[Edit: redefined $\theta$ for simplicity and introduced $a$ for generality.]
Let $a$ be the radius of the pillar ($a=1$ in the example). The length of the rope is $L=\pi a.$
Let the length of the rope (assumed taut) not in contact with the pillar be $l(\theta)$, where $\theta$ is the angle between a vertical line descending from the pillar centre (O) and a line between O and the rope's point of attachment. Then $$ l = a\theta $$ $\theta$ is also the angle of the rope measured from the horizontal.
The area swept out as the rope turns anticlockwise through angle increment $d\theta$ is $$ dA = \tfrac{1}{2}l^2 d\theta $$ The total area swept out in the lower right quadrant is therefore $$ A = \int dA = \int_0^{\pi}\tfrac{1}{2}a^2\theta^2 d\theta = \tfrac{1}{6}\pi^3a^2 $$ Doubling up for the other side and adding the semicircular area at the top gives a total area of $$ 2 (\tfrac{1}{6}\pi^3a^2) + \tfrac{1}{2}\pi(\pi a)^2 = \tfrac{5}{6}\pi^3a^2 $$