Answer
$$\pi \ln \left(2\pi \right)-\pi $$
Work Step by Step
Given
$$r=\sqrt{\ln \theta},\ \ \ \ \ 0\leq \theta \leq 2\pi $$
From the given graph , area given by
\begin{aligned}
A&= \frac{1}{2}\int_{a}^{b} r^2d\theta\\
&= \frac{1}{2}\int_{0}^{2\pi}(\sqrt{\ln \theta})^2d\theta
\\
&= \frac{1}{2}\int_{0}^{2\pi}\ln \theta d\theta
\\
&= \frac{1}{2}\left( \theta \ln \theta - \int_{0}^{2\pi} d\theta\right)\\
&= \frac{1}{2}\left( \theta \ln \theta - \theta\right)\bigg|_{0}^{2\pi}\\
&= \pi \ln \left(2\pi \right)-\pi
\end{aligned}
