# area of the $n$-sphere

The area of $S^{n}$ the unit $n$-sphere (or hypersphere) is the same as the total solid angle it subtends at the origin. To calculate it, consider the following integral

 $I(n)=\int_{\mathbb{R}^{n+1}}e^{-\sum_{i=1}^{n+1}x_{i}^{2}}\,d^{n+1}x.$

Switching to polar coordinates we let $r^{2}=\sum_{i=1}^{n+1}x_{i}^{2}$ and the integral becomes

 $I(n)=\int_{S^{n}}d\Omega\int_{0}^{\infty}r^{n}e^{-r^{2}}\,dr.$

The first integral is the integral over all solid angles and is exactly what we want 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 $\Gamma(x)$:

 $I(n)/A(n)=\frac{1}{2}\int_{0}^{\infty}t^{\frac{n-1}{2}}e^{-t}\,dt=\frac{1}{2}% \Gamma\left(\frac{n+1}{2}\right).$

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

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

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

Finally, we can solve for the area

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

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 $n\geq 0$. The first few special cases are

• $n=0$

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

• $n=1$

$\Gamma(1)=1$, hence $A(1)=2\pi$ (this is the familiar result for the circumference of the unit circle);

• $n=2$

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

• $n=3$

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

• $n=4$

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

Title area of the $n$-sphere AreaOfTheNsphere 2013-03-22 13:47:06 2013-03-22 13:47:06 CWoo (3771) CWoo (3771) 14 CWoo (3771) Derivation msc 51M05 VolumeOfTheNSphere AreaOfASphericalTriangle AreaOfSphericalZone