Arc Length and Surface Area

A road sign says the mountain village is 12 km away — but that's as the crow flies. Your car's odometer, patiently counting wheel-turns through every hairpin bend, clocks 31 km by the time you arrive. The odometer is measuring something the map cannot: the true length of a curve, hairpins and all. It does it the only way anyone can — by adding up an enormous number of tiny, essentially straight steps.

That is precisely a job for the integral. We have already sliced regions into strips to get areas and solids into shells to get volumes. The same strategy measures the length of a curve — and, as a bonus, the surface area swept out when the curve is spun about an axis. As ever, the whole game is finding the right infinitesimal element and adding up.

For a smooth curve y = f(x) on [a, b], the arc length turns out to be

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.

Those square roots look intimidating, but they hide something you have known since your first geometry lesson: Pythagoras, applied to steps too small to see.

The arc-length element, from a tiny triangle

Zoom in on any smooth curve far enough and it stops looking curved — that is what smooth means. So chop the curve into a huge number of tiny pieces, each essentially a short straight segment, measure each with Pythagoras, and add them all up. Here is that plan, carried out honestly.

Step 1 — zoom in on a tiny piece of curve. Over an infinitesimal step the curve is essentially straight. Call its length ds (the "s" is traditional for distance along a path). 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. For a right triangle, however small:

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

Step 3 — factor out dx. We want to integrate with respect to x, so pull a factor of dx^2 out of the square root — and watch the derivative dy/dx appear all by itself:

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

Read \sqrt{1 + f'(x)^2} as a local stretch factor: it says how much longer the slanted path is than the flat step beneath it. Where the curve is level (f' = 0) the factor is exactly 1 — the curve is no longer than its shadow. Where the slope is 1 the factor is \sqrt{2} \approx 1.414, the familiar diagonal-of-a-square stretch. The steeper the curve, the bigger the factor — and it is never less than 1, because a hypotenuse is never shorter than its leg.

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.

That is the whole derivation. Everything difficult about arc length lives in one square root, and the square root is nothing but Pythagoras whispered infinitely many times.

The tiny right triangle, live

Below is the curve y = \tfrac23 x^{3/2} with the little right triangle drawn at a moving point: a horizontal leg dx, a vertical leg dy = f'(x)\,dx, and the slanted hypotenuse ds riding along the curve. Slide x and watch the readout: near the origin the curve is nearly flat, the slope is tiny, and the stretch factor \sqrt{1 + f'(x)^2} hugs 1. Push out to the right, the curve rears up, and the same horizontal step buys you a visibly longer hypotenuse.

The integral \int_a^b \sqrt{1+f'(x)^2}\,dx simply drags this triangle from a to b, totting up the hypotenuses as it goes — exactly what your odometer does, one wheel-turn at a time.

Sanity check: a straight line first

Before trusting a new formula on a curve, make it prove itself on something we can measure without calculus. Take the straight line y = \tfrac34 x from x = 0 to x = 4 — it runs from (0,0) to (4,3), so plain Pythagoras says its length is \sqrt{4^2 + 3^2} = \sqrt{25} = 5. Now the formula's turn.

Step 1 — differentiate. f'(x) = \tfrac34, a constant — a line has the same slope everywhere.

Step 2 — the stretch factor.

\sqrt{1 + f'(x)^2} = \sqrt{1 + \tfrac{9}{16}} = \sqrt{\tfrac{25}{16}} = \tfrac54.

Step 3 — integrate. A constant integrand makes this a one-liner:

L = \int_0^4 \tfrac54\,dx = \tfrac54 \cdot 4 = 5. \checkmark

Perfect agreement — and it had to be: for a straight line the "tiny triangle" is just the whole 3–4–5 triangle shrunk down, and every ds is \tfrac54 of its dx. The formula is Pythagoras with delusions of grandeur — and that is exactly why we believe it on genuine curves, where the slope, and hence the stretch factor, changes from point to point.

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

Now a real curve — the one in the figure above.

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. Two quick reality checks: it is longer than the straight horizontal run of 3 (any curve must beat its shadow), and it is longer than the straight segment joining its endpoints (0,0) and (3, 2\sqrt3), which is \sqrt{9 + 12} = \sqrt{21} \approx 4.58 — a curve is always at least as long as the chord it spans. Both pass. That habit — bracketing an arc length between the horizontal run and something plausible — catches most slips before they cost you marks.

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

The arc length of y = f(x) is \displaystyle L = \int_a^{b} \sqrt{1 + f'(x)^2}\,dx, built from the element ds = \sqrt{1 + f'(x)^2}\,dx (Pythagoras: ds^2 = dx^2 + dy^2).
Revolving it about the x-axis gives a surface area \displaystyle S = \int_a^{b} 2\pi\, f(x)\sqrt{1 + f'(x)^2}\,dx — circumference 2\pi f(x) times the slant element ds.
About the y-axis instead, the radius is x: \displaystyle S = \int_a^{b} 2\pi\, x\sqrt{1 + f'(x)^2}\,dx.

