In short

The length of a curve y = f(x) from x = a to x = b is L = \int_a^b \sqrt{1 + [f'(x)]^2}\,dx. For a parametric curve x = x(t), y = y(t), the length is L = \int_\alpha^\beta \sqrt{[x'(t)]^2 + [y'(t)]^2}\,dt. Both come from the same Pythagorean strip. If you rotate the curve about the x-axis, the surface area of the resulting shape is S = 2\pi \int y\,ds where ds is the arc length element.

Take the parabola y = x^2 from x = 0 to x = 1. How long is it?

The straight-line distance from (0, 0) to (1, 1) is \sqrt 2 \approx 1.414. But the parabola bends upward, so the actual path along the curve is longer. How much longer?

You cannot just "measure" a curve the way you measure a straight line. Curves bend, and every bit of bend adds to the length. What you need is a way to break the curve into pieces so small that each piece is almost straight — approximate each piece with a line segment, measure the segments, and add up.

That is the entire idea behind arc length. Once you see it, the formula is just Pythagoras's theorem in disguise.

The arc length formula

Pick a point (x, f(x)) on the curve. Move a tiny amount dx to the right. The new point is (x + dx, f(x + dx)). The horizontal change is dx. The vertical change is f(x + dx) - f(x) \approx f'(x)\,dx.

The tiny piece of curve from the first point to the second is almost a straight line — a hypotenuse of a right triangle with legs dx (horizontal) and f'(x)\,dx (vertical). By Pythagoras, the length of that hypotenuse is

ds = \sqrt{(dx)^2 + (f'(x)\,dx)^2} = \sqrt{1 + [f'(x)]^2}\,dx

That is the arc length element. Integrate it from a to b to get the total length.

Arc Length Formula

The length of the curve y = f(x) from x = a to x = b is

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

provided f is differentiable and the derivative is continuous on [a, b].

Reading the formula. The \sqrt{1 + [f'(x)]^2} is the length of the hypotenuse of a right triangle with horizontal leg 1 and vertical leg f'(x). Multiplying by dx gives the length of the tiny piece, and integrating adds up all the tiny pieces.

Try it on the parabola y = x^2 on [0, 1]. Here f'(x) = 2x, so

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

This is a standard integral. With the substitution 2x = \tan\theta (or by recognising the form \int\sqrt{1 + u^2}\,du):

\int_0^1 \sqrt{1 + 4x^2}\,dx = \frac{1}{2}\left[x\sqrt{1 + 4x^2} + \frac{1}{2}\sinh^{-1}(2x)\right]_0^1
= \frac{1}{2}\left[\sqrt{5} + \frac{1}{2}\ln(2 + \sqrt{5})\right]
\approx \frac{\sqrt{5}}{2} + \frac{\ln(2 + \sqrt{5})}{4} \approx 1.118 + 0.361 \approx 1.479

The parabola has length about 1.479 — roughly 4.5\% longer than the straight line between its endpoints. The curve is not much bendier than a line over this range, and the numbers agree with intuition.

The parabola $y = x^2$ from $(0, 0)$ to $(1, 1)$ (solid) and the straight line connecting the same points (dashed). The arc is $\sqrt 2 \approx 1.414$ long if you go straight; following the curve makes it about $1.479$. Every bit of bend adds a little length.

Why the formula is just Pythagoras

Rewrite the arc length element in a way that makes the geometry transparent:

ds = \sqrt{(dx)^2 + (dy)^2}

This looks familiar — it is the Pythagorean theorem for a right triangle with legs dx and dy. It is the length of a straight-line segment between two points that are a horizontal distance dx and a vertical distance dy apart.

From this one expression, the three forms of the arc length formula fall out depending on what you factor out:

All three are the same formula. They differ only in which variable you use as the parameter for the integration.

