The Volume of the n-ball

Introduction

This post is an extension of my Medium article: The Math Behind “The Curse of Dimensionality”. If you have any questions or feedback, feel free to leave a comment below or reach out on LinkedIn.

Here, I derive the formula for the volume of the n-ball using integral calculus and Wallis integrals.

An n-ball is a generalization of the ordinary “ball” to arbitrary dimensions. It is defined as the set of points in the n-dimensional Euclidean space that are at a fixed distance from a central point (we will condider 0, the origin). The n-ball has important applications, particulary in machine learning and data analysis.

The equation of an n-ball is given by:

x_{1}^{2} + x_{2}^{2} + \dots + x_{N}^{2} \leq r^{2}

where $r$ is the radius of the ball.

Derivation of the volume formula

One of the most important properties of the n-ball is its volume. The volume of an n-ball of radius $r$ is given by:

V_{N} (r) = \frac{π^{(N + 1) / 2}}{Γ ((N + 1) / 2)} r^{N + 1}

where $Γ$ is the gamma function. In the rest of this post, we will derive this formula.

We define the unit (radius equal to 1) n-ball as the following space: $B_{n} = {x_{1}, \dots, x_{m}; \sum_{i = 1}^{n} x_{i}^{2} \leq 1}$

We note $V_{n} = V_{n} (1)$ the volume of this ball. And more generally $V_{n} (R)$ the volume of the n-ball with radius $R$ . From now on, we will use $V_{n}$ to refer to the volume of the unit n-ball:

V_{n} = \int_{x \in B_{n}} d x_{1} d x_{2} \dots d x_{n}

First, note that the volume of the n-ball of radius $R$ is $R^{n} V_{n}$ : simply use the change of variable $y_{i} \leftarrow x_{i} / R$ in the expression of $V_{n} (R)$ .

Now, let’s simplify $V_{n}$ :

\begin{aligned} V_{n} & = \int_{x_{1}^{2} + \dots + x_{n}^{2} \leq 1} d x_{1} d x_{2} \dots d x_{n} \\ = \int_{x_{1}^{2} \leq 1} (\int_{x_{2}^{2} + \dots + x_{n}^{2} \leq 1 - x_{1}^{2}} d x_{2} \dots d x_{n}) d x_{1} \\ = \int_{x_{1}^{2} \leq 1} V_{n - 1} (\sqrt{1 - x_{1}^{2}}) d x_{1} \end{aligned}

We can now replace the expression of the volume of the $n - 1$ ball of radius $\sqrt{1 - x_{1}^{2}}$ with the previous relation:

\begin{aligned} V_{n} & = \int_{x_{1}^{2} \leq 1} V_{n - 1} (\sqrt{1 - x_{1}^{2}}) d x_{1} \\ = V_{n - 1} \int_{x_{1}^{2} \leq 1} {(\sqrt{1 - x_{1}^{2}})}^{n - 1} d x_{1} \\ = V_{n - 1} \int_{- 1}^{1} {(\sqrt{1 - x^{2}})}^{n - 1} d x \end{aligned}

We use the change of variable $x = \cos (θ)$ (so $d x = - \sin (θ) d θ$ ) to simplify the integral:

\begin{aligned} V_{n} & = V_{n - 1} \int_{- 1}^{1} {(\sqrt{1 - x^{2}})}^{n - 1} d x \\ = - V_{n - 1} \int_{π}^{0} \sin^{n - 1} (θ) \sin (θ) d θ \\ = V_{n - 1} \int_{0}^{π} \sin^{n} (θ) d θ \\ = 2 V_{n - 1} \int_{0}^{π / 2} \sin^{n} (θ) d θ \\ = I_{n} V_{n - 1} \end{aligned}

We note $I_{n} = 2 \int_{0}^{π / 2} \sin^{n} (θ) d θ$ .

Note that $\int_{0}^{π / 2} \sin^{n} (θ) d θ$ is a famous integral in mathematics: it is called the Wallis integral and often denoted by $W_{n}$ . It can be derived using integration by parts (I will prove this formula in the appendix). Depending on the parity of $n$ , the integral can be expressed as:

$W_{2 p} = \frac{π}{2} \frac{(2 p)!}{2^{2 p} (p!)^{2}}$ and $W_{2 p + 1} = \frac{2^{2 p} (p!)^{2}}{(2 p + 1)!}$

Using our recursive relation, we know that: $V_{n} = I_{n} I_{n - 1} \dots I_{2} V_{1}$ and $V_{1} = V_{1} (1) = 2$ (the length of the segment $[- 1, 1]$ ).

We are going to make use of a very useful property:

I_{2 p} I_{2 p + 1} = 4 W_{2 p} W_{2 p + 1} = \frac{π}{2} \frac{(2 p)!}{2^{2 p} (p!)^{2}} \frac{2^{2 p} (p!)^{2}}{(2 p + 1)!} = \frac{2 π}{2 p + 1} = \frac{π}{p + 1 / 2}

We will also need the Gamma function. All you need to know is these 2 expressions:

Γ (n) = (n - 1)!

and

Γ (n + 1 / 2) = (n - 1 / 2) \times \dots \times 1 / 2 \times π^{1 / 2}

Now, we can easily compute the volume $V_{n}$ by grouping successive terms, depending again on the parity of $n$ .

