Arc length and surface area: VCE Specialist Mathematics Unit 4

Syllabus dot point

“The use of definite integrals to find the arc length of a curve and the surface area of a solid of revolution, in Cartesian form $y = f(x)$ and in parametric form $x = x(t)$ , $y = y(t)$ , and the setting up of the appropriate integral”

7 min readUpdated 2 Jun 2026

Quick answer

The arc length of $y = f(x)$ from $x = a$ to $x = b$ is $L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx$ . For a parametric curve from $t = t_1$ to $t = t_2$ it is $L = \int_{t_1}^{t_2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt$ . The surface area generated by rotating $y = f(x)$ about the $x$ -axis is $S = \int_a^b 2\pi y\,\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx$ , that is, the arc-length element multiplied by the circumference $2\pi y$ of the circle each point traces. Get the derivative right and square it before integrating.

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What this dot point is asking
Arc length in Cartesian form
Arc length in parametric form
Surface area of revolution
Examples in context
Try this
Exam-style questions

What this dot point is asking

VCAA wants you to set up and evaluate definite integrals for the length of a curve (arc length) and for the surface area generated when a curve is rotated about an axis, working from either a Cartesian equation $y = f(x)$ or a parametric description $x = x(t)$ , $y = y(t)$ . The emphasis is on building the correct integral; evaluation may use a calculator.

Arc length in Cartesian form

Approximate a curve by tiny straight pieces. A small step $dx$ produces a small rise $dy$ , and by Pythagoras the length of the piece is $\sqrt{dx^2 + dy^2} = \sqrt{1 + (dy/dx)^2}\,dx$ . Summing (integrating) over the interval gives

L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx.

The integrand is built from the gradient, so the first task is to differentiate, then square, then add $1$ under the root.

Arc length in parametric form

When $x$ and $y$ are functions of a parameter $t$ , the small length element is $\sqrt{dx^2 + dy^2} = \sqrt{(dx/dt)^2 + (dy/dt)^2}\,dt$ , giving

L = \int_{t_1}^{t_2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt.

This is the symmetric form: differentiate both coordinates with respect to $t$ , square each, add, take the root, and integrate over the parameter range. It also gives the distance travelled by a particle whose position is $(x(t), y(t))$ .

Surface area of revolution

Rotating the curve about the $x$ -axis sweeps each point through a circle of radius $y$ and circumference $2\pi y$ . Multiplying the arc-length element by this circumference and integrating gives the surface area:

S = \int_a^b 2\pi y\,\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx.

In parametric form the same idea gives $S = \int_{t_1}^{t_2} 2\pi y\,\sqrt{(dx/dt)^2 + (dy/dt)^2}\,dt$ . For rotation about the $y$ -axis, the radius is $x$ , so the circumference factor becomes $2\pi x$ .

Worked example

Length of a parametric arc

Find the length of the curve $x = t^2$ , $y = \frac{2}{3}t^3$ for $0 \le t \le 1$ .

Step 1: Differentiate both parametric components

\frac{dx}{dt} = 2t, \qquad \frac{dy}{dt} = 2t^2.

The arc-length formula requires the derivatives of both $x$ and $y$ with respect to the parameter $t$ , not with respect to each other. Differentiating each coordinate separately is the first step before building the integrand.

Step 2: Square and add to form the integrand

\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = (2t)^2 + (2t^2)^2 = 4t^2 + 4t^4 = 4t^2(1 + t^2).

The parametric arc-length formula uses the sum of the squared derivatives under a square root. Factoring out $4t^2$ will simplify the root in the next step.

Step 3: Take the square root of the simplified expression

For $t \ge 0$ ,

\sqrt{4t^2(1 + t^2)} = 2t\sqrt{1 + t^2}.

Because $t \ge 0$ on the given interval, $\sqrt{4t^2} = 2t$ without needing an absolute value. This clean factoring makes the integral tractable by substitution.

Step 4: Evaluate the arc-length integral

L = \int_0^1 2t\sqrt{1 + t^2}\,dt.

Substitute $u = 1 + t^2$ , $du = 2t\,dt$ . When $t = 0$ , $u = 1$ ; when $t = 1$ , $u = 2$ :

L = \int_1^2 \sqrt{u}\,du = \left[\frac{2}{3}u^{3/2}\right]_1^2 = \frac{2}{3}\left(2^{3/2} - 1\right) = \frac{2}{3}\left(2\sqrt{2} - 1\right).

