Answer
$$\iint \limits_{R} \frac{\tan \theta}{\sqrt{1-t^{2}}} d A
=\frac{\pi \ln 2}{6} $$
Work Step by Step
Given$$\iint \limits_{R} \frac{\tan \theta}{\sqrt{1-t^{2}}} d A, \quad R=\left\{(\theta, t) | 0 \leqslant \theta \leqslant \pi / 3,0 \leqslant t \leqslant \frac{1}{2}\right\}$$
So, we have
\begin{aligned}I&= \iint \limits_{R} \frac{\tan \theta}{\sqrt{1-t^{2}}} dA \\
&=\int_{0}^{\pi / 3} \int_{0}^{1 / 2} \frac{\tan \theta}{\sqrt{1-t^{2}}} d t d \theta\\
&=\left[\int_{0}^{1 / 2} \frac{d t}{\sqrt{1-t^{2}}}\right]\left[\int_{0}^{\pi / 3} \tan \theta d \theta\right]\\
&=\left[\int_{0}^{1 / 2} \frac{d t}{\sqrt{1-t^{2}}}\right]\left[\int_{0}^{\pi / 3} \frac{\sin t}{\cos t} \theta d \theta\right]\\
&=[\arcsin t]_{0}^{1 / 2} \cdot[-\ln |\cos \theta|]_{0}^{\pi / 3}\\
&=\left[\arcsin \frac{1}{2}-\arcsin 0\right] \cdot\left[-\ln \left|\cos \frac{\pi}{3}\right|+\ln |\cos 0|\right]\\
&=\left[\frac{\pi}{6}-0\right] \cdot[-\ln \frac{1}{2}+0]\\
&=\left[\frac{\pi}{6} \right] \cdot[\ln 2 ]\\
&=\frac{\pi \ln 2}{6}
\end{aligned}