Solid angle subtended at the centre of the sphere:

Let us consider a sphere of radius r,

Now we want to calculate the solid angle at the centre O of the sphere by the total surface area of the sphere.

Let’s consider, an elementary area dA on the surface of the sphere.

So the solid angle subtended at the centre O by the surface area dA of the sphere is given by,

\( d\omega=\frac{dA\ \cos\alpha}{r^2}=\frac{dA}{r^2} \)

Where, \( \alpha \) is the angle between the area vector \( \vec{dA} \) and the direction of \( \vec{r} \). Here \( \alpha=0 \), therefore \( \cos\alpha=\cos\ 0=1 \).

There are three area elements \( {dA}_r,\ {dA}_{\theta} \) and \( {dA}_{\phi} \), out of which only the area element \( {dA}_r \) subtends a solid angle at the centre \( O \).

[To know about the three types of area elements (CLICK HERE)]

[To know, why only the area element \( {dA}_r \) subtends a solid angle at the centre O (CLICK HERE) ]

Again we know that,

\( {dA}_r=r^2\ \sin\theta\ d\theta\ d\phi \)

So the solid angle is given by,

\( d\omega=\displaystyle{\frac{ r^2\ \sin\theta\ d\theta\ d\phi }{r^2}}=\sin\theta\ d\theta\ d\phi \)

In order to get the solid angle subtended at the centre by the total surface area of the sphere, we have to integrate \( d\omega \) with respect to \( \theta \) from 0 to \( \pi \) and with respect to \( \phi \) from 0 to \( 2\pi \).

\( \displaystyle{\omega=\int_{0}^{\pi}\sin\theta\ d\theta\int_{0}^{2\pi}d\phi} \)

\( or,\ \displaystyle{\omega=[-\cos\theta]_{0}^{\pi}\ [\phi]_{0}^{2\pi}} \)

\( or,\ \omega=(1+1)2\pi=4\pi\ steradian \)