Seven½ volumes of n-balls

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

1. Contents
2. Elementary inductions
3. Multivariable calculus and n-spheres
4. Without integration
5. Generating functions
6. 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.