Answer
$A=\pi\,\ln\,2\pi-\pi+\frac{1}{2}$
Work Step by Step
Setting up the integral using the formula for the area of a region in polar coordinates, we get:
$$\frac{1}{2}\int_{1}^{2\pi}(\sqrt{ln\,\theta})^2\,d\theta$$
We can rewrite this integral as:
$$\frac{1}{2}\int_{1}^{2\pi}ln\,\theta\,d\theta$$
Solving the integral, we get:
$$\frac{1}{2}(\theta\,\ln\,\theta-\theta)\bigg\rvert_{1}^{2\pi}=\frac{1}{2}\bigg((2\pi\,\ln\,2\pi-2\pi)-(-1)\bigg)=\pi\,\ln\,2\pi-\pi+\frac{1}{2}$$
