What is the arc-length integral (Cartesian)?

A short piece of curve has length ds = dx^2 + dy^2. Dividing inside the root by dx^2 gives the Cartesian arc length:

What is the arc-length integral (parametric)?

For a parametric curve, divide inside the root by dt^2 instead:

What is surface area of revolution?

Rotating an arc through 2π generates a surface. Each band has area 2π(radius)\,ds, where the radius is the distance from the axis. About the x-axis the radius is y:

What is the strategy that makes these tractable?

These integrals are often awkward unless the expression under the root simplifies. Many exam curves are designed so that 1 + (dy/dx)^2 becomes a perfect square, removing the root cleanly. Always expand and look for that before reaching for a substitution.

What is wrong radius for the surface?

About the x-axis the radius is y; about the y-axis it is x. Using the wrong one gives the wrong surface.

What is sign of the square root?

sqrt(()^2) is the positive value on the interval; check the sign of the simplified expression.

Write the Cartesian arc-length formula for y = f(x) between x = a and x = b. [1 mark]

For the parametric curve x = t, y = t^2, write the integrand (x)^2 + (y)^2. [1 mark]

SEAB Original-style practice: Find the length of the curve y = 13(x^2 + 2)^3/2 from x = 0 to x = 3.

Arc length is _a^b sqrt(1 + ((dy)/(dx))^2)\,dx. Differentiate: dydx = 13·32(x^2 + 2)^1/2· 2x = x(x^2 + 2)^1/2. So (dydx)^2 = x^2(x^2 + 2) = x^4 + 2x^2, and $1 + ((dy)/(dx))^2 = x^4 + 2x^2 + 1 = (x^2 + 1)^2.$ Therefore sqrt(1 + (dy/dx)^2) = x^2 + 1 (positive on the interval), and $L = _0^3 (x^2 + 1)\,dx = [(x^3)/(3) + x]_0^3 = (9 + 3) - 0 = 12.$ Markers reward the arc-length formula, the derivative, recognising the perfect square (x^2+1)^2, and the value 12.

SEAB Original-style practice: The arc of the curve y = sqrt(x) from x = 0 to x = 1 is rotated through 2π about the x-axis. Set up the integral for the surface area generated, and simplify the integrand.

The surface area of revolution about the x-axis is S = _a^b 2π ysqrt(1 + ((dy)/(dx))^2)\,dx. For y = x^1/2, dydx = 12x^-1/2, so (dydx)^2 = 14x and $1 + ((dy)/(dx))^2 = 1 + (1)/(4x) = (4x + 1)/(4x).$ Hence sqrt(1 + (dy/dx)^2) = sqrt(4x + 1)2sqrt(x), and with y = sqrt(x): $S = _0^1 2(x)·(sqrt(4x + 1))/(2sqrt(x))\,dx = π_0^1 sqrt(4x + 1)\,dx.$ Markers reward the surface-area formula, the derivative and its square, simplifying so the sqrt(x) cancels, and the tidy integrand (4x + 1).

SingaporeFurther MathsSyllabus dot point

How do we compute the length of a curve and the area of a surface of revolution by integration?

Calculate the arc length of a curve and the area of a surface of revolution for curves given in Cartesian or parametric form

A focused answer to the H2 Further Mathematics outcome on arc length and surfaces of revolution. The arc-length integral in Cartesian and parametric form, the surface-area-of-revolution formula, and worked applications.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-06

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

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to compute two geometric quantities by integration: the length of an arc of a curve, and the area of the surface generated when an arc is rotated about an axis. You should handle curves given in Cartesian form $y = \mathrm{f}(x)$ and in parametric form $x = \mathrm{f}(t)$ , $y = \mathrm{g}(t)$ , and choose the right formula for the axis of rotation.

The answer

The arc-length integral (Cartesian)

A short piece of curve has length $\mathrm{d}s = \sqrt{\mathrm{d}x^2 + \mathrm{d}y^2}$ . Dividing inside the root by $\mathrm{d}x^2$ gives the Cartesian arc length:

L = \int_a^b \sqrt{1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}\,\mathrm{d}x.

The arc-length integral (parametric)

For a parametric curve, divide inside the root by $\mathrm{d}t^2$ instead:

L = \int_{t_1}^{t_2} \sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2}\,\mathrm{d}t.

Surface area of revolution

Rotating an arc through $2\pi$ generates a surface. Each band has area $2\pi(\text{radius})\,\mathrm{d}s$ , where the radius is the distance from the axis. About the $x$ -axis the radius is $y$ :

S = \int 2\pi y\,\mathrm{d}s = \int_a^b 2\pi y\sqrt{1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}\,\mathrm{d}x.

About the $y$ -axis the radius is $x$ , so use $2\pi x$ instead. In parametric form replace $\mathrm{d}s$ by the parametric arc-length element.

The strategy that makes these tractable

These integrals are often awkward unless the expression under the root simplifies. Many exam curves are designed so that $1 + (\mathrm{d}y/\mathrm{d}x)^2$ becomes a perfect square, removing the root cleanly. Always expand and look for that before reaching for a substitution.