Numerically $2\sqrt{2} \approx 2.828$ , so $L \approx \frac{2}{3}(1.828) \approx 1.219$ . The curve runs from $(0, 0)$ to $(1, \frac{2}{3})$ ; a straight-line distance of $\sqrt{1 + \frac{4}{9}} \approx 1.20$ , slightly less than the arc length, as expected since a curve is at least as long as the chord.

Common exam traps

Forgetting to square the derivative: The integrand has $\left(\frac{dy}{dx}\right)^2$ under the root, not $\frac{dy}{dx}$ .
Dropping the $+1$: The Cartesian arc-length integrand is $\sqrt{1 + (\dots)^2}$ . The $1$ comes from the $dx^2$ term.
Wrong radius for the surface: Rotation about the $x$ -axis uses radius $y$ (factor $2\pi y$ ); about the $y$ -axis uses radius $x$ .
Mismatched limits: In parametric problems integrate over $t$ with $t$ -limits, not $x$ -limits.

Examples in context

Example 1. Arc length of $y = \frac{2}{3}x^{3/2}$ from $0$ to $3$ : with $\frac{dy}{dx} = x^{1/2}$ , the integrand is $\sqrt{1 + x}$ .

Example 2. Rotating $y = \sqrt{x}$ , $0 \le x \le 4$ , about the $x$ -axis gives $S = \int_0^4 2\pi\sqrt{x}\,\sqrt{1 + \frac{1}{4x}}\,dx$ .

Try this

Q1. Write the Cartesian arc-length integral for $y = f(x)$ from $a$ to $b$ . [1 mark]

Cue. $\int_a^b \sqrt{1 + (f'(x))^2}\,dx$ .

Q2. Set up the parametric arc-length integral for $x = \cos t$ , $y = \sin t$ , $0 \le t \le \pi$ . [2 marks]

Cue. $\int_0^\pi \sqrt{\sin^2 t + \cos^2 t}\,dt = \int_0^\pi 1\,dt = \pi$ .

Q3. State the surface-area integrand for rotating $y = f(x)$ about the $x$ -axis. [1 mark]

Cue. $2\pi y\sqrt{1 + (f'(x))^2}$ .

Exam-style questions

Questions in the style of VCAA 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.

Practice3 marks

A curve is given by

y = \dfrac{2}{3}x^{3/2}

for

0 \le x \le 3

. Write down a definite integral for the arc length, simplify the integrand, and evaluate the arc length exactly.

Show worked answer →

The derivative is $\dfrac{dy}{dx} = x^{1/2}$ , so $\left(\dfrac{dy}{dx}\right)^2 = x$ .

Arc length $L = \displaystyle\int_0^3 \sqrt{1 + x}\,dx$ .

Antidifferentiate: $\int (1 + x)^{1/2}\,dx = \dfrac{2}{3}(1 + x)^{3/2}$ . Evaluate from $0$ to $3$ :

$L = \dfrac{2}{3}\left[(1 + 3)^{3/2} - (1 + 0)^{3/2}\right] = \dfrac{2}{3}\left(4^{3/2} - 1\right) = \dfrac{2}{3}(8 - 1) = \dfrac{14}{3}$ .

Markers reward the squared derivative, the simplified integrand $\sqrt{1 + x}$ , and the exact value $\frac{14}{3}$ .

Practice5 marks

A curve is given parametrically by

x = t - \sin t

y = 1 - \cos t

for

0 \le t \le 2\pi

(one arch of a cycloid). (a) Show that the arc-length integrand simplifies to

\sqrt{2 - 2\cos t}

. (b) Hence calculate the exact length of one arch, using the identity

1 - \cos t = 2\sin^2\!\left(\dfrac{t}{2}\right)

Show worked answer →

(a) $\dfrac{dx}{dt} = 1 - \cos t$ and $\dfrac{dy}{dt} = \sin t$ . Then $\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt}\right)^2 = (1 - \cos t)^2 + \sin^2 t = 1 - 2\cos t + \cos^2 t + \sin^2 t = 2 - 2\cos t$ . So the integrand is $\sqrt{2 - 2\cos t}$ .

(b) Using $1 - \cos t = 2\sin^2\!\left(\frac{t}{2}\right)$ , the integrand is $\sqrt{4\sin^2\!\left(\frac{t}{2}\right)} = 2\left|\sin\frac{t}{2}\right| = 2\sin\frac{t}{2}$ for $0 \le t \le 2\pi$ .

$L = \displaystyle\int_0^{2\pi} 2\sin\frac{t}{2}\,dt = \left[-4\cos\frac{t}{2}\right]_0^{2\pi} = -4\cos\pi + 4\cos 0 = 4 + 4 = 8$ .

Markers reward the parametric integrand, the half-angle simplification, and the exact length $8$ .