If we denote the Cartesian coordinate of a $$n$$-dimensional Euclidean space by a $$n$$-vector $$\mathbf{x}$$, then a $$n$$-ball centered at the origin with radius $$r$$ is the set of points that satisfy

$||\mathbf{x}||_2 \le r$

Taking this definition, a 1-ball is the line segment $$[-r,r]$$, which its “volume” is $$2r$$ and has no surface area. A 2-ball is a circle with radius $$r$$, which has the “surface” area of $$2\pi r$$ and “volume” of $$\pi r^2$$.

Let us denote the volume of an $$n$$-ball with radius $$r$$ by $$V_n(r)$$, and consider that $$n$$-ball is defined by

\begin{align} ||\mathbf{x}||_2 = x_1^2 + x_2^2 + \cdots + x_{n-1}^2 + x_n^2 &\le r \\ x_1^2 + x_2^2 + \cdots + x_{n-1}^2 &\le r - x_n^2 \end{align}

Obviously

$0 \le r - x_n^2 \\ -r \le x_n \le r.$

Therefore, we can find $$V_n(r)$$ recursively:

\begin{align} V_n(r) &= \int_{-r}^r V_{n-1}(\sqrt{r^2-x_n^2})dx_n \\ &= \int_{-r}^r V_{n-1}(r\sqrt{1-u^2})d(ru) \\ &= r\int_{-1}^1 V_{n-1}(r\sqrt{1-u^2})du. \\ \end{align}

## Proportionality of Volume

Firstly, we prove $$V_n(r)$$ is proportional to $$r^n$$ by induction:

Consider $$n=1$$, $$V_1(r)=2r=rV_1(1)$$. If we have $$V_k(r)=r^kV_k(1)$$, then

\begin{align} V_{k+1}(r) &= r\int_{-1}^1 V_k(r\sqrt{1-u^2})du \\ &= r\int_{-1}^1 (r\sqrt{1-u^2})^kV_k(1)du \\ &= r^{k+1}\int_{-1}^1 (\sqrt{1-u^2})^kV_k(1)du \\ &= r^{k+1}\int_{-1}^1 V_k(\sqrt{1-u^2})du \\ &= r^{k+1}V_{k+1}(1). \end{align}

Thus

$V_n(r) = r^n V_n(1).$

## Formula for Volume

\begin{align} V_n(1) &= \int_{-1}^1 V_{n-1}(\sqrt{1-u^2})du \\ &= \int_{-1}^1 (\sqrt{1-u^2})^{n-1}V_{n-1}(1)du \\ &= V_{n-1}(1)\int_{-1}^1 (\sqrt{1-u^2})^{n-1}du. \end{align}

If we denote for $$n>1$$,

$I_n = \int_{-1}^1 (\sqrt{1-u^2})^n du \\$

then

\begin{align} V_n(1) &= V_{n-1}(1)I_{n-1} \\ &= V_{n-2}(1)I_{n-2}I_{n-1} \\ &= V_1(1)I_1I_2\cdots I_{n-1} \\ &= 2I_1I_2\cdots I_{n-1}, \end{align}

and

$V_n(r) = 2r^nI_1I_2\cdots I_{n-1},$

valid for $$n\ge 1$$ if we specially define $$I_0=1$$, and which

$I_n = \int_{-1}^1 (\sqrt{1-u^2})^n du = \int_{-\pi/2}^{\pi/2} \cos^{n+1}t dt = \int_0^{\pi} \sin^{n+1}t dt = \frac{\Gamma(\frac{n}{2}+1)}{\Gamma(\frac{n+3}{2})}\sqrt{\pi}$

where the Gamma function is defined as

$\Gamma(z) = \int_0^\infty e^{-t}t^{z-1} dt,$

with the property that $$z\Gamma(z) = \Gamma(z+1)$$, and $$\Gamma(n)=(n-1)!$$ for positive integers $$n$$. But we can also consider $$I_n$$ without resolving to the Gamma function. Consider that

\begin{align} I_n &= \int_0^{\pi} \sin^{n+1}t dt \\ &= \int_0^{\pi} \sin^n t \sin t dt \\ &= [(-\cos t)\sin^n t]_0^{\pi} + \int_0^{\pi} n\cos^2 t\sin^{n-1} t dt \\ &= n\int_0^{\pi} (1-\sin^2 t)\sin^{n-1} t dt \\ &= n[I_{n-2}-I_n] \\ \textrm{therefore}\quad I_n &= \frac{n}{n+1} I_{n-2}. \end{align}