Parametric arc length

A parametric curve is one where both x and y are functions of some parameter t:

x = x(t), \quad y = y(t), \quad t \in [\alpha, \beta]

The point (x(t), y(t)) traces out a curve as t increases. The length of that curve is

L = \int_\alpha^\beta \sqrt{[x'(t)]^2 + [y'(t)]^2}\,dt

Why this version is important. Some curves are awkward or impossible to write as y = f(x). A circle, for example — the upper and lower halves have different formulas, and at the rightmost point x = 1 the derivative dy/dx is infinite. But parametrically, a unit circle is just

x(t) = \cos t, \quad y(t) = \sin t, \quad t \in [0, 2\pi]

and there is nothing awkward anywhere. The derivatives are x'(t) = -\sin t and y'(t) = \cos t, so

\sqrt{[x'(t)]^2 + [y'(t)]^2} = \sqrt{\sin^2 t + \cos^2 t} = 1

and the circumference is

L = \int_0^{2\pi} 1\,dt = 2\pi

One line, no trickery, no singularities. You have just derived the circumference formula.

Parametric Arc Length

For a parametric curve x = x(t), y = y(t) with t \in [\alpha, \beta], the length of the curve is

L = \int_\alpha^\beta \sqrt{[x'(t)]^2 + [y'(t)]^2}\,dt

The integrand \sqrt{[x'(t)]^2 + [y'(t)]^2} is the speed of a point moving along the curve with parameter t representing time.

The speed interpretation is genuinely useful. If you think of t as time and (x(t), y(t)) as the position of a particle, then (x'(t), y'(t)) is the velocity vector and its magnitude is the speed. Arc length is the total distance travelled, which is the integral of speed over time.

Surface area of revolution

Now a new question. Take the curve y = f(x) from x = a to x = b, and rotate it about the x-axis. This sweeps out a surface in three dimensions — the outer skin of a solid of revolution. What is the area of that surface?

Consider a tiny piece of the curve ds long, at height y above the x-axis. When this piece rotates about the x-axis, it traces out a thin band — almost a cylinder with radius y and height ds. If you unroll the band, it is a rectangle with length 2\pi y (the circumference) and width ds. Its area is

dS = 2\pi y\,ds

Integrating along the curve:

S = \int 2\pi y\,ds

This is the surface area of revolution. The formula takes different forms depending on how you write ds:

Surface Area of Revolution

When the curve y = f(x) on [a, b] is rotated about the x-axis, the surface area of the resulting solid is

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

For a parametric curve x = x(t), y = y(t) with t \in [\alpha, \beta], rotated about the x-axis,

S = 2\pi \int_\alpha^\beta y(t)\sqrt{[x'(t)]^2 + [y'(t)]^2}\,dt

Rotating about the y-axis instead gives S = 2\pi \int x\,ds.

The intuition. Multiply the length of the curve by "how far each point on the curve travels as it rotates." A point at height y travels a circle of circumference 2\pi y when rotated about the x-axis. The surface area is the sum of all these circle-circumferences over the length of the curve.

Test it on the sphere. A sphere of radius R is the surface of revolution of the upper semicircle y = \sqrt{R^2 - x^2} rotated about the x-axis.

y = \sqrt{R^2 - x^2}, \quad \frac{dy}{dx} = \frac{-x}{\sqrt{R^2 - x^2}}
1 + \left(\frac{dy}{dx}\right)^2 = 1 + \frac{x^2}{R^2 - x^2} = \frac{R^2}{R^2 - x^2}
\sqrt{1 + (dy/dx)^2} = \frac{R}{\sqrt{R^2 - x^2}}

So

S = 2\pi \int_{-R}^{R} \sqrt{R^2 - x^2} \cdot \frac{R}{\sqrt{R^2 - x^2}}\,dx = 2\pi R \int_{-R}^{R} dx = 4\pi R^2

The \sqrt{R^2 - x^2} cancels perfectly, leaving a constant integrand. The surface area of a sphere of radius R is 4\pi R^2 — the formula from middle school, rederived from calculus in four lines.

Worked examples

Example 1: Arc length of the catenary

The curve y = \cosh x on [-1, 1] is called a catenary — it is the shape of a hanging cable. Find its length.

Step 1. Compute the derivative.

f(x) = \cosh x, \quad f'(x) = \sinh x

Using the definitions \cosh x = \frac{e^x + e^{-x}}{2} and \sinh x = \frac{e^x - e^{-x}}{2}.

