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

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

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

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:

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 $2\pi x$.

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