The volume of a $d$ dimensional hypersphere of radius $r$ is given by:
$$V(r,d)=\frac{(\pi r^2)^{d/2}}{\Gamma\left(\frac{d}{2}+1\right)}$$
What intrigues me about this, is that $V\to 0$ as $d\to\infty$ for any fixed $r$. How can this be? For fixed $r$, I would have thought adding a dimension would make the volume bigger, but apparently it does not. Anyone got a good explanation?
Best Answer
I suppose you could say that adding a dimension "makes the volume bigger" for the hypersphere, but it does so even more for the unit you measure the volume with, namely the unit cube. So the numerical value of the volume does go towards zero.
Really, of course, it is apples to oranges because volumes of different dimensions are not commensurable -- it makes no sense to compare the area of the unit disk with the volume of the unit sphere.
All we can say is that in higher dimensions, a hypersphere is a successively worse approximation to a hypercube (of side length twice the radius). They coincide in dimension one, and it goes downward from there.