Why: the catenary has the special property that its derivative is \sinh x, and a key identity links \cosh^2 x - \sinh^2 x = 1. That identity will make the square root collapse.

Step 2. Set up the arc length integrand.

1 + [f'(x)]^2 = 1 + \sinh^2 x = \cosh^2 x

using \cosh^2 x - \sinh^2 x = 1. So

\sqrt{1 + [f'(x)]^2} = \sqrt{\cosh^2 x} = \cosh x

(the square root is unambiguous because \cosh x > 0 always).

Why: the identity 1 + \sinh^2 = \cosh^2 turns the ugly \sqrt{1 + \sinh^2 x} into the clean \cosh x. The catenary is designed for this — it is one of the few curves whose arc length integral is elementary.

Step 3. Integrate.

L = \int_{-1}^{1}\cosh x\,dx = [\sinh x]_{-1}^{1} = \sinh 1 - \sinh(-1) = 2\sinh 1

Why: \sinh x is the antiderivative of \cosh x, and it is an odd function, so \sinh(-1) = -\sinh 1.

Step 4. Evaluate numerically.

\sinh 1 = \frac{e - e^{-1}}{2} = \frac{2.71828 - 0.36788}{2} \approx 1.1752

So L \approx 2(1.1752) = 2.3504.

Why: the closed form is 2\sinh 1; the decimal is only for comparison. The straight-line distance between (-1, \cosh(-1)) and (1, \cosh 1) is just 2, so the catenary is about 17.5\% longer than the straight line.

Result: L = 2\sinh 1 \approx 2.3504.

The catenary $y = \cosh x$ on $[-1, 1]$. Its shape — the natural curve of a hanging chain under gravity — has the special property that $1 + (dy/dx)^2 = \cosh^2 x$, so the arc length integral closes up in elementary form. The length is $2\sinh 1 \approx 2.35$.

The catenary comes up whenever a flexible cable hangs under its own weight — suspension bridges, telephone wires, the arch of the Golden Gate. The arc length formula applied to this shape gives a clean answer; most curves do not cooperate this nicely, and the resulting integral must be done numerically.

Example 2: Lateral surface of a cone

A cone has circular base of radius R and height h. Find its lateral surface area (the sloped part, not the base) using the surface-of-revolution formula.

Step 1. Set up the generating curve.

The cone is the surface of revolution of the line segment from (0, 0) to (h, R) rotated about the x-axis. The line is y = (R/h)\,x for x \in [0, h].

Why: the cone is generated by rotating a straight line, so the curve f(x) = (R/h)\,x is just a linear function. Its derivative is constant.

Step 2. Compute the arc length element.

f'(x) = \frac{R}{h}, \quad \sqrt{1 + [f'(x)]^2} = \sqrt{1 + \frac{R^2}{h^2}} = \frac{\sqrt{h^2 + R^2}}{h}

Why: the slant height of the cone is \ell = \sqrt{h^2 + R^2}, and the factor \ell/h is this slant height divided by the axial length — geometrically, it is the reciprocal of the cosine of the cone's half-angle.

Step 3. Set up the surface area integral.

S = 2\pi \int_0^h y\sqrt{1 + (y')^2}\,dx = 2\pi \int_0^h \frac{R}{h}x \cdot \frac{\sqrt{h^2 + R^2}}{h}\,dx
= \frac{2\pi R\sqrt{h^2 + R^2}}{h^2}\int_0^h x\,dx

Why: everything except x is constant, so it comes out of the integral. The remaining \int_0^h x\,dx is elementary.

Step 4. Evaluate.

\int_0^h x\,dx = \frac{h^2}{2}
S = \frac{2\pi R\sqrt{h^2 + R^2}}{h^2} \cdot \frac{h^2}{2} = \pi R\sqrt{h^2 + R^2} = \pi R \ell

where \ell = \sqrt{h^2 + R^2} is the slant height of the cone.

Why: the h^2 cancels perfectly, and you are left with \pi R \ell — the formula that you have seen in geometry textbooks.

