The volume of an $n$-sphere of radius $R$ is
$V_n(R) = \frac{\pi^{n/2} R^n}{\Gamma(\frac{n}{2}+1)}$
and the surface area is
$S_n(R) = \frac{2\pi^{n/2}R^{n-1}}{\Gamma(\frac{n}{2})} = \frac{n\pi^{n/2}R^{n-1}}{\Gamma(\frac{n}{2}+1)} = \frac{d V_n(R)}{dR}$
What is the intuition for this relationship between the volume and surface area of an $n$-sphere? Does it relate to the fact that the $n$-sphere is the most compact shape in $n$ dimensions, or is that merely a coincidence?
Are there other shapes for which it holds, or is this the limiting case of a relationship of inequality? For example, an $n$-cube of side $R$ has volume $V^c_n(R)=R^n$ and surface area $S^c_n(R)=2nR^{n-1}$, so that
$\frac{dV^c_n(R)}{dR} = n R ^{n-1}$
and we have $S^c_n = 2 dV^c_n/dR$.
Edit: As pointed out by Rahul Nahrain in the comments below, if we define $R$ to be the half-side length of the unit cube rather than the side length, then we have the relationship $S^c_n(R)=\frac{d}{dR} V^c_n(R)$, exactly as for the sphere. Is there a sense in which relationships like this can be stated for a large class of shapes?