If $n = 2 p$ : $\begin{aligned} V_{2 p} & = I_{2 p} I_{2 p - 1} \dots I_{2} V_{1} \\ = I_{2 p} (I_{2 p - 1} I_{2 p - 2}) \dots (I_{3} I_{2}) \times 2 \\ = π \times \frac{(2 p)!}{2^{2 p} (p!)^{2}} \times \frac{π}{p - 1 / 2} \times \frac{π}{p - 3 / 2} \times \dots \times \frac{π}{3 / 2} \times \frac{π}{1 / 2} \\ = \frac{(2 p)!}{2^{p} (p!)^{2}} \times \frac{π^{p}}{(2 p - 1) (2 p - 3) \dots (1)} \\ = \frac{π^{p}}{2^{p}} \times \frac{(2 p)!}{(p!)^{2}} \times \frac{(2 p) (2 p - 2) \dots (4) (2)}{(2 p!)} \\ = \frac{π^{p}}{2^{p}} \times \frac{(2 p)!}{(p!)^{2}} \times \frac{2^{p} p!}{(2 p)!} \\ = \frac{π^{p}}{p!} \\ = \frac{π^{n / 2}}{Γ (\frac{n}{2} + 1)} \end{aligned}$

If $n = 2 p + 1$ , it is a bit less complicated:

\begin{aligned} V_{2 p + 1} & = I_{2 p + 1} I_{2 p} I_{2 p - 1} \dots I_{2} V_{1} \\ = (I_{2 p + 1} I_{2 p}) (I_{2 p - 1} I_{2 p - 2}) \dots (I_{3} I_{2}) \times 2 \\ = \frac{π}{p + 1 / 2} \times \frac{π}{p - 1 / 2} \times \frac{π}{p - 3 / 2} \times \dots \times \frac{π}{3 / 2} \times \frac{1}{1 / 2} \\ = \frac{π^{p + \frac{1}{2}}}{(p + 1 / 2) (p - 1 / 2) \dots (1 / 2) π^{\frac{1}{2}}} \\ = \frac{π^{p + \frac{1}{2}}}{Γ (p + \frac{1}{2} + 1)} \\ = \frac{π^{n / 2}}{Γ (\frac{n}{2} + 1)} \end{aligned}

So we obtained a single formula for the volume of the n-ball, regardless of the parity of $n$ :

V_{n} = V_{n} (1) = \frac{π^{n / 2}}{Γ (\frac{n}{2} + 1)}

And by extension, the volume of the n-ball of radius $R$ is:

V_{n} (R) = R^{n} V_{n} (1) = R^{n} \frac{π^{n / 2}}{Γ (\frac{n}{2} + 1)}

Note that there is a very easy way to derive the area of the surface of n-ball using this expression. We simply need to derive the volume of the n-ball of radius $R$ with respect to $R$ and we obtain the surface area of the n-ball of radius $R$ :

\frac{d}{d R} V_{n} (R) = n R^{n - 1} V_{n} (1) = n R^{n - 1} \frac{π^{n / 2}}{Γ (\frac{n}{2} + 1)}

Paricularly, the surface area of the n-ball of radius $1$ is:

S_{n} = n \frac{π^{n / 2}}{Γ (\frac{n}{2} + 1)} = \frac{2 π^{n / 2}}{Γ (\frac{n}{2})}

Appendix: Derivation of the Wallis integrals

The Wallis integrals are defined as:

W_{n} = \int_{0}^{π / 2} \sin^{n} (θ) d θ

We will derive a recursive formula using integration by parts.

\begin{aligned} W_{n} & = \int_{0}^{π / 2} \sin^{n} (θ) d θ \\ = \int_{0}^{π / 2} \sin^{n - 2} (θ) (1 - \cos^{2} (θ)) d θ \\ = W_{n - 2} - \int_{0}^{π / 2} \sin^{n - 2} (θ) \cos^{2} (θ) d θ \\ = W_{n - 2} - ({[\frac{\sin^{n - 1} (θ)}{n - 1} \cos (θ)]}_{0}^{π / 2} - \frac{1}{n - 1} \int_{0}^{π / 2} - \sin^{n - 1} (θ) \sin (x) d θ) \\ = W_{n - 2} - \frac{1}{n - 1} W_{n} \end{aligned}

We conclude that:

W_{n} = \frac{n - 1}{n} W_{n - 2}

We can now distinguish 2 cases depending on the parity of $n$ :

If $n = 2 p$ , we have:

W_{2 p} = \frac{2 p - 1}{2 p} W_{2 p - 2} = \frac{2 p - 1}{2 p} \frac{2 p - 3}{2 p - 2} \dots \frac{1}{2} W_{0} = \frac{2 p - 1}{2 p} \frac{2 p - 3}{2 p - 2} \dots \frac{1}{2} \frac{π}{2} = \frac{π}{2} \frac{(2 p)!}{2^{2 p} (p!)^{2}}

We obtained the last equality by multiplying the denominator and the numerator by $2 p (2 p - 2) \dots 2$ to make $(2 p)!$ appear. We then factorized each term of the denominator by 2.

Similarly, if $n = 2 p + 1$ , we have:

W_{2 p + 1} = \frac{2 p}{2 p + 1} W_{2 p - 1} = \frac{2 p}{2 p + 1} \frac{2 p - 2}{2 p - 1} \dots \frac{2}{3} W_{1} = \frac{2 p}{2 p + 1} \frac{2 p - 2}{2 p - 1} \dots \frac{2}{3} \frac{2}{1} = \frac{2^{2 p} (p!)^{2}}{(2 p + 1)!}

We conclude that

W_{2 p} = \frac{π}{2} \frac{(2 p)!}{2^{2 p} (p!)^{2}}

and

W_{2 p + 1} = \frac{2^{2 p} (p!)^{2}}{(2 p + 1)!}