Consider the value $b$ with $b \in (0, 1)$ and let $D_b$ be the region to the right side of the line $x = b$, but also inside the circle $x^2 + y^2 = 1$ and call $f(b) = \text{Area of }D_b$, by converting to polar coords, calculate $f(b)$. Also calculate $\lim_{b \to 0} f(b)$ and $\lim_{b \to 1} f(b)$
I see that $ x^2 + y^2 = r^2 \implies r^2 = 1$
So we should have $r = 1$ for the polar bound.
$x = b \implies r\cos(\theta) = b \implies \theta = \arccos(b)$
But I am unsure how to proceed.
So $|\theta|\le \arccos b$. And the upper limit on $r$ is $1$. What's the lower limit? If you fix a $\theta$, $r$ starts when $r\cos\theta = b$, so $r=b\sec\theta$. Now set up your integral.