It is easy to see that $$I_1=\pi/2$$, $$I_2=4/3$$, $$I_3=3\pi/8$$, and

$I_nI_{n-1} = \frac{n}{n+1}I_{n-2}\frac{n-1}{n}I_{n-3} = \frac{n-1}{n+1}I_{n-2}I_{n-3}$

thus

\begin{align} I_{2k}I_{2k-1}\cdots I_1 &= \frac{2k-1}{2k+1}(I_{2k-2}I_{2k-3})^2I_{2k-4}I_{2k-5}\cdots I_1 \\ &= \frac{2k-1}{2k+1}(\frac{2k-3}{2k-1})^2(I_{2k-4}I_{2k-5})^3I_{2k-6}I_{2k-7}\cdots I_1 \\ &= \frac{1}{2k+1}\frac{(2k-3)^2}{2k-1}(I_{2k-4}I_{2k-5})^3I_{2k-6}I_{2k-7}\cdots I_1 \\ &= \frac{1}{2k+1}\frac{1}{2k-1}\cdots\frac{1}{7}\frac{3^{k-1}}{5}(I_2I_1)^k \\ &= \frac{1}{2k+1}\frac{1}{2k-1}\cdots\frac{1}{7}\frac{3^{k-1}}{5}(\frac{2\pi}{3})^k \\ &= \frac{1}{2k+1}\frac{1}{2k-1}\cdots\frac{1}{7}\frac{1}{5}\frac{1}{3}(2\pi)^k \\ &= \frac{2^k k!}{(2k+1)!} (2\pi)^k \\ &= \frac{2^{2k} k!}{(2k+1)!}\pi^k \end{align}

and

\begin{align} I_{2k-1}I_{2k-2}\cdots I_1 &= \frac{2k-2}{2k}(I_{2k-3}I_{2k-4})^2I_{2k-5}I_{2k-6}\cdots I_1 \\ &= \frac{2k-2}{2k}(\frac{2k-4}{2k-2})^2(I_{2k-5}I_{2k-6})^3I_{2k-7}I_{2k-8}\cdots I_1 \\ &= \frac{1}{2k}\frac{(2k-4)^2}{2k-2}(I_{2k-5}I_{2k-6})^3I_{2k-7}I_{2k-8}\cdots I_1 \\ &= \frac{1}{2k}\frac{1}{2k-2}\cdots\frac{4^{k-2}}{6}(I_3I_2)^{k-1}I_1 \\ &= \frac{1}{2k}\frac{1}{2k-2}\cdots\frac{4^{k-2}}{6}(\frac{\pi}{2})^k \\ &= \frac{1}{2k}\frac{1}{2k-2}\cdots\frac{2^{k-4}}{6}\pi^k \\ &= \frac{1}{k}\frac{1}{k-1}\cdots\frac{1}{3}\frac{\pi^k}{4} \\ &= \frac{1}{k!}\frac{\pi^k}{2}. \end{align}

Therefore we have

\begin{align} V_{2k+1}(r) &= 2r^{2k+1}\frac{2^{2k}k!}{(2k+1)!}\pi^k = \frac{2^{2k+1}k!}{(2k+1)!}\pi^kr^{2k+1} \\ V_{2k}(r) &= 2r^{2k}\frac{1}{k!}\frac{\pi^k}{2} = \frac{1}{k!}\pi^kr^{2k}. \end{align}

Indeed, we can also prove by induction that

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

And the surface area of the $$n$$-ball is obtained by $$\frac{d}{dr}V_n(r)$$.

