Quick way to explain the concept of a solid angle

I was struggling to explain the concept of a solid angle to a student. I found that the following approach, by analogy with plane angles, succeeded.

In the case of a plane angle $d \theta$ (in radians) subtended by an arc length $d a$ , the following relationship holds:

$\frac{d \theta}{2 \pi} = \frac{da}{2 \pi r} \qquad \qquad \qquad (1)$

On the left-hand side we have the ratio of $d \theta$ to the circumference of a unit circle, $2 \pi$ . On the right-hand side, we have the ratio of the arc length $d a$ to the circumference of a circle of radius $r$ , $2 \pi r$ . So, the idea is that $d \theta$ is a measure of the part of the circumference of the unit circle at the apex that is covered by the projection of $d a$ on to this unit circle. From (1) we get

$d \theta = \frac{da}{r} \qquad \qquad \qquad (2)$

(or the familiar formula for calculating arc lengths, $da = r d \theta$ ).

A solid angle is simply a 3D analogue of this.

Instead of a unit circle at the apex, we have a unit sphere with surface area $4 \pi$ . The solid angle $d \Omega$ (in steradians) is subtended by the area element $d A$ , and these satisfy the following relationship:

$\frac{d \Omega}{4 \pi} = \frac{d A}{4 \pi r^2} \qquad \qquad \qquad (3)$

On the left-hand side we have the ratio of $d \Omega$ to the surface area of the unit sphere, $4 \pi$ . On the right-hand side, we have the ratio of the area element $d A$ to the surface area of a sphere of radius $r$ , $4 \pi r^2$ . So, the idea is that $d \Omega$ is a measure of the part of the surface area of the unit sphere at the apex that is covered by the projection of $d A$ on to this unit sphere. From (3), we get the familiar formula for solid angles,

$d \Omega = \frac{d A}{r^2} \qquad \qquad \qquad (4)$

It is possible to solve some simple problems involving solid angles purely by symmetry considerations. For example, to work out the solid angle subtended by one of the faces of a cube, we note that the cube has six faces of equal area and together they must account for the entire surface area of the unit sphere at the apex. Therefore, a single cube face must account for one-sixth of the surface area of the unit sphere at the apex, so we have

$\frac{d \Omega}{4 \pi} = \frac{1}{6}$