area of the $n$-sphere

The area of $S^n$ the unit $n$-sphere (or hypersphere) is the same as the totalsolid angle it subtends at the origin. To calculate it, consider the followingintegral

$$I(n)={\int }_{{ℝ}^{n+1}}{e}^{-{\sum }_{i=1}^{n+1}{x}_i^2}{d}^{n+1}x.$$

Switching to polar coordinates we let $r^2={\sum }_{i=1}^{n+1}{x}_i^2$ and theintegral becomes

$$I(n)={\int }_{S^n}𝑑\mathrm{\Omega }{\int }_0^{\mathrm{\infty }}{r}^n{e}^{-r^2}𝑑r.$$

The first integral is the integral over all solid angles and is exactly what wewant to evaluate. Let us denote it by $A(n)$. With the change of variable$t=r^2$, the second integral can be evaluated in terms of the gamma function$\mathrm{\Gamma }(x)$:

$$I(n)/A(n)=\frac{1}{2}{\int }_0^{\mathrm{\infty }}{t}^{\frac{n-1}{2}}{e}^{-t}𝑑t=\frac{1}{2}\mathrm{\Gamma }\left(\frac{n+1}{2}\right).$$

We can also evaluate $I(n)$ directly in Cartesian coordinates:

$$I(n)={\left[{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2}𝑑x\right]}^{n+1}={\pi }^{\frac{n+1}{2}},$$

where we have used the standard Gaussian integral ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2}𝑑x=\sqrt{\pi }$.

Finally, we can solve for the area

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

If the radius of the sphere is $R$ and not $1$, the correct area is$A(n)R^n$.

Note that this formula works only for $n\ge 0$. The first few special casesare

• $\mathrm{\Gamma }(1/2)=\sqrt{\pi }$, hence $A(0)=2$ (in this case, thearea just counts the number of points in $S^0=\{+1,-1\}$);

• $\mathrm{\Gamma }(1)=1$, hence $A(1)=2\pi$ (this is the familiar resultfor the circumference of the unit circle);

• $\mathrm{\Gamma }(3/2)=\sqrt{\pi }/2$, hence $A(2)=4\pi$ (this is thefamiliar result for the area of the unit sphere);

• $\mathrm{\Gamma }(2)=1$, hence $A(3)=2{\pi }^2$;

• $\mathrm{\Gamma }(5/2)=3\sqrt{\pi }/4$, hence $A(4)=8{\pi }^2/3$.