Volume of spheres in higher dimensions? What is the volume of spheres in higher dimensions?
 A: Let $\kappa_n$ be the volume of the unit $n$-sphere $B_n$ and write the points of $B_n$ in the form $(x,y,{\bf z})$ with ${\bf z}\in{\mathbb R}^{n-2}$. For given $(x,y)\in B_2$ one has $|{\bf z}|^2\leq 1-r^2$, where $r:=\sqrt{x^2+y^2}$. These ${\bf z}$ fill an $(n-2)$-sphere of radius $\sqrt{1-r^2}$, and the $(n-2)$-dimensional volume of this sphere amounts to $\kappa_{n-2}(1-r^2)^{(n-2)/2}$. Therefore we get $$\kappa_n=\int_{B_2}\kappa_{n-2}(1-r^2)^{(n-2)/2}{\rm d}(x,y)=2\pi \kappa_{n-2}\int_0^1 (1-r^2)^{(n-2)/2}\>r\>dr={2\pi\over n}\>\kappa_{n-2}\ .$$
By means of this this recursion formula and using the known values $\kappa_1=2$, $\kappa_2=\pi$ one easily obtains the $\Gamma$-formula quoted in other answers.
A: $$V_n = \frac{\pi^{n/2}}{\Gamma(n/2+1)}$$
As posted earlier here.
A: Did you try the obvious?
A: Start with an unrelated integral
$$I = \int_{-\infty}^\infty \cdots \int_{-\infty}^\infty e^{(-x_1^2+x_1^2+\cdots x_n^2) } dx_1 dx_2 \cdots dx_n = \pi^{n/2}$$
This should be easy to easy if you notice that this can be factored as a product of n integrals each in the variable $x_i$ and that $\int_{-\infty}^\infty e^{-x^2}dx =\sqrt{\pi}$
Now try to calculate the same integrate by making a change to polar coordinates. These new coordinates will have one $r$ coordinate and $n-1$ angular coordinates $\{\theta_i\}$
$$I = \int e^{(-x_1^2+x_1^2+\cdots x_n^2) } dV_n =$$
To express the volume element in the new coordinates you can either use inference from lower dimensions (or if you're a physicist, a dimensional argument) and say that $$V = C_n r^n \Rightarrow dV_n = nC_n r^{n-1}dr$$
or you can explicitly compute the Jacobian to show that $|J(r,\theta_1,\cdots,\theta_n)|= r^{n-1} \cos^{n-2}\theta_1\cos^{n-3}\theta_2 \cdots \cos \theta_{n-2} $
so
$$I = \int_0^\infty e^{-r^2} nC_nr^{n-1} dr = C_n\frac{n}{2} \Gamma (\frac{n}{2}) = C_n\left(\frac{n}{2}\right)! $$
$$ C_n = \frac{\pi^{n/2}}{\left(\frac{n}{2}\right)!} $$
$$V_n =  \frac{\pi^{n/2}}{\left(\frac{n}{2}\right)!}  r^n$$
on a sidenote, you also get the surface area from $dV_n = S_n(r) dr$
$$S_n(r)= \frac{(2\pi)^{n/2}}{\Gamma (\frac{n}{2})} r^{n-1}$$
