# Volumes of n-balls: what is so special about n=5?

1. The volume of an $n$-dimensional ball of radius $1$ is given by the classical formula

For small values of $n$, we have

It is not difficult to prove that $V_n$ assumes its maximal value when $n=5$.

Question. Is there any non-analytic (i.e. geometric, probabilistic, combinatorial…) demonstration of this fact? What is so special about $n=5$?

2. I also have a similar question concerning the $n$-dimensional volume $S_n$ (“surface area”) of a unit $n$-sphere. Why is the maximum of $S_n$ attained at $n=7$ from a geometric point of view?

note: the question has also been asked on MathOverflow for those curious to other answers.

If you compare the volume of the sphere to that of its enclosing hyper-cube you will find that this ratio continually diminishes. The enclosing hyper-cube is 2 units in length per side if $R=1$. Then we have:
The reason for this behavior is how we build hyper-spheres from low dimension to high dimensions. Think for example extending $S_1$ to $S_2$. We begin with a segment extending from $-1$ to $+1$ on the $x$ axis. We build a 2 sphere by sweeping this sphere out along the $y$ axis using the scaling factor $\sqrt{1-y^2}$. Compare this to the process of sweeping out the respective cube where the scale factor is $1$. So now we only occupy approximately $3/4$ of the enclosing cube (i.e. square for $n=2$). Likewise for $n=3$, we sweep the circle along the $z$ axis using the scaling factor, loosing even more volume compared to the cylinder if we had not scaled the circle as it was swept. So as we extend $S_{n-1}$ to get $S_n$ we start with the diminished volume we have and loose even more as we sweep out into the $n^{th}$ dimension.