Volume of a $l_2$ ball of radius $r$

Question

What is the volume of a ball $B \subset \mathbb{R}^n$, where $B = \{ \theta : ||\theta||_2 \leq r\}$ where $|| \cdot ||_2$ is the $l_2$-norm?

The $l_2$-norm is given by $||x||_2 := \left( \sum_{i=1}^d |x_i|^2 \right)^{1/2}$.

Answer 1

The volume of an $l_2$ ball in $\mathbb{R}^n$ is given by

$$ V_n(R)={\pi^{n\over2}\over\Gamma\left({n\over2}+1\right)}R^n $$

We can actually derive this formula using some intuition and the properties of Beta function. Before jumping into $\mathbb{R}^n$ case, let's look at the cases for $\mathbb{R}$, $\mathbb{R}^2$, and $\mathbb{R}^3$. Since $V_2(R)$ gives us the area of a circle with radius of $R$, we know that

$$ V_2(R)=\int_{-R}^R 2\sqrt{R^2-x^2}\mathrm{d}x=\pi R^2 $$

$$ V_3(R)=\int_{-R}^R \pi\left(\sqrt{R^2-x^2}\right)^2\mathrm{d}x={4\over3}\pi R^3 $$

For consistency, we define $V_1(R)=2R$, and we can discover that we are always integrating $V_{n-1}(R)$ to obtain $V_n(R)$. As a result, we can find a one-dimensional recursive formula for $V_n$:

$$ V_n(R)=\int_{-R}^RV_{n-1}\left(\sqrt{R^2-x^2}\right)\mathrm{d}x $$

We know that the volume of an $n$-ball with radius $R$ is always proportional to $R^n$, so we can transform the above identity into the following:

$$ V_n(R)=V_{n-1}(1)\int_{-R}^R(R^2-x^2)^{n-1\over 2}\mathrm{d}x $$

Now, it is time to introduce substitution: $x=uR$, $\mathrm{d}x=R\mathrm{d}u$.

$$ \begin{aligned} V_n(R)&=V_{n-1}(1)\int_{-1}^1(R^2-u^2R^2)^{n-1\over2}R\mathrm{d}u \\ &=V_{n-1}(1)R^n\int_{-1}^1(1-u^2)^{n-1\over2}\mathrm{d}u \\ &=2V_{n-1}(1)R^n\int_0^1(1-u^2)^{n-1\over2}\mathrm{d}u \end{aligned} $$

In order to utilize Beta function, we perform another substitution where $t=u^2\Rightarrow u=\sqrt{t}$ and $\mathrm{d}u={t^{-1/2}\over2}\mathrm{d}t$.

$$ \begin{aligned} V_n(R)&=2V_{n-1}(1)\int_0^1{t^{-1/2}\over2}(1-t)^{n-1\over2}\mathrm{d}t \\ &=V_{n-1}(1)\int_0^1t^{{1\over2}-1}(1-t)^{{n+1\over2}-1}\mathrm{d}t \end{aligned} $$

Now, we introduce Beta function $B(x,y)=\int_0^1t^{x-1}(1-t)^{y-1}\mathrm{d}t$.

$$ V_n(R)=R^nV_{n-1}(1)R^nB\left({1\over2},{n+1\over2}\right) $$

If we define the "volume coefficient" $M_k\equiv V_k(1)$, we obtain the following relation for $M_n$ and $M_{n-1}$ by setting $R=1$ in the above equation:

$$M_n=B\left({1\over2},{n+1\over2}\right)M_{n-1}$$

Using another definition for beta function $B(x,y)={\Gamma(x)\Gamma(y)\over\Gamma(x+y)}$, we can express $M_n$ in terms of Gamma functions:

$$M_n={\Gamma(1/2)\Gamma\left({n-1\over2}+1\right)\over\Gamma\left({n\over2}+1\right)}M_{n-1}$$

It can be shown that $\Gamma(1/2)=\sqrt\pi$, so

$$M_n=\sqrt\pi\cdot{\Gamma\left({n-1\over2}+1\right)\over\Gamma\left({n\over2}+1\right)}M_{n-1}$$

Now, we can try to apply the recursive formula for multiple times and obtain a closed form for $M_n$ due to massive cancellation.

$$ \begin{aligned} M_n &=\sqrt\pi\cdot{\Gamma\left({n-1\over2}+1\right)\over\Gamma\left({n\over2}+1\right)}\cdot\sqrt\pi\cdot{\Gamma\left({n-2\over2}+1\right)\over\Gamma\left({n-1\over2}+1\right)}\cdot\sqrt\pi\cdot{\Gamma\left({n-3\over2}+1\right)\over\Gamma\left({n-2\over2}+1\right)}\cdots \sqrt\pi\cdot{\Gamma\left({3-2\over2}+1\right)\over\Gamma\left({3-1\over2}+1\right)}M_2 \\ &={\pi^{n-2\over2}M_2\over\Gamma\left({n\over2}+1\right)} \end{aligned} $$

Given $M_2=V_2(1)=\pi$, we obtain

$$M_n={\pi^{n/2}\over\Gamma\left({n\over2}+1\right)}$$

Using the definition $V_n(R)=V_n(1)R^n=M_nR^n$, we can get the volume formula of a ball in $\mathbb{R}^n$ with $l_2$ metric:

$$V_n(R)={\pi^{n\over2}\over\Gamma\left({n\over2}+1\right)}R^n$$