### Vanishing Hyperballs

Spherical ball contained within a cubic region

We all know that the area of a disk — the interior of a circle — is ${\pi r^2}$ where ${r}$ is the radius. Some of us may also remember that the volume of a ball — the interior of a sphere — is ${\frac{4}{3}\pi r^3}$.

The unit disk and ball have equations in two and three variables:

$Disk: x^2+y^2<1$

$Ball: x^2+y^2+z^2<1$.

These are special cases of an ${n}$-ball, the region ${\mathbb{B}_n}$ of ${n}$-dimensional space comprising all points whose distance from the origin is less than 1:

$n-Ball: x_1^2+x_2^2+ . . . x_n^2<1$.

or ${\mathbb{B}_n = \{ x\in \mathbb{R}^n : ||x|| < 1 \}}$. For ${n>3}$, we call ${\mathbb{B}_n}$ a hyperball. A 1-ball is just a line segment. A 2-ball is a disk and a 3-ball is the interior of a sphere. The surface of the n-ball ${\mathbb{B}_n}$ is the (n-1)-sphere ${\mathbb{S}_{n-1}}$

Volumes and Hyper-volumes

We denote the hyper-volume of the n-ball ${\mathbb{B}_n}$ by ${V_n}$ and its surface area by ${S_{n-1}}$. These quantities can be defined recursively, starting from ${n=0}$ or ${n=1}$ and iterating to higher dimensions. But there are also closed form solutions. The volume of the ${n}$-ball is

$V_n=\frac{\pi^{n/2} r^n}{\Gamma(\frac{n}{2}+1)}$

where ${\Gamma}$ is the gamma function. We recall some properties of ${\Gamma(n)}$:

$\Gamma(\frac{1}{2}) = \sqrt{\pi} , \Gamma(1) = 1 \quad and \quad \Gamma(x+1)=x\Gamma(x)$

Using these properties, we can express ${V_n}$ without evaluation of integrals using Euler’s definition of the gamma function. We get

$Even\ Dimension: \quad V_{2n} = \frac{\pi^n r^{2n}}{n!}$

$Odd\ Dimension: \quad V_{2n+1} = \frac{2^{2n+1}n!\pi^n r^{2n+1}}{(2n+1)!}$

Vanishing Hyper-volumes

Hyper-volume of B_n for n<= 8 [Table from Wikipedia article “

A table of the values of ${V_n}$ for ${1\le 8}$ is shown here. We note from this table that the volume of ${\mathbb{B}_n}$ is proportional to ${r^n}$. But the constant of proportionality behaves in a curious way. The coefficient increases with dimension for ${n\le 5}$, but for higher dimensions it decreases. Indeed, as we shall see, this coefficient tends to zero as ${n\rightarrow\infty}$.

The expressions above allow us to relate the hyper-volumes for different dimensions:

$\frac{V_{2n+2}}{V_{2n}} = \frac{\pi r^2}{n+1}$

$\frac{V_{2n+1}}{V_{2n-1}} = \frac{\pi r^2}{n+\textstyle{\frac{1}{2}}}$

It is a simple matter to confirm that the familiar values for low dimensions satisfy these relationships.

Hyperballs and Hypercubes

We may wonder how we can compare hyper-volumes in different dimensional spaces. But it is straightforward to compare the hyperball in ${n}$ dimensions with a fixed cube in the same space. The hyperball ${\mathbb{B}_n}$ of unit radius, and diameter 2, fits within a cube of edge-length 2, touching it at the centre of each face. The volume of this cube is ${2^n}$, and we may compare the volume ${V_n}$ of the n-ball to this.

With ${r=1}$, the above equations become

$\frac{V_{2n+2}}{V_{2n}} = \frac{\pi}{n+1}\quad\quad\quad \frac{V_{2n+1}}{V_{2n-1}} = \frac{\pi}{n+\textstyle{\frac{1}{2}}}$

Combining these, we see the effect of an increase in dimension by 2 is

$V_{n+2} \sim \frac{\pi}{n}\times V_n$

It is abundantly clear that, as ${n}$ increases, ${V_n}$ becomes smaller, rapidly approaching zero for large ${n}$. This is in marked contrast to the surrounding cube, whose volume ${2^n}$ doubles with every additional dimension. For growing ${n}$, the hyperball occupies a smaller and smaller proportion of the associated cube.