## Alternative Derivation

Below is an alternative derivation by Greg Huber (1982). “Gamma function derivation of n-sphere volumes”. Am. Math. Monthly 89(5):301–302 (PDF)

Consider

\begin{align} \Gamma(z) &=\int_0^{\infty} e^{-t}t^{z-1} dt \\ &= \int_0^{\infty} e^{-u^2}u^{2z-2} d(u^2) \\ &= 2 \int_0^{\infty} e^{-u^2}u^{2z-1} du. \\ \Gamma(\tfrac{1}{2}) &= 2\int_0^{\infty} e^{-u^2} du = \sqrt{\pi}, \end{align}

for the reason that $$\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi\sigma^2}}\exp(-\frac{x^2}{2\sigma^2}) du = 1$$.

Denote the surface of $$n$$-ball by $$\omega_{n-1}(r)$$, we can write the volume as the integration of surface area as the radius expands from zero:

$V_n(r) = \int_0^r \omega_{n-1}(1)u^{n-1} du = \frac{r^n}{n}\omega_{n-1}(1)$

or the volume as the integration of all the points in the ball:

$V_n(r) = \iint\cdots\int_{||\mathbf{x}||\le R} dx_1dx_2\cdots dx_n.$

Consider the density integration on an infinitely large $$n$$-ball, where the density at point $$\mathbf{x}$$ is $$e^{-r^2}$$ with $$r$$ defined by

$r^2 = x_1^2 + x_2^2 + \cdots + x_n^2.$

Using the first integral we have (assumed radius 1)

$\int_0^{\infty} \omega_{n-1}r^{n-1}e^{-r^2}dr = \omega_{n-1}\int_0^\infty e^{-r^2}r^{n-1}dr = \frac{1}{2}\omega_{n-1}\Gamma(\frac{n}{2}).$

But using the second integral we have

\begin{align} & \iint\cdots\int_{||\mathbf{x}||\le \infty} e^{-r^2} dx_1dx_2\cdots dx_n \\ =&\iint\cdots\int_{||\mathbf{x}||\le \infty} e^{-x_1^2-x_2^2-\cdots-x_n^2} dx_1dx_2\cdots dx_n \\ =&\iint\cdots\int_{||\mathbf{x}||\le \infty} e^{-x_1^2}dx_1 e^{-x_2^2}dx_2\cdots e^{-x_n^2}dx_n \\ =&\left(\int_{-\infty}^{\infty} e^{-x^2}dx\right)^n \\ =&\pi^{n/2}. \end{align}

Because they shall be equal, we have

\begin{align} \frac{1}{2}\omega_{n-1}\Gamma(\frac{n}{2}) &= \pi^{n/2} \\ \omega_{n-1} &= \frac{2\pi^{n/2}}{\Gamma(\frac{n}{2})} \\ \omega_{n-1}(r) &= \frac{2\pi^{n/2}r^{n-1}}{\Gamma(\frac{n}{2})} \\ V_n = \frac{1}{n}\omega_{n-1} &= \frac{2\pi^{n/2}}{n\Gamma(\frac{n}{2})} \\ &= \frac{\pi^{n/2}}{\frac{n}{2}\Gamma(\frac{n}{2})} \\ &= \frac{\pi^{n/2}}{\Gamma(\frac{n}{2}+1)} \\ V_n(r) &= \frac{\pi^{n/2}r^n}{\Gamma(\frac{n}{2}+1)}. \end{align}

We can also verify that $$\frac{d}{dr}V_n(r) = \omega_{n-1}(r)$$.

## Properties of n-ball

If we consider the $$n$$-ball with unit diameter instead of unit radius, its volume is $$V_n(1/2)$$. We can see that, due to the exponential nature in the denominator,

$\lim_{n\to\infty} V_n(\tfrac{1}{2}) = 0.$

Therefore, the higher the dimension, the denser we can pack the balls. I first learned this result from Professor Raymond Yeung.