Total Solid Angle Subtended by Spherical Surface

Theorem
The total solid angle subtended by a spherical surface is $4 \pi$.

Proof

 * Angle-subtended-by-spherical-surface.png

Let $\d S$ be an element of a surface $S$.

Let $\mathbf n$ be a unit normal on $\d S$ positive outwards.

From a point $O$, let a conical pencil touch the boundary of $S$.

Let $\mathbf r_1$ be a unit vector in the direction of the position vector $\mathbf r = r \mathbf r_1$ with respect to $O$.

Let spheres be drawn with centers at $O$ of radii $1$ and $r$.

Let $\d \omega$ be the area cut from the sphere of radius $1$.

We have:
 * $\dfrac {\d \omega} {1^2} = \dfrac {\d S \cos \theta} {r^2}$

where $\theta$ is the angle between $\mathbf n$ and $\mathbf r$.

Thus $\d \omega$ is the solid angle subtended by $\d S$ at $O$.

Hence by definition of solid angle subtended:


 * $\ds \Omega = \iint_S \frac {\mathbf {\hat r} \cdot \mathbf {\hat n} \rd S} {r^2}$