variable of integration must run along the axis of rotation.

Rotating about the y-axis

Rotating about the yy-axis makes the radius the horizontal distance xx, so you express xx as a function of yy and integrate over yy:

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

The limits cc and dd are now yy-values. Match the variable of integration to the axis of rotation.

Rotating the region between two curves

The washer (annulus) idea

If the region between an outer curve R(x)R(x) and an inner curve r(x)r(x) is rotated about the xx-axis, each slice is an annulus, so V=π∫ab([R(x)]2−[r(x)]2) dxV = \pi\int_a^b \big([R(x)]^2 - [r(x)]^2\big)\,dx. Subtract the squares, not the radii.

Setting limits

The limits are where the region begins and ends along the axis of rotation, often the intersection points of the bounding curves with each other or with the axis. A sketch confirms them.

Matching the integral to the axis

The single most important set-up decision is matching the variable of integration to the axis of rotation. For rotation about the xx-axis, the discs are stacked along xx, so integrate π∫[f(x)]2 dx\pi\int [f(x)]^2\,dx with xx-limits. For rotation about the yy-axis, the discs are stacked along yy, so express the radius xx as a function of yy and integrate π∫[x(y)]2 dy\pi\int [x(y)]^2\,dy with yy-limits. Mixing the two, for instance integrating in xx for a yy-axis rotation, gives a meaningless answer. A clear sketch with a single representative disc drawn perpendicular to the axis makes the correct choice obvious.

Rotation about a line other than an axis

A harder variant rotates the region about a horizontal or vertical line that is not a coordinate axis, such as y=1y = 1 or x=3x = 3. The method is the same, but the radius becomes the distance from that line to the curve rather than from the axis. Rotating y=f(x)y = f(x) about the line y=cy = c uses a radius of ∣f(x)−c∣|f(x) - c|, so the volume is π∫ab[f(x)−c]2 dx\pi\int_a^b [f(x) - c]^2\,dx. Reading the radius as a distance, rather than just as the function value, is the key adjustment, and it also covers the washer case where an inner and outer distance are each measured from the same line.

Worked example

Find the volume when y=xy = \sqrt{x} for 0≤x≤40 \le x \le 4 is rotated about the xx-axis.

Step 1: Identify the disc radius and set up the integral

When a curve is rotated about the xx-axis, each thin vertical slice sweeps out a disc whose radius equals the yy-value of the curve at that position. The volume formula sums infinitely many disc areas πr2\pi r^2 across the interval.

The radius of each disc is r=f(x)=xr = f(x) = \sqrt{x}, so [f(x)]2=x[f(x)]^2 = x. The volume is

V=π∫04(x)2 dx=π∫04x dx.V = \pi\int_0^4 (\sqrt{x})^2\,dx = \pi\int_0^4 x\,dx.

Step 2: Evaluate the integral

We integrate xx with respect to xx using the power rule, then apply the limits of integration.

V=π[x22]04=π⋅162=8π.V = \pi\Big[\frac{x^2}{2}\Big]_0^4 = \pi\cdot\frac{16}{2} = 8\pi.

Final answer: The volume of the solid of revolution is 8π8\pi cubic units.

Common traps
Forgetting to square the radius
The integrand is π[f(x)]2\pi[f(x)]^2, not πf(x)\pi f(x).
Subtracting radii instead of squares
For a region between two curves, subtract R2−r2R^2 - r^2, never (R−r)2(R - r)^2.
Mismatched variable and axis
Rotation about the yy-axis means integrate in yy with xx as a function of yy.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2022-style7 marks
Calculator-assumed. The region between y=x2y = x^2 and y=2xy = 2x is rotated about the xx-axis. Find the volume of the solid generated.
Show worked answer →

A washer (annulus) volume between two curves.

Intersections: x2=2xx^2 = 2x gives x=0x = 0 and x=2x = 2. On [0,2][0, 2] the line y=2xy = 2x is the outer radius R=2xR = 2x and the parabola y=x2y = x^2 is the inner radius r=x2r = x^2.

Volume =π∫02[(2x)2−(x2)2] dx=π∫02(4x2−x4) dx=π[4x33−x55]02= \pi\displaystyle\int_0^2 \big[(2x)^2 - (x^2)^2\big]\,dx = \pi\displaystyle\int_0^2 (4x^2 - x^4)\,dx = \pi\left[\dfrac{4x^3}{3} - \dfrac{x^5}{5}\right]_0^2.

At x=2x = 2: 323−325=160−9615=6415\dfrac{32}{3} - \dfrac{32}{5} = \dfrac{160 - 96}{15} = \dfrac{64}{15}. So the volume is 64π15\dfrac{64\pi}{15} cubic units.

Markers reward identifying the outer and inner radii, subtracting the squares (not the radii), the integration, and the volume 64π15\dfrac{64\pi}{15}.

WACE 2024-style5 marks
Calculator-free. The region bounded by y=x2y = x^2, the yy-axis and y=4y = 4 is rotated about the yy-axis. Find the volume of the solid generated.
Show worked answer →

A rotation about the yy-axis.

Rotation about the yy-axis means the radius is xx, expressed in terms of yy. From y=x2y = x^2, x2=yx^2 = y. The region runs from y=0y = 0 to y=4y = 4.

Volume =π∫04x2 dy=π∫04y dy=π[y22]04=π×8=8π= \pi\displaystyle\int_0^4 x^2\,dy = \pi\displaystyle\int_0^4 y\,dy = \pi\left[\dfrac{y^2}{2}\right]_0^4 = \pi \times 8 = 8\pi cubic units.

Markers reward expressing x2=yx^2 = y, integrating in yy with the correct limits, and the volume 8π8\pi.

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