Arc Length & Solid of Revolution

Arc length and x-axis revolution formulas, evaluated numerically with Simpson integration.

Worked example — Arc Length & Solid of Revolution

Method	500

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Arc Length & Solid of Revolution

A curve cannot be measured directly, so arc length is defined as a limit rather than a formula: replace the curve with a chain of straight chords, add their lengths, then let the chords shrink towards nothing. Everything below is that one idea plus Pythagoras.

What each symbol means

f(x): the curve, given as a function of x — one y for every x. That restriction is why the presets offer a semicircle and never a full circle: a circle stands two y values tall above most of its width, so no single f can trace it.
f'(x): the slope, and the only thing the integrand ever consults. How high the curve sits never enters — the flat function 1 across [0, 2] has slope 0 everywhere and length exactly 2, the width of the interval and nothing more.
[a, b]: the interval, which is part of the answer rather than a detail of it. The parabola runs 4.646784 over [0, 2] and 1.478943 over [0, 1] — not half, because a steeper stretch of curve packs more length into the same width.
n: the number of slices Simpson is given. Alone among these, it says nothing about the curve: it decides only how well the integral is approximated, and the Error estimate row is what reports whether it was enough.

Where the formula comes from

Take two points on the curve a small step Δx apart. The straight line between them has length √(Δx² + Δy²). That is Pythagoras and there is nothing else in it.
Pull Δx out of the root: √(Δx² + Δy²) = Δx·√(1 + (Δy/Δx)²). The shape of the final answer is already visible.
Now let Δx shrink. The ratio Δy/Δx becomes the derivative f′(x), so each vanishing chord contributes √(1 + f′(x)²) dx.
Add them all: L = ∫ₐᵦ √(1 + f′(x)²) dx, the line the readout prints. Step 1 is on screen as well, as the Chord length row — and for a straight line the two agree exactly. Set the function to x on [0, 1] and both read 1.414214, because there one chord already is the whole curve.

How to read what you see

The Chord length row is step 1 with a single segment, so it can never be the larger of the two: 4.472136 against the parabola’s 4.646784, and the gap between them is how much the curve bends. Read Error estimate carefully, because it is not a distance from the truth — it prints |L₂ₙ − Lₙ|, the disagreement between n slices and 2n, so it says whether the method has settled, not whether it landed in the right place.

Assumes: That f can be differentiated across [a, b], and that it is a function of x in the first place. Simpson’s rule assumes more than the arc-length formula does: that the integrand is smooth enough to be matched by parabolas across each pair of slices.
Breaks when: Choose frac(x) on [0, 3] and set the integration steps to 400. The graph is three rising segments separated by three vertical drops, and at every drop there is no derivative for the formula to use. The readout answers regardless: 29.23057, which is neither the 4.242641 you get by ignoring the drops nor the 7.242641 you get by counting them — it is the integrator differencing across a jump. The only thing that gives it away is the Error estimate row beneath it, 12.49396, nearly half the answer it is reporting on. A single corner, by contrast, costs almost nothing: |x| on [−1, 1] returns 2.827046 against the exact 2√2 = 2.828427, because one bad point inside an integral is one point too few to matter.

The same curve, the same n — and one answer is eleven times better 🖖

The semicircle and sphere presets use the identical function on the identical interval with the identical 800 slices. Yet the arc length comes out 3.1705 against π = 3.1416, wrong by 0.9%, while the surface area comes out 12.5559 against 4π = 12.5664, wrong by only 0.08%. The difference is in the integrand: √(1 + f′²) spikes to about 71 at x = ±1, where a semicircle turns vertical, and the arc-length integral has nothing to damp it. The surface integral carries a factor of |f(x)|, which is exactly 0 at those endpoints — so it multiplies the worst part of the curve by nothing.

Spinning a curve into a solid 🖖

The second mode of this tool acts like a lathe or a potter's wheel: take the curve y=f(x) and spin it around the x-axis to sweep out a 3D shape. To find its volume, the tool slices the solid into thin discs, each a circle of radius f(x), and adds them up. Spin the line y=x from 0 to 3 and you get a perfect cone of volume 9π — no geometry formula needed, just the integral.

