# Seven½ volumes of n-balls

Posted by in math

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?

## 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.