Arc Length and Surface Area

We have measured areas and volumes by integrating. The same idea measures the length of a curve and the surface area swept out when that curve is revolved. The trick, as ever, is to find the right infinitesimal element and add up.

For a smooth curve y = f(x) on [a, b], the arc length is

L = \int_a^{b} \sqrt{1 + f'(x)^2}\,dx,

and revolving that curve about the x-axis sweeps a surface of area

S = \int_a^{b} 2\pi\, f(x)\sqrt{1 + f'(x)^2}\,dx.

The arc-length element, from a tiny triangle

That \sqrt{1 + f'(x)^2} is just the Pythagorean theorem, applied to a speck of the curve.

Step 1 — zoom in on a tiny piece of curve. Over an infinitesimal step the curve is essentially straight. Call its length ds. Moving along it, the x-coordinate changes by dx and the y-coordinate by dy — a tiny right triangle with legs dx and dy and hypotenuse ds.

Step 2 — Pythagoras on the triangle.

ds^2 = dx^2 + dy^2.

Step 3 — factor out dx. Pull dx^2 out from under nothing yet — divide and multiply so the derivative dy/dx appears:

ds = \sqrt{dx^2 + dy^2} = \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\;dx = \sqrt{1 + f'(x)^2}\,dx.

Step 4 — add up all the little hypotenuses. Integrate the element ds across the interval:

L = \int ds = \int_a^{b} \sqrt{1 + f'(x)^2}\,dx.

For the surface of revolution, each element ds spins into a thin band — a frustum — of radius f(x), so its area is the circumference 2\pi f(x) times the slant length ds; integrating gives the S above.

Worked example: length of y = \tfrac{2}{3}x^{3/2} on [0, 3]

Step 1 — differentiate. The power rule gives

f(x) = \tfrac{2}{3}x^{3/2} \;\Longrightarrow\; f'(x) = \tfrac{2}{3}\cdot\tfrac{3}{2}x^{1/2} = x^{1/2} = \sqrt{x}.

Step 2 — form 1 + f'(x)^2. Squaring the square root cleans up beautifully:

1 + f'(x)^2 = 1 + \big(\sqrt{x}\big)^2 = 1 + x.

Step 3 — set up the integral. (This is exactly why the curve was chosen — the radical collapses to \sqrt{1 + x}, which actually integrates.)

L = \int_0^{3} \sqrt{1 + x}\,dx.

Step 4 — integrate and evaluate. The antiderivative of (1+x)^{1/2} is \tfrac{2}{3}(1+x)^{3/2}:

L = \left[\frac{2}{3}(1 + x)^{3/2}\right]_0^{3} = \frac{2}{3}\Big(4^{3/2} - 1^{3/2}\Big) = \frac{2}{3}(8 - 1) = \frac{14}{3}.

The curve is 14/3 \approx 4.67 units long — a touch longer than the straight horizontal run of 3, exactly as a curve should be.

Let f be continuously differentiable on [a, b].

The example above was rigged: \tfrac23 x^{3/2} was chosen precisely so that 1 + f'(x)^2 became a perfect, integrable 1 + x. Such curves are rare birds. The square root in \sqrt{1 + f'(x)^2} almost never has an elementary antiderivative.

The most innocent-looking curve breaks it. The length of the parabola y = x^2 on [0, 1] is

L = \int_0^{1} \sqrt{1 + 4x^2}\,dx,

which needs a hyperbolic substitution and produces an \operatorname{arcsinh} term; the circumference of an ellipse has no closed form at all (it births the "elliptic integrals"). In practice you reach for numerical integration — Simpson's rule, say — and get the length to as many digits as you like. The geometry is exact and honest; it is merely antidifferentiation that runs out of road. That is the everyday face of calculus: setting up the right integral is the real skill — evaluating it is sometimes a job for a computer.

The tiny right triangle, live

Below is the curve y = \tfrac23 x^{3/2} with a small right triangle drawn at the moving point: a horizontal leg dx, a vertical leg dy = f'(x)\,dx, and the slanted hypotenuse ds riding along the curve. Slide x: the readout reports the local slope f'(x) and the element ds = \sqrt{1 + f'(x)^2}\,dx — steeper curve, longer hypotenuse.