I'm currently reading a paper and try to understand this one formula. The problem is: In an n dimensional space. A cone with half-angle $\theta$ is given (the top of the cone is in the origin). We are interested in the solid angle of the cap cut out by this cone of the unit sphere.

According to this paper, the procedure is: Sum up all contributions of ring-shaped elements of area. The formula follwing from this procedure is

$\Omega(\theta_1) = \frac{(n-1)\pi^{(n-1)/2}}{\Gamma(\frac{n+1}{2})} \int_0^{\theta_1} (sin \theta)^{n-2} d\theta$

Obviously, the first fraction is the surface of the unit (n-1)-ball. What I don't understand are 2 points:

  1. Why does he choose the (n-1) surface instead of the n surface? (Later, the calculates the fraction of this solid angle by dividing by the n-ball surface)
  2. Where does this integral come from?

I hope somebody does have an idea...

If anyone is interested, the paper I'm reading is 'Probability of Error for Optimal Codes in a Gaussian Channel' by Claude Shannon


The integral represents how much of the surface is cut out. Since $\theta$ is measured at the center of the sphere, and it is a unit sphere, $sin\theta$ will give the radius at the surface.

Consider n=3:

The first fraction will evaluate to $2\pi$, the circumference of the unit circle (1-sphere).

Notice that $\Omega(\theta_1) = \int_0^{\theta_1} \!2\pi r \,d\theta$.

Consider n=4:

The first fraction will evaluate to $4\pi$, the surface area of the unit 2-sphere.

Notice that $\Omega(\theta_1) = \int_0^{\theta_1} \!4\pi r^2 \, d\theta$.

So yes, you do sum up all contributions of ring-shaped elements of area. In an $n$-dimensional space, each element of area will be that of a $(n-2)$-sphere. Multiply the element of area for the unit $(n-2)$-sphere by the radius raised to the $(n-2)$th power, and integrate the product over $\theta$.

  • $\begingroup$ Thank you very much, I do understand it now :) $\endgroup$ – disaster Oct 30 '13 at 16:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.