volume of the $n$-sphere

The volume contained inside $S^n$, the $n$-sphere (or hypersphere), isgiven by the integral

$$V(n)={\int }_{{\sum }_{i=1}^{n+1}{x}_i^2\le 1}{d}^{n+1}x.$$

Going to polar coordinates ($r^2={\sum }_{i=1}^{n+1}{x}_i^2$) this becomes

$$V(n)={\int }_{S^n}𝑑\mathrm{\Omega }{\int }_0^1{r}^n𝑑r.$$

The first integral is the integral over all solid angles subtended by thesphere and is equal to its area$A(n)=\frac{2{\pi }^{\frac{n+1}{2}}}{\mathrm{\Gamma }\left(\frac{n+1}{2}\right)}$,where $\mathrm{\Gamma }(x)$ is the gamma function.The second integral is elementary and evaluates to${\int }_0^1{r}^n𝑑r=1/(n+1)$.

Finally, the volume is

$$V(n)=\frac{{\pi }^{\frac{n+1}{2}}}{\frac{n+1}{2}\mathrm{\Gamma }\left(\frac{n+1}{2}\right)}=\frac{{\pi }^{\frac{n+1}{2}}}{\mathrm{\Gamma }\left(\frac{n+3}{2}\right)}.$$

If the sphere has radius $R$ instead of $1$, then the correct volume is$V(n)R^{n+1}$.

Note that this formula works for $n\ge 0$. The first few cases are

• $\mathrm{\Gamma }(3/2)=\sqrt{\pi }/2$, hence $V(0)=2$ (this is thelength of the interval $[-1,1]$ in $ℝ$);

• $\mathrm{\Gamma }(2)=1$, hence $V(1)=\pi$ (this is the familiarresult for the area of the unit circle);

• $\mathrm{\Gamma }(5/2)=3\sqrt{\pi }/4$, hence $V(2)=4\pi /3$(this is the familiar result for the volume of the sphere);

• $\mathrm{\Gamma }(3)=2$, hence $V(3)={\pi }^2/2$.