Spinning the curve: surface of revolution

Now revolve the curve about the x-axis, like a lathe turning a table leg. Each little element ds sweeps out a thin band — a sliver of a cone, called a frustum. Unrolled, that band is (to first order) a rectangle: its length is the circumference of the circle it travelled, 2\pi f(x), and its width is the slant length ds — not dx! The band lies tilted in space, so its true width is the hypotenuse, not the horizontal step. Hence

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

Worked example: a cone, from first principles

The formula should reproduce geometry we already trust. A cone of base radius r and height h is what you get by revolving the line f(x) = \tfrac{r}{h}x on [0, h] about the x-axis. School geometry says its curved surface is \pi r l, where l = \sqrt{h^2 + r^2} is the slant height. Let's see.

Step 1 — the stretch factor. f'(x) = r/h, constant, so

\sqrt{1 + f'(x)^2} = \sqrt{1 + \frac{r^2}{h^2}} = \frac{\sqrt{h^2 + r^2}}{h} = \frac{l}{h}.

Step 2 — assemble and integrate. The constant factors stroll out of the integral:

S = \int_0^{h} 2\pi \cdot \frac{r}{h}x \cdot \frac{l}{h}\,dx = \frac{2\pi r l}{h^2}\int_0^{h} x\,dx = \frac{2\pi r l}{h^2}\cdot\frac{h^2}{2} = \pi r l. \checkmark

The calculus recovers the classical formula exactly. Try dropping the slant factor l/h and you get \pi r h instead — visibly wrong for any squat cone, which is a nice memory hook for why the ds matters.

One more classic for your collection: revolving the semicircle y = \sqrt{r^2 - x^2} on [-r, r] gives, after a very satisfying cancellation inside the square root, a constant integrand 2\pi r — so the sphere's area is 2\pi r \cdot 2r = 4\pi r^2. The cancellation means every slice of a sphere of equal thickness has equal surface area, whether cut near the pole or the equator. Archimedes discovered this ("the hat-box theorem") two millennia before calculus, and was so proud he asked for a sphere in its cylinder to be carved on his tomb.

When the integral fights back

You may have noticed the examples above were suspiciously cooperative. That is no accident — and knowing why is part of mastering this topic.

Two traps catch nearly everyone:

Most arc-length integrals cannot be done by hand. The example \tfrac23 x^{3/2} was rigged so that 1 + f'(x)^2 became a perfect, integrable 1 + x. Such curves are rare birds. Even the innocent parabola y = x^2 on [0,1] leads to \int_0^1 \sqrt{1 + 4x^2}\,dx, which needs a hyperbolic substitution and delivers an \operatorname{arcsinh} term (L = \tfrac{\sqrt5}{2} + \tfrac14\operatorname{arcsinh} 2 \approx 1.479). The circumference of an ellipse has no closed form at all — the struggle to compute it forced Legendre, Abel and Jacobi to invent a whole new branch of mathematics, the elliptic integrals. In practice you set up the integral exactly and then hand it to numerical integration — Simpson's rule, say — for as many digits as you like. Setting up the right integral is the examinable skill; evaluating it is sometimes a job for a computer.
The surface-area integrand needs BOTH factors. Writing \int 2\pi f(x)\,dx (forgetting the slant) or \int \sqrt{1+f'(x)^2}\,dx (that's just arc length) are the two standard ways to lose the marks. The band's area is circumference × slant width: 2\pi f(x) and \sqrt{1 + f'(x)^2}\,dx, always together. Check yourself with the cone: only the full integrand returns \pi r l.

Take y = 1/x for x \ge 1 and revolve it about the x-axis: an infinitely long trumpet, narrowing forever — Gabriel's Horn. Its volume (by discs) is

V = \pi\int_1^{\infty} \frac{1}{x^2}\,dx = \pi \;-\; \text{finite!}

But its surface area is S = 2\pi\int_1^{\infty} \tfrac{1}{x}\sqrt{1 + \tfrac{1}{x^4}}\,dx \ge 2\pi\int_1^{\infty}\tfrac{dx}{x} = \infty — infinite, because the square-root factor is always at least 1 and \int 1/x\,dx diverges. So the horn holds exactly \pi units of paint, yet no finite amount of paint could coat its inside wall. Torricelli announced the finite volume in 1643 — before calculus proper existed — and the philosophical uproar (Hobbes called it madness) helped convince mathematicians that infinity needed careful rules, not intuition. The resolution of the "paint paradox" is that mathematical paint has zero thickness; real paint doesn't, and a real coat of any thickness would eventually plug the throat of the horn entirely.

It depends on your ruler — and not in a small way. Measure Britain's coast with 100 km steps and you get about 2,800 km; with 50 km steps, about 3,400 km; keep shrinking the ruler and the answer keeps growing, without settling down. Lewis Fry Richardson noticed this in the 1950s (Spain and Portugal disagreed about their shared border's length by hundreds of kilometres — each had used a different ruler), and Benoît Mandelbrot turned it into the founding example of fractal geometry.

Why doesn't our formula rescue us? Because L = \int\sqrt{1+f'(x)^2}\,dx assumes the curve is smooth: zoom in and it flattens into a straight hypotenuse. A coastline never flattens — every zoom reveals fresh bays inside bays, wiggles on wiggles, so the "tiny straight piece" at the heart of our derivation never arrives. Smoothness isn't decoration in the theorem statement; it is the load-bearing wall. Arc length is a property of ideal curves — real geography plays by wilder rules.