# Volume of $n$-hyper-sphere derivation

I am trying to prove the curse of dimensionality and on Wikipedia https://en.wikipedia.org/wiki/Curse_of_dimensionality The volume of Hypersphere is given as $$\frac{2r^d \pi^{d/2}}{d\;\Gamma (d/2)}$$ Can anybody help me that how we can derive this formula? as far as I know, the volume for $$N$$-dimensional hypersphere is $$\frac{r^n\pi^{n/2}}{\;\Gamma (n/2+1)}$$

• You only need the $r^n$ term to demonstrate the curse of dimensionality – Henry Mar 24 '20 at 17:33
• In most texts I've seen, your formula is for the measure of the $n$-ball, not the $n$- or $n+1$-sphere. (In other words, the sphere is the surface, not the bulk.) – J.G. Mar 24 '20 at 17:34
• @Henry can you show me how? – christopher Mar 24 '20 at 17:39
• $99\%$ or the volume of a hyperball radius is between $\sqrt[n]{0.01}r$ and $r$ from the origin. For large $n$ you find $\sqrt[n]{0.01}r$ is close to $r$; e.g. for $n=10000$ it is about $0.99954r$ – Henry Mar 24 '20 at 17:45

The Gamma function has the property $$\Gamma(z+1)=z\Gamma(z)$$, so $$\frac {r^{n}\pi ^{{n/2}}}{\Gamma (n/2+1)} = \frac {r^{n}\pi ^{{n/2}}}{\frac n2\Gamma (n/2)} = \frac {2r^{n}\pi ^{{n/2}}}{n\Gamma (n/2)}.$$ The two formulas you have therefore agree.
• I once derived the volume of a sphere by using an integral of the cross-section circles making thin disks of $\delta x$ thickness. Then I did the same thing using a cross-section of a 3D sphere instead to get the volume of a 4D sphere. Is that the same thing your answer shows? – Nεo Pλατo Mar 24 '20 at 18:45