The trumpet you can fill but never paint 🖖

Revolve y=1/x from x=1 to infinity around the x-axis and you get Gabriel's Horn. Its volume converges to exactly π, yet its surface area is infinite. This "painter's paradox" means the horn holds a finite π units of paint, but no finite amount could ever coat its inner wall. Evangelista Torricelli found this in the 1640s, before calculus formally existed, and even he thought it was absurd.

Practice

Check yourself

Predict the answer first, then use the controls above to find out. Reveal only after you have committed to a guess — that is what makes it practice.

Load the sphere preset and set the interval to a = 0, b = 0.5 — a band cut from the sphere’s equator up to half its radius. The surface area reads 3.141593. Now slide the band up to a = 0.4, b = 0.9, keeping its height at 0.5. Predict the new surface area before you press anything.

Show answer

Still 3.141593. Every band of height 0.5 encloses exactly π of surface, wherever you cut it — [−0.25, 0.25] and [−0.9, −0.4] print the same figure. Volume does nothing of the kind: those bands hold 1.439897, 0.87441 and 1.538071. Near the pole each strip of surface is shorter around but tilted more steeply, and the two effects cancel exactly, so only the height survives. Archimedes proved this and asked for a sphere inside a cylinder on his tomb — band for band, the sphere has the same surface as the cylinder that just contains it.
The third insight above claims Gabriel’s Horn has finite volume and infinite surface area, but gives you no numbers. Get them: choose f(x) = 1/x, set a = 1 and the mode to Solid of revolution, and read both figures at b = 10, then 100, then 1000.

Show answer

Volume goes 2.827433 → 3.110179 → 3.153626, closing on π and gaining less each decade. Surface area goes 15.1775 → 29.6451 → 44.1739 — up by roughly 14.5 every time you multiply b by ten, and it will do that forever. Far out along the horn the surface integrand is essentially 2π/x, whose integral is 2π ln b, and 2π ln 10 = 14.47. Something that gains a fixed amount per decade has no limit; something that gains less each decade has one. Distrust that last volume, though: 3.153626 is above π, which is impossible, because 2000 uniform steps cannot resolve the steep part near x = 1 across an interval that long. The overshoot is the grid, not the horn.

Problems solved in full

The arc length of y = x² from 0 to 2 in closed form 5 steps
The arc length of y = x² from 0 to 2 is 4.646784. Derive it in closed form — then explain why you were lucky. This is the parabola arc state.
1. Set it up. The derivative of x² is 2x, so the integrand carries 4x² and nothing worse.
  
  $L = \int_{0}^{2} 1 + (f^{'} (x))^{2} d x = \int_{0}^{2} 1 + 4 x^{2} d x$
2. The antiderivative of √(1 + 4x²) is standard — one algebraic term and one inverse hyperbolic sine, which is what a quadratic under a root always produces.
  
  $\int 1 + 4 x^{2} d x = \frac{x}{2} 1 + 4 x^{2} + \frac{1}{4} arsinh (2 x) + C$
3. Evaluate at the ends. The lower limit contributes nothing, and the two pieces at x = 2 come to 4.1231 and 0.5237.
  
  $L = [\frac{x}{2} 1 + 4 x^{2} + \frac{1}{4} arsinh 2 x]_{0}^{2} = 17 + \frac{1}{4} arsinh 4 = 4.1231 + 0.5237 = 4.646784$
4. Compare with the straight line between the endpoints. The curve is only 3.9% longer than the chord over this range, which is why numerical arc length looks easy until it is not.
  
  $chord = (2 - 0)^{2} + (4 - 0)^{2} = 20 = 4.472136; \frac{L}{chord} = 1.0391$
5. Now try to repeat the trick one power up. The integrand becomes a square root of a quartic, and that is the boundary between elementary and elliptic.
  
