In a pale imitation of the Victor Guillemin style, here are seven ways to calculate the volume of an n-ball, in seven not-so-easy but somewhat spoiler-ful exercises. What’s your favorite? And can you shed any light on negative dimensions or the generating function in the last method?

## Contents

- Contents
- Elementary inductions
- Multivariable calculus and n-spheres
- Without integration
- Generating functions
- Special values, just for fun

## Elementary inductions

If you weren’t satisfied with being confined to dimension 3 when you learned the formula \[ |D^3| = \frac{4}{3} \pi \] for the volume of the unit 3-ball, then I suspect you eventually figured out that you could:

- Method 1
- Integrate over a 2-disc and induct from dimensions 0 and 1. (Walkthrough on Wikipedia)

If you were particularly bored in your calculus class, maybe you challenged yourself to use only one base case:

- Method 2
- Starting from a 1-disc instead, finish off the resulting integral with a trigonometric substitution and a reduction formula. (Walkthrough by Raulston)

## Multivariable calculus and n-spheres

There’s a variant of Method 2 for the analogue of surface area, in which you calculate \[ |S^{n+1}| = \int_{x=-1}^1 |S^n| (1-x^2)^{\frac{1}{2}n} \, ds \] where \(ds = (1-x^2)^{-1/2} \, dx\) is the element of arc length for \(y = \sqrt{1-x^2}\). This is halfway to a purely spherical-coordinates method, which a little multivariable calculus lets you access:

- Method 3
- Carefully write down, in spherical coordinates, the element of \(n\)-volume in the unit \(n\)-sphere (the boundary \(S^n\) of the unit \((n+1)\)-ball). Then integrate to find \(|S^n|\). (Walkthrough on Wikipedia, using Beta functions instead of a reduction formula).

Or if you were bored in a probability or stats class, maybe you pulled some shenanigans with Gaussian integrals and the radius coordinate:

- Method 4
- Using spherical symmetry, write \(\left(\int_{-\infty}^{\infty} e^{-x^2} \, dx\right)^n\) as a product of \(|S^{n-1}|\) and a 1-variable integral that can be written as a value of the Gamma function. Then compute its value for all \(n\) by using polar coordinates in the case \(n=2\). (Walkthrough on Wikipedia)

A high-power variant of this method is to compute the Laplace transform of \(f(y) = |D^n| y^{n/2} = \int_{|x|^2 \leq y} dx\) and is detailed by Lasserre (paywalled).

## Without integration

Wikipedia’s a great source for these walkthroughs, and it even includes mention of unit \(L^p\)-balls (with sources!). But let’s go off the trail a little bit. Our first stop goes way back—to Archimedes’s use of Cavalieri’s principle to avoid explicitly using calculus:

- Method 5
- Calculate that over each point of \(D^n\), the 2-dimensional cross-sections of \(D^{n+2}\) and the cone on \(D^n \times S^1\) (with vertex at the origin) have the same area. Then note that \(|S^1 \times X| = 2\pi |X|\) and that passing to a height-1 cone divides volume by \(n+2\) (i.e. the dimension of the ambient space). (Walkthrough by Makarov, paywalled)

As with Method 2, there’s an \(S^n\) variant that deserves mentioning. (To recover the same recurrence as above, use the fact that \(D^{n+2}\) is the cone on \(S^{n+1}\) to relate their volumes.)

- Method 5½
- Calculate that projecting to the equatorial disc \(f: S^n \to D^n\) multiplies n-volume by a factor of \((1-|f(x)|^2)^{-1/2}\). Since the \(S^1\)-shaped strip of \(S^{n+1}\) over \(f(x) \in D^n\) has radius \((1-|f(x)|^2)^{1/2}\), conclude that \(|S^{n+1}| = 2\pi|D^n|\).

If you’re willing to think only in even dimensions (i.e. \(D^{2n}\)), here’s a super-cute method exploiting complex numbers that I first heard of from my advisor:

- Method 6
- Let \((D^2)^n\) denote the product of unit discs in \(\mathbb{C}^n\) (so its volume is \(\pi^n\)), and let \(E = \{ 0 < |z_1| < \cdots < |z_n| < 1\} \subseteq (D^2)^n\) be a fundamental domain under the action of permuting the coordinates. Check that \[ \begin{align*} f: z &\mapsto w \in \mathbb{C}^n \\ w_1 &= z_1 \\ w_i &= z_i \sqrt{1 - \left| \frac{z_{i-1}}{z_i} \right|^2} \end{align*} \] defines a volume-preserving diffeomorphism from \(E\) to a full-measure subset of \(D^{2n}\). (Walkthrough by Hijab, paywalled)

## Generating functions

Method 5½ produces the recurrence relation \(|S^{n+1}| = 2\pi |D^n|\). Generating functions are a tool for solving recurrence relations; it’s overkill in this case, but you can do it:

- Method 7
- Let \( b(r) = \sum_{n=0}^\infty |D^n| r^n \). Show that \(b’(r) = \sum_{n=0}^\infty |S^n| r^n\), and use the recurrence above to conclude \(b’ = 2 + 2\pi r b\). This is a linear ODE; solve it to get \[b(r) = e^{\pi r^2} \left(1 + 2\int_0^r e^{-\pi t^2}\, dt\right),\] and expand it as a power series to recover the coefficients \(|D^n|\). (Walkthrough by Badger, paywalled)

Cook remarks that there ought to be some way to explain the occurrence of the Gaussian in the formula for \(b(r)\); you can get part of the way there with the following variant.

- Method 7½
- The Laplace transform of \(f(x) = e^{-x^2}\) is \[ \mathcal{L}\{f\}(y) = \int_0^\infty e^{-x^2} e^{-xy} \, dx . \] Expand \(e^{-xy}\) as a power series, and use the techniques of Method 4 to write the \(y^n\) coefficient of \(\mathcal{L}\{f\}(y)\) in terms of \(|S^n|\). (Yes, you can swap the integral and the sum since the sum of integrals converges.)

The wrinkle here is that \(|S^n|\) ends up
in the *denominators* of the coefficients,
while a substitution in the formula for \(b(r)\) shows that
\[ b(r) = \mathcal{L}\{x \mapsto e^{-\pi x}\}(-2\pi r) . \]
Alas, I don’t see how to get more directly from this Laplace transform
to \(b(r)\); but on the bright side, equating the coefficients
in the two variants provides a roundabout way of proving
the Gamma function’s duplication formula at integers,
\[ \Gamma(n)\Gamma\left(n+\frac{1}{2}\right) =
2^{1-2n} \sqrt{\pi} \, \Gamma(2n) . \]

## Special values, just for fun

Trivial coefficients occur with \(|D^0| = 1\), \(|D^2| = \pi\), and \(|S^5| = \pi^3\), which makes me think the Raspberry Pi Foundation missed a minor branding opportunity earlier this year. Or if you’re a tauist, you also get \(|S^1| = \tau\). For something approaching a tauist spiritual experience, go back and replace all occurrences of \(\pi\) with \(\frac{1}{2}|S^1|\).

Maybe you noticed that the Gamma function is still defined at negative half-integers, which goads you into asking whether \[ |S^{-2}| = \frac{2 \pi^{\frac{-2+1}{2}}}{\Gamma\left(\frac{-2+1}{2}\right)} = -\frac{1}{\pi} \] has any meaning. Despite having stumbled over negative-dimensional spheres in a quest to convince myself that (-2)-categories are a thing, I’m leaning toward: probably not—at least, not in any sense that looks obviously to me like a moral analogue of volume. But if you’ve got ideas, I’m listening.