How to Find the Area Covered by a Solid Angle Using the Half-Angle?

[Gear300]

For a sphere, the relation between steradians and the area they cover is O = A/(r^2), in which O is the measure of the solid angle, A is the area it covers, and r is the radius. If I were instead given the half-angle of the steradian...meaning that if there was a central axis running through the solid angle, connecting the surface of the sphere to its center, the half-angle would simply be the angle between the central axis and the edge of the solid angle...then how would I find the area covered by the solid angle in respect to the half angle.

 

[StatusX]

You can integrate over the sphere using spherical coordinates by:

$$\int_0^{2 \pi} d \phi \int_0^\pi d\theta R^2 \sin\theta$$

The region you're talking about then corresponds to the range $0\leq \phi < 2 \pi$ and $0 \leq \theta < \theta_0$, where $\theta_0$ is the half angle. So you can find the area by evaluating:

$$A = \int_0^{2 \pi} d \phi \int_0^{\theta_0} d\theta R^2 \sin\theta$$

 

[Gear300]

...wait...wut? I sort of half get what you're saying. I was using earlier the integral with the sine in it...but where did the integral of dphi pop out of?

 

[StatusX]

Look up spherical coordinates.