Result: The lateral surface area of a cone is \pi R \ell, where \ell = \sqrt{h^2 + R^2} is the slant height.

Cross-section of a cone with $R = 3$, $h = 4$. The line $y = (R/h)x$ generates the cone when rotated about the x-axis. The slant height is $\ell = \sqrt{R^2 + h^2} = 5$. Lateral surface area: $\pi R \ell = 15\pi$.

You have rederived two of the three standard cone formulas: lateral surface \pi R \ell and (from the previous article on volumes) volume \frac{1}{3}\pi R^2 h. The third — total surface \pi R(\ell + R) — just adds the base.

Common confusions

Going deeper

If you can set up arc length and surface area integrals in the standard forms, you have the essentials. The rest of this section explores the deeper structure — why ds is a single object, what arc length has to do with geometry, and where the elliptic integrals come from.

The arc length element as a 1-form

All three arc length formulas come from the same object:

ds^2 = dx^2 + dy^2

This is the metric of the Euclidean plane — it says how to measure infinitesimal distances. Different parameterisations of the curve let you factor ds in different ways, but the object itself is coordinate-independent.

In more advanced geometry you will encounter curved surfaces, where the metric is not dx^2 + dy^2 but something more complicated like ds^2 = g_{11}\,du^2 + 2g_{12}\,du\,dv + g_{22}\,dv^2. The coefficients g_{ij} form the metric tensor, and the arc length formula in general is \int\sqrt{g_{ij}\,du^i\,du^j}. The flat-plane version you just learned is a special case where g_{11} = g_{22} = 1 and g_{12} = 0.

Arc length as a reparametrisation

Given any curve (x(t), y(t)) with t \in [\alpha, \beta], you can reparameterise by arc length: introduce a new variable s defined by

s(t) = \int_\alpha^t \sqrt{[x'(\tau)]^2 + [y'(\tau)]^2}\,d\tau

Then s runs from 0 to L as t runs from \alpha to \beta. Writing the curve as (X(s), Y(s)) for s \in [0, L], you have

\sqrt{[X'(s)]^2 + [Y'(s)]^2} = 1

A curve parameterised by arc length has unit speed — it moves one unit of distance per unit of parameter. This is the canonical parameterisation, and it makes many formulas in differential geometry cleaner (curvature, torsion, the Frenet-Serret equations).

Why the ellipse fights back

The arc length of an ellipse \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 leads to the integral

L = 4a\int_0^{\pi/2}\sqrt{1 - e^2\sin^2\theta}\,d\theta

where e = \sqrt{1 - b^2/a^2} is the eccentricity. This integral does not close up in elementary functions — it defines the complete elliptic integral of the second kind, written E(e). The circumference of the ellipse is 4a\,E(e).

The elliptic integrals were studied intensively in the 18th and 19th centuries because they appear everywhere — pendulum periods, planetary orbits, and of course ellipses themselves. They are the gateway to a whole branch of mathematics called elliptic functions, which turned out to be central to number theory, cryptography, and even the proof of Fermat's Last Theorem.

For practical purposes, a very good approximation is the Ramanujan formula:

L \approx \pi\bigl[3(a + b) - \sqrt{(3a + b)(a + 3b)}\bigr]

which is accurate to within 0.04\% for all eccentricities. The Indian mathematician Srinivasa Ramanujan found this formula in 1914 by manipulating series expansions of the elliptic integral — another example of a hand-crafted approximation that out-performs what most numerical methods can do.

The coarea formula

There is a beautiful generalisation of the surface-of-revolution formula to arbitrary surfaces. The coarea formula relates the area of a surface in space to an integral over level sets. Stated loosely: if you have a function f on a region and you integrate the arc length of each level set against df, you recover the volume. The surface-of-revolution case is the simplest example — each "level set" is a circle (one for each x), and summing 2\pi y times the arc length element gives the total surface.

Where this leads next

You now know how to measure lengths along curves and areas of surfaces of revolution. These are the one-dimensional and two-dimensional pieces of the same puzzle that multivariable calculus will turn into a full theory of integration over curves and surfaces.