The correct answer is (c) 4π^2/3
The explanation: Total power radiated \(P_{rad}= ∯ w_r.a_n\overline{d}l \)
\(=\iint_{\theta = 0}^{\pi} w_r.r^2 sin\theta d\theta d\emptyset\)
\(=\iint_{\theta = 0}^π\frac{4sin\theta}{3r^2}r^2sin\theta d\theta d \emptyset\)
\(=\iint_{\theta = 0}^π \frac{4}{3}sin^2\theta d\theta d\emptyset=\iint_{\theta=0}^π\frac{4}{3}(\frac{1-cos2\theta}{2})d\theta d\emptyset\)
\(=\frac{4}{3}(\frac{1}{2})(π)(2π)=\frac{4}{3}\) π^2.