| Version current |
Version 10 |
| The area of $S^n$ the unit $n$-sphere (or hypersphere) is the same as the total |
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 |
solid angle it subtends at the origin. To calculate it, consider the following |
| integral |
integral |
| \[ |
\[ |
| I(n) = \int_{\R^{n+1}} e^{-\sum_{i=1}^{n+1} x_i^2}\, d^{n+1} x. |
I(n) = \int_{\R^{n+1}} e^{-\sum_{i=1}^{n+1} x_i^2}\, d^{n+1} x. |
| \] |
\] |
| Switching to polar coordinates we let $r^2=\sum_{i=1}^{n+1} x_i^2$ and the |
Switching to polar coordinates we let $r^2=\sum_{i=1}^{n+1} x_i^2$ and the |
| integral becomes |
integral becomes |
| \[ |
\[ |
| I(n) = \int_{S^n} d\Omega \int_{0}^{\infty} r^{n} e^{-r^2}\, dr. |
I(n) = \int_{S^n} d\Omega \int_{0}^{\infty} r^{n} e^{-r^2}\, dr. |
| \] |
\] |
| The first integral is the integral over all solid angles and is exactly what we |
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 |
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
|
$t=r^2$, the second integral can be evaluated in terms of the Gamma function
|
| $\Gamma(x)$: |
$\Gamma(x)$: |
| \[ |
\[ |
| I(n)/A(n) = \frac{1}{2}\int_0^\infty t^{\frac{n-1}{2}} e^{-t}\, dt |
I(n)/A(n) = \frac{1}{2}\int_0^\infty t^{\frac{n-1}{2}} e^{-t}\, dt |
| = \frac{1}{2}\Gamma\left(\frac{n+1}{2}\right). |
= \frac{1}{2}\Gamma\left(\frac{n+1}{2}\right). |
| \] |
\] |
| We can also evaluate $I(n)$ directly in Cartesian coordinates: |
We can also evaluate $I(n)$ directly in Cartesian coordinates: |
| \[ |
\[ |
| I(n) = \left[ \int_{-\infty}^\infty e^{-x^2}\, dx \right]^{n+1} |
I(n) = \left[ \int_{-\infty}^\infty e^{-x^2}\, dx \right]^{n+1} |
| = \pi^{\frac{n+1}{2}}, |
= \pi^{\frac{n+1}{2}}, |
| \] |
\] |
| where we have used the standard Gaussian integral $\int_{-\infty}^\infty |
where we have used the standard Gaussian integral $\int_{-\infty}^\infty |
| e^{-x^2}\, dx = \sqrt{\pi}$. |
e^{-x^2}\, dx = \sqrt{\pi}$. |
|
|
| Finally, we can solve for the area |
Finally, we can solve for the area |
| \[ A(n) = \frac{2\pi^{\frac{n+1}{2}}}{\Gamma\left(\frac{n+1}{2}\right)}. \] |
\[ A(n) = \frac{2\pi^{\frac{n+1}{2}}}{\Gamma\left(\frac{n+1}{2}\right)}. \] |
| If the radius of the sphere is $R$ and not $1$, the correct area is |
If the radius of the sphere is $R$ and not $1$, the correct area is |
| $A(n)R^{n}$. |
$A(n)R^{n}$. |
|
|
| Note that this formula works only for $n\ge0$. The first few special cases |
Note that this formula works only for $n\ge0$. The first few special cases |
| are |
are |
| \begin{itemize} |
\begin{itemize} |
| \item[$n=0$] $\Gamma(1/2)=\sqrt{\pi}$, hence $A(0)=2$ (in this case, the |
\item[$n=0$] $\Gamma(1/2)=\sqrt{\pi}$, hence $A(0)=2$ (in this case, the |
| area just counts the number of points in $S^0=\{+1,-1\}$); |
area just counts the number of points in $S^0=\{+1,-1\}$); |
| \item[$n=1$] $\Gamma(1)=1$, hence $A(1)=2\pi$ (this is the familiar result |
\item[$n=1$] $\Gamma(1)=1$, hence $A(1)=2\pi$ (this is the familiar result |
| for the circumference of the unit circle); |
for the circumference of the unit circle); |
| \item[$n=2$] $\Gamma(3/2)=\sqrt{\pi}/2$, hence $A(2)=4\pi$ (this is the |
\item[$n=2$] $\Gamma(3/2)=\sqrt{\pi}/2$, hence $A(2)=4\pi$ (this is the |
| familiar result for the area of the unit sphere); |
familiar result for the area of the unit sphere); |
| \item[$n=3$] $\Gamma(2)=1$, hence $A(3)=2\pi^2$; |
\item[$n=3$] $\Gamma(2)=1$, hence $A(3)=2\pi^2$; |
| \item[$n=4$] $\Gamma(5/2)=3\sqrt{\pi}/4$, hence $A(4)=8\pi^2/3$. |
\item[$n=4$] $\Gamma(5/2)=3\sqrt{\pi}/4$, hence $A(4)=8\pi^2/3$. |
| \end{itemize} |
\end{itemize} |