Worked example

Find the length of the parametric curve $x = \cos^3 t$ , $y = \sin^3 t$ for $0 \leq t \leq \dfrac{\pi}{2}$ (one quarter of an astroid).

Step 1: Differentiate with respect to t

\frac{\mathrm{d}x}{\mathrm{d}t} = -3\cos^2 t\sin t, \qquad \frac{\mathrm{d}y}{\mathrm{d}t} = 3\sin^2 t\cos t.

Step 2: Form the sum of squares

\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2 = 9\cos^4 t\sin^2 t + 9\sin^4 t\cos^2 t = 9\cos^2 t\sin^2 t(\cos^2 t + \sin^2 t).

Step 3: Simplify with the Pythagorean identity

Since $\cos^2 t + \sin^2 t = 1$ , this is $9\cos^2 t\sin^2 t$ , so

\sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2} = 3\cos t\sin t = \tfrac{3}{2}\sin 2t,

taking the positive root on $0 \leq t \leq \tfrac{\pi}{2}$ .

Step 4: Integrate

L = \int_0^{\pi/2}\tfrac{3}{2}\sin 2t\,\mathrm{d}t = \tfrac{3}{2}\left[-\tfrac{1}{2}\cos 2t\right]_0^{\pi/2} = \tfrac{3}{4}\left(-\cos\pi + \cos 0\right) = \tfrac{3}{4}(1 + 1) = \tfrac{3}{2}.

Step 5: State the result

The quarter astroid has length $\dfrac{3}{2}$ .

Examples in context

Example 1. Length of a hanging cable. A cable hangs as a catenary $y = a\cosh\dfrac{x}{a}$ , for which $1 + (y')^2 = \cosh^2\dfrac{x}{a}$ is a perfect square. The arc-length integral then gives the cable length exactly, a standard engineering calculation.

Example 2. Surface area of a sphere. Rotating the semicircle $y = \sqrt{r^2 - x^2}$ about the $x$ -axis and applying the surface formula recovers the area $4\pi r^2$ , confirming the classical result by integration.

Try this

Q1. Write the Cartesian arc-length formula for $y = \mathrm{f}(x)$ between $x = a$ and $x = b$ . [1 mark]

Cue. $L = \displaystyle\int_a^b \sqrt{1 + (y')^2}\,\mathrm{d}x$ .

Q2. State the surface-area integral for an arc rotated about the $x$ -axis. [1 mark]

Cue. $S = \displaystyle\int_a^b 2\pi y\sqrt{1 + (y')^2}\,\mathrm{d}x$ .

Q3. For the parametric curve $x = t$ , $y = t^2$ , write the integrand $\sqrt{(\dot{x})^2 + (\dot{y})^2}$ . [1 mark]

Cue. $\dot{x} = 1$ , $\dot{y} = 2t$ , so the integrand is $\sqrt{1 + 4t^2}$ .

Exam-style practice questions

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

Original6 marksFind the length of the curve

y = \dfrac{1}{3}(x^2 + 2)^{3/2}

from

x = 0

x = 3

Show worked answer →

Arc length is $\displaystyle\int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx$ .

Differentiate: $\dfrac{dy}{dx} = \dfrac{1}{3}\cdot\dfrac{3}{2}(x^2 + 2)^{1/2}\cdot 2x = x(x^2 + 2)^{1/2}$ .

So $\left(\dfrac{dy}{dx}\right)^2 = x^2(x^2 + 2) = x^4 + 2x^2$ , and

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

Therefore

\sqrt{1 + (dy/dx)^2} = x^2 + 1

(positive on the interval), and

L = \int_0^3 (x^2 + 1)\,dx = \left[\frac{x^3}{3} + x\right]_0^3 = (9 + 3) - 0 = 12.

Markers reward the arc-length formula, the derivative, recognising the perfect square $(x^2+1)^2$ , and the value $12$ .

Original6 marksThe arc of the curve

y = \sqrt{x}

from

x = 0

x = 1

is rotated through

2\pi

about the

x

-axis. Set up the integral for the surface area generated, and simplify the integrand.

Show worked answer →

The surface area of revolution about the $x$ -axis is $\displaystyle S = \int_a^b 2\pi y\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx$ .

For $y = x^{1/2}$ , $\dfrac{dy}{dx} = \dfrac{1}{2}x^{-1/2}$ , so $\left(\dfrac{dy}{dx}\right)^2 = \dfrac{1}{4x}$ and

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

Hence

\sqrt{1 + (dy/dx)^2} = \dfrac{\sqrt{4x + 1}}{2\sqrt{x}}

, and with

y = \sqrt{x}

S = \int_0^1 2\pi\sqrt{x}\cdot\frac{\sqrt{4x + 1}}{2\sqrt{x}}\,dx = \pi\int_0^1 \sqrt{4x + 1}\,dx.

Markers reward the surface-area formula, the derivative and its square, simplifying so the $\sqrt{x}$ cancels, and the tidy integrand $\pi\sqrt{4x + 1}$ .