A $\pi$ day $\pi$cnic on $\pi$-land

You are planning a picnic on the remote tropical island of $\pi$-land. The island’s shape is a perfect semi-disk with two beaches, as illustrated below: Semicircular Beach (along the northern semicircular edge of the disk) and Diametric Beach (along the southern diameter of the disk).

If you pick a random spot on $\pi$-land for your picnic, what is the probability that it will be closer to Diametric Beach than to Semicircular Beach? (Unlike the illustrative diagram above, assume the beaches have zero width.)

The local $\pi$-landers typically measure everything from the midpoint of the Diametric Beach, so let's assume that point is the origin of our $xy$-plane, with Diametric Beach coinciding with the $x$-axis. In this case, assuming that $\pi$-land has a radius of $R,$ then the entire area of $\pi$-land is $A = \frac{1}{2} \pi R^2.$ At any point $(x,y)$ on the $\pi$-land the distance to the Diametric Beach is given by $d_D (x,y) = y,$ while the distance to the Semicircular Beach is $d_S (x,y) = R - \sqrt{x^2 + y^2}.$ So the region of $\pi$-land that is closer to Diametric Beach than to Semicircular Beach is given by

$$\begin{align*} \Omega &= \left\{ (x,y) \in \mathbb{R}^2_+ \mid x^2 + y^2 \leq R^2, d_D(x,y) \leq d_S(x,y) \right\} \\ &= \left\{ (x,y) \in \mathbb{R}_+^2 \mid y \leq R - \sqrt{x^2+ y^2} \right\} \\ &= \left\{ (x,y) \in \mathbb{R}^2_+ \mid x^2 + y^2 \leq (R-y)^2 \right\} \\ &= \left\{ (x,y) \in \mathbb{R}^2_+ \mid x^2 \leq R^2 -2Ry \right\} \\ &= \left\{ (x,y) \in \mathbb{R}^2_+ \mid y \leq \frac{R^2 - x^2}{2R} \right\} \end{align*}$$

The area of $\Omega$ is given by the integral

$$A_\Omega = \int_{-R}^R \frac{R^2 - x^2}{2R} \,dx = 2 \left[ \frac{Rx}{2} - \frac{x^3}{6R} \right]^{x=R}_{x=0} = 2 \left( \frac{R^2}{2} - \frac{R^2}{6} \right) = \frac{2R^2}{3}.$$ Therefore, the probability of randomly choosing a $\pi$-land picnic spot closer to Diametric Beach than to Semicircular Beach, that is, in $\Omega,$ is given by $$p = \frac{A_\Omega}{A} = \frac{ \frac{2R^2}{3} }{ \frac{\pi R^2}{2} } = \frac{4}{3\pi} \approx 0.424413181578....$$