Areas between curves and volumes of revolution in QCE Specialist Mathematics Unit 4

Syllabus dot point

“Determine areas between curves and volumes of solids of revolution generated by rotating a region about the x-axis or the y-axis, setting up the correct definite integral and evaluating it”

QCESpecialist MathematicsUnit 4: Further calculus, and…8 min readUpdated 2 Jun 2026

Quick answer

The area between curves with $f \geq g$ is $\int_a^b (f - g)\, dx$ , upper minus lower, with limits set by intersection points. Rotating $y = f(x)$ about the $x$ -axis gives $V = \pi\int_a^b [f(x)]^2\, dx$ ; rotating $x = g(y)$ about the $y$ -axis gives $V = \pi\int_c^d [g(y)]^2\, dy$ . For a region between an outer radius $R$ and inner radius $r$ , use the washer $V = \pi\int ([R]^2 - [r]^2)\, dx$ , subtracting squares of radii, not the square of the difference.

Jump to a section

What this dot point is asking
The answer
Exam-style questions

What this dot point is asking

QCAA wants you to use definite integrals to measure areas enclosed between curves and the volumes of solids formed by rotating a region about an axis. The assessable skill is setting up the correct integral, with the right integrand, the right limits, and the right axis, then evaluating it. Drawing the region first prevents most errors.

The answer

Area between two curves

If $f(x) \geq g(x)$ on $[a, b]$ , the area enclosed between them is

A = \int_a^b \big(f(x) - g(x)\big)\, dx,

the upper curve minus the lower curve. Find the intersection points to fix the limits $a$ and $b$ . If the curves cross within the interval, split the integral at each crossing and integrate the positive difference on each piece.

Volume of revolution about the x-axis

Rotating the region under $y = f(x)$ from $x = a$ to $x = b$ about the $x$ -axis sweeps out a solid of circular cross-sections of radius $f(x)$ . Each thin disc has volume $\pi [f(x)]^2\, dx$ , so

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

The integrand is the radius squared; the $\pi$ is the area of the circular face.

Volume of revolution about the y-axis

Rotating the region between $x = g(y)$ and the $y$ -axis from $y = c$ to $y = d$ about the $y$ -axis gives

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

Here you express $x$ as a function of $y$ and integrate with respect to $y$ . The structure mirrors the $x$ -axis case with the roles of the variables swapped.

Region between two curves rotated about an axis

When the rotated region lies between an outer curve $R(x)$ and an inner curve $r(x)$ , the cross-section is an annulus (washer):

V = \pi \int_a^b \Big( [R(x)]^2 - [r(x)]^2 \Big)\, dx.

You subtract the squares of the radii, not the square of the difference. Squaring each radius separately is essential.

Choosing the variable of integration

Rotation about the $x$ -axis integrates in $x$ with radius given by a $y$ -value; rotation about the $y$ -axis integrates in $y$ with radius given by an $x$ -value. Match the variable of integration to the axis of rotation.

Setting up the integral from a sketch

The reliable workflow is always the same: sketch the region, mark the limits where the boundary curves meet, decide the axis of rotation, and then identify the radius. For an $x$ -axis rotation the radius is the vertical distance from the axis to the curve, namely the $y$ -value; for a $y$ -axis rotation it is the horizontal distance, the $x$ -value, so you must first rearrange the curve into the form $x = g(y)$ . Getting the limits from the intersection points and the radius from the correct distance accounts for most of the marks; the integration itself is usually routine.

Areas where curves cross

When the two curves cross inside the interval, the upper and lower roles swap, and a single integral of $f - g$ would let positive and negative pieces cancel. Find every intersection in the interval, split the region at each one, and integrate the positive difference (upper minus lower) on each piece separately, then add the parts. Equivalently, integrate the absolute value of the difference. For example, the area between $y = \sin x$ and the $x$ -axis over $[0, 2\pi]$ must be split at $x = \pi$ , because the curve dips below the axis on $(\pi, 2\pi)$ .

