Chapter 6, Problem 6.1P

To determine

The proof to show that the solid angle subtended about P gives ${\Omega }_{\text{tot}}={\displaystyle \oint d\Omega }$ which is equal to $4\pi$.

Answer to Problem 6.1P

The total solid angle is $4\pi$.

Explanation of Solution

Write the expression for solid angle subtended at the centre due to a small area.

    $d\Omega =\sin \theta d\theta$        (1)

Conclusion:

Integrate Equation (1) to get ${\Omega }_{\text{tot}}$

    $\begin{array}{c}{\Omega }_{\text{tot}}={\displaystyle \oint d\Omega }\\ ={\displaystyle \underset{0}{\overset{2\pi }{\int }}{\displaystyle \underset{0}{\overset{\pi }{\int }}\sin \theta d\theta }}d\varphi \\ ={\displaystyle \underset{0}{\overset{2\pi }{\int }}d\varphi }{\displaystyle \underset{0}{\overset{\pi }{\int }}\sin \theta d\theta }\\ ={\left[\varphi \right]}_0^{2\pi }{\left[-\cos \theta \right]}_0^{\pi }\end{array}$

Solving further,

    $\begin{array}{c}{\Omega }_{\text{tot}}=\left(2\pi -0\right)\left[-\left(\cos \pi -\cos 0\right)\right]\\ =\left(2\pi \right)\left(2\right)\\ =4\pi \end{array}$

Thus, the total solid angle is $4\pi$.