  $f = x^{3} : \int 1 + 9 x^{4} d x — elliptic, not elementary$
Answer

√17 + ¼ arsinh 4, and the luck is that the parabola is one of almost nothing that works. Arc length asks you to integrate √(1 + f′²), and a square root of a polynomial is elementary only in a handful of cases. Move up one power to y = x³ and the integrand is √(1 + 9x⁴), which is elliptic and has no expression in elementary functions. So is the arc length of an ellipse — that is where elliptic integrals got their name, and it is why the perimeter of an ellipse has no formula while its area has a trivial one. The parabola escapes because its derivative is linear, so 1 + f′² is a quadratic, and quadratics under a square root are exactly what the arsinh substitution was invented for.

$L = 17 + \frac{1}{4} arsinh 4 = 4.646784$
Surface 12.55591 and volume 4.18879 of the unit semicircle 6 steps
Revolve the unit semicircle and the panel gives surface 12.55591 and volume 4.18879. One of those is exact to six figures and the other is 0.083% wrong. Work out which, and why. This is the sphere by revolution state.
1. Write down the function and its derivative. The derivative is where the trouble is going to come from, and it is visible already: the denominator vanishes at both ends.
  
  $f (x) = 1 - x^{2}, f^{'} (x) = \frac{- x}{1 - x ^{2}}, 1 + (f^{'})^{2} = \frac{1}{1 - x ^{2}}$
2. Now form the surface integrand. The factor that blows up and the factor that vanishes are reciprocals, and they cancel completely.
  
  $2 π f 1 + (f^{'})^{2} = 2 π 1 - x^{2} \cdot \frac{1}{1 - x ^{2}} = 2 π — a constant$
3. Integrating a constant over an interval of length 2 needs no calculus. The answer is 4π, and it does not depend on how the sphere was sliced — which is the content of Archimedes' result.
  
  $S = \int_{- 1}^{1} 2 π d x = 4 π = 12.566371 (Archimedes: sphere = its cylinder’s side)$
4. The volume is an ordinary polynomial integral, and it is the one Simpson's rule gets right to the last digit it prints.
  
  $V = \int_{- 1}^{1} π (1 - x^{2}) d x = π [x - \frac{x ^{3}}{3}]_{- 1}^{1} = \frac{4 π}{3} = 4.18879$
5. Compare the printed surface with 4π. The shortfall is small but it is not rounding — it is a systematic miss concentrated at the two poles.
  
  $printed S = 12.55591 : \frac{4 π - 12.55591}{4 π} = 0.083% low, while V is exact to 6 figures$
6. The cause is in step 1. Exact cancellation is not available to a routine that only samples values.
  
  $f^{'} (\pm 1) \to \infty : the integrand is constant only in exact arithmetic; sampled, the poles misbehave$
Answer

The volume is exact and the surface is short by 0.083%, because the surface integrand is a constant the integrator cannot see. Work it by hand and something remarkable happens: √(1 + f′²) is 1/√(1 − x²), and it cancels the √(1 − x²) in front of it exactly, leaving 2π. The integrand is flat, the integral is 4π, and that is Archimedes' theorem — a sphere has the same surface as the side of the cylinder that just contains it, which he had carved on his tombstone. Numerically, though, f′ blows up at x = ±1, and Simpson's rule is sampling a function that is 2π everywhere except at two points where the arithmetic goes to pieces. The volume integrand is π(1 − x²), an honest polynomial, and Simpson integrates polynomials exactly. The panel's own error estimate, 0.005, tells you which of the two answers to trust.

$S = 4 π exactly, because the integrand is 2 π; V = \frac{4 π}{3}$

References (1)

Insight block 3 — Gabriel's Horn, and how strange it looked at the time: P. Mancosu and E. Vailati, "Torricelli's Infinitely Long Solid and Its Philosophical Reception in the Seventeenth Century." Isis 82(1), 50–70, 1991.

By the 42Math team · Updated 2026-07-15

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

parabola arc - Arc length of y=x^2 on [0,2] is longer than the straight chord.
semicircle arc - Semicircle arc on [-1,1] approaches pi as expected.
sphere by revolution - Revolving a semicircle around x-axis yields sphere-like 4pi and 4pi/3 values.
cone by revolution - Revolving y=x on [0,3] produces a cone with analytic volume 9pi.