A worked link to trigonometric integration

Volumes of revolution frequently produce integrands like $\sin^2 x$ or $\cos^2 x$ after squaring the radius, which cannot be integrated directly. The power-reducing identities $\sin^2 x = \tfrac{1 - \cos 2x}{2}$ and $\cos^2 x = \tfrac{1 + \cos 2x}{2}$ convert them into terms that integrate immediately. Recognising that a squared trigonometric radius will need a double-angle reduction is the key planning step in those questions, and it ties this dot point directly to the trigonometric-integration techniques in the same unit.

Worked example

The region bounded by $y = \sqrt{x}$ , the $x$ -axis and $x = 4$ is rotated about the $x$ -axis. Find the volume.

Rotating about the $x$ -axis creates circular cross-sections of radius equal to the $y$ -value, so we use the disc formula $V = \pi\int [f(x)]^2\,dx$ .

Step 1: Identify the radius and limits

The boundary curve is $y = \sqrt{x}$ , so the disc radius at position $x$ is $f(x) = \sqrt{x}$ . The region runs from $x = 0$ (where the curve meets the origin) to $x = 4$ (the right boundary).

Step 2: Set up the disc integral

Squaring the radius simplifies the integrand from a square root to a linear term:

V = \pi \int_0^4 \big[\sqrt{x}\big]^2\, dx = \pi \int_0^4 x\, dx.

Step 3: Evaluate the integral

Integrate $x$ and apply the limits:

V = \pi \left[\frac{x^2}{2}\right]_0^4 = \pi\left(\frac{16}{2} - 0\right) = 8\pi.

Step 4: Check the scale

The solid is a paraboloid tapering to a point at the origin, with a circular face of radius $f(4) = 2$ at $x = 4$ . The enclosing cylinder would have volume $\pi (2)^2 (4) = 16\pi$ , so $8\pi \approx 25.1$ is less than $16\pi \approx 50.3$ , which is exactly what we expect for a solid that tapers.

Final answer: The volume is $8\pi$ cubic units.

Common trap

The most penalised error in volumes is forgetting to square the radius, integrating $\pi\int f(x)\, dx$ instead of $\pi\int [f(x)]^2\, dx$ . Always square. For washers, students often write $\pi\int (R - r)^2\, dx$ , but the correct form subtracts the squares, $\pi\int ([R]^2 - [r]^2)\, dx$ . For areas, integrating without checking which curve is on top can give a negative value; take upper minus lower and split at any crossing.

Exam-style questions

Questions in the style of QCAA 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.

Practice5 marks

Paper 2 (complex familiar). The region between the

x

-axis and

y = 1 + \sin 2x

for

0 \leq x \leq \tfrac{\pi}{2}

is rotated about the

x

-axis. Determine the exact volume of the solid of revolution, in simplest form.

Show worked answer →

Disc formula: $V = \pi \displaystyle\int_0^{\pi/2} (1 + \sin 2x)^2\, dx.$

Expand: $(1 + \sin 2x)^2 = 1 + 2\sin 2x + \sin^2 2x$ . Use $\sin^2 2x = \tfrac{1 - \cos 4x}{2}$ , so the integrand is $\tfrac{3}{2} + 2\sin 2x - \tfrac{1}{2}\cos 4x.$

Integrate: $V = \pi\Big[\tfrac{3}{2}x - \cos 2x - \tfrac{1}{8}\sin 4x\Big]_0^{\pi/2}.$

At $x = \tfrac{\pi}{2}$ : $\tfrac{3\pi}{4} - \cos\pi - \tfrac{1}{8}\sin 2\pi = \tfrac{3\pi}{4} + 1$ . At $x = 0$ : $-1$ . So $V = \pi\big[(\tfrac{3\pi}{4} + 1) - (-1)\big] = \pi\big(\tfrac{3\pi}{4} + 2\big)$ cubic units.

Markers reward the disc setup, the double-angle reduction, the antiderivative, and the exact evaluation.

Practice4 marks

Paper 1 (technique). Determine the area enclosed between the curves

y = x^2

and

y = 2x