What are area between two curves?

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

What is check the scale?

The solid is a paraboloid bounded by a disc of radius 2 at x = 4. Its volume 8π ≈ 25.1 is less than the enclosing cylinder π (2)^2 (4) = 16π, as expected for a solid tapering to a point at the origin.

QLDSpecialist MathematicsSyllabus dot point

Topic 1: Integration and applications of integration

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

A focused answer to the QCE Specialist Mathematics Unit 4 dot point on areas and volumes. Covers the area between two curves, the disc formula for rotation about each axis, and rotating a region bounded by two curves, with a verified worked example and the squaring trap.

Generated by Claude Opus 4.76 min answerUpdated 2026-05-29

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

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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.

Worked example

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

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

Set up the disc integral.

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

Evaluate.

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

Check the scale. The solid is a paraboloid bounded by a disc of radius $2$ at $x = 4$ . Its volume $8\pi \approx 25.1$ is less than the enclosing cylinder $\pi (2)^2 (4) = 16\pi$ , as expected for a solid tapering to a point at the origin.

Exam-style practice questions

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

2022 QCAA5 marksThe region between the x-axis and the curve of the function y = (1 + sin 2x) for 0 less than or equal to x less than or equal to pi/2 is rotated about the x-axis to form a solid of revolution. Determine the volume of this solid. Express your answer in simplest form.

Show worked answer →

Use the disc formula V = pi * integral of y^2 dx from 0 to pi/2.

Set up [1 mark]: V = pi * integral from 0 to pi/2 of (1 + sin 2x)^2 dx.

Expand the integrand [1 mark]: (1 + sin 2x)^2 = 1 + 2 sin 2x + sin^2(2x).

Apply the double-angle identity sin^2(2x) = (1 - cos 4x)/2 [1 mark], so the integrand becomes 3/2 + 2 sin 2x - (1/2) cos 4x.

Integrate [1 mark]: pi * [ (3/2)x - cos 2x - (1/8) sin 4x ] from 0 to pi/2.

Evaluate [1 mark]: at x = pi/2, the bracket is 3pi/4 - cos(pi) - (1/8) sin(2pi) = 3pi/4 + 1; at x = 0 it is -1.
V = pi[(3pi/4 + 1) - (-1)] = pi(3pi/4 + 2) cubic units.