The volume of a d dimensional hypersphere of radius r is given by:

V(r,d)=(πr2)d/2Γ(d2+1)

What intrigues me about this, is that V→0 as d→∞ 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?

**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.

**Attribution***Source : Link , Question Author : probabilityislogic , Answer Author : hmakholm left over Monica*