Here I present a (fairly common) derivation for the volume of an n sphere.
Where an n-sphere is defined by
$S^n=\left\{x \epsilon \mathbb{R}^{n+1}: ||x||)=r\right\}$
We first start by calling on what we already know;
$V_{1}=2 \pi r$ for $S^0$ $V=\pi r^2$ for $S^{1}$ and $V=\frac{4}{3} \pi r^3$ for $S^2$ These can both be obtained most easily via usual methods of integrating in polar and spherical coordinates, respectively.
By inspection - and intuition - we should think that the volume is proportional to $r^{n+1}$ i.e some constant times r to the power of the dimension we are in.
Therefore we should expect that the formula for the volume of a sphere in
m-dimensions looks a little like
$V_{m}=\nu_{m} r^m$
where $\nu_{m}$ is the constant of proportionality that we are after.
Hence the volume element (in m-dimensions and spherical coordinates) is given by; $dV_{m}= m \nu_{m} r^{m-1} dr$
Now this seems like a leap of faith, or unjustified - but let's take a look at some Gaussian Integrals;
$$I_m=\int \! e^{-\left(x_{1}^2+...+x_m^{2}\right)} dx_1....dx_m=$$ where the integral is over all of $\mathbb{R}^m$
From our knowledge of a standard Gaussian Integral;
$I=\int_{- \infty}^{\infty} \! e^{-x^2} dx=\sqrt{\pi}$
and recognising the fact that the $x_i$ are independent, we can factorise the integral;
$$I_m=\int \! e^{-x_1^2} dx_1...\int \! e^{-x_m^2} dx_m = I^m $$
Therefore $$I_m=\sqrt{\pi}^m = \pi^\frac{m}{2}$$
Now, transforming into m-dimensional spherical coordinates so that;
$$x_1^2+....+x_m^2=r^2$$
Which gives $$I_m=\int \! e^{-r^2} dV_{m} $$
and using our definition for the spherical volume element we got before;
$$\int_{0}^{\infty} \! e^{-r^2} m\nu_m r^{m-1} dr=\pi^\frac{m}{2}$$
Taking out the variables which aren't dependent on r from the integrand we get;
$$I_m=m\nu_m \int_{0}^{\infty} \! e^{-r^2} r^{m-1} dr =\pi^\frac{m}{2}$$
You math geeks out there should recognise the form of the integral as somewhat close to the gamma function;
$$\Gamma(z+1)= \int_{0}^{\infty} u^{z}e^{-u} du $$
Now, substituting $$u=r^2$$ then $$\frac{du}{dr}=2r$$ or in terms of u; $$\frac{du}{dr}=2\sqrt{u}$$
$$V_m=\nu_m m \int_{0}^{\infty} e^{-u}\left(\sqrt{u}\right)^{m-1}\frac{1}{2\sqrt{u}} du$$
$$\pi^{\frac{m}{2}}=\frac{\nu_m m}{2} \int_{0}^{\infty}e^{-u} \left(\sqrt{u}\right)^{m-2} du$$
$$=\frac{\nu_m m}{2} \int_{0}^{\infty}e^{-u} u^{\frac{m}{2}-1} du$$
Which we can almost assimilate with the above form of the Gamma function;
A simple substitution provides us with;
$$\frac{m}{2}-1=z \rightarrow \frac{m}{2}=z+1$$
Hence $$\Gamma(z+1)= \int_{0}^{\infty} u^{z}e^{-u} du $$ becomes;
$$\Gamma(\frac{m}{2})=\int_{0}^{\infty} u^{\frac{m}{2}-1} e^{-u} du$$
Therefore; $$\pi^{\frac{m}{2}}=\frac{ \nu_m m \Gamma(\frac{m}{2}) }{2}$$
So putting it all together, we recover our unknown coefficient $\nu_m$;
$$\nu_m= \frac{\pi^\frac{m}{2}}{\frac{m}{2}\Gamma(\frac{m}{2})}$$
Or you can obtain an equivalent formula by looking at $$\Gamma(\frac{m}{2}+1)= \int_{0}^{\infty} u^{\frac{n}{2}} e^{-u} du $$ and integrating by parts to recognise that;
$$\Gamma(\frac{m}{2}+1)=\frac{m}{2}\Gamma(\frac{m}{2})$$
Hence putting it all together, we recover the Volume of an n-dimensional sphere of radius r as;
$$V_m=\frac{\pi^{\frac{m}{2}} r^m}{\frac{m}{2} \Gamma(\frac{m}{2})}$$
= $$\frac{\pi^\frac{m}{2} r^m}{\Gamma(\frac{m}{2}+1)}$$
Having done all that, it could be an interesting idea to see how the volume of the unit n-sphere changes as we increase the number of dimensions;
In dimension 2; $$V=\pi$$
In dimension 3; $$V=\frac{4}{3}\pi$$
.
.
.
and so on.
Below is a plot of volume vs dimension for the unit sphere - constructed using the volume formula we obtained above;
I find it quite amazing that there is a maximum! That is that there is a point (around n=5) that when you start adding dimensions to the sphere; its volume decreases!
Final Word:
It may not seem like there are any real world applications for the use of n-dimensional spheres, but there actually are! In statistical mechanics, a system of N independent particles may be confined to the surface or within a volume of an N dimensional sphere in phase space (due to energy considerations)- where here we don't think of N as spatial dimensions, but degrees of freedom. So when N becomes large (as it does in the thermodynamic limit) we are required to use the above formula and it is just as well that the volume decreases to zero as N becomes large, otherwise it would be indicating that the thermodynamic properties (which in general depend on the statistical behaviour of the particles) of our ensemble have would become infinite - which is an indication that something has gone wrong in the math!
