Expected absolute y-coordinate in a disc

The quantity $\mathbb{E}[|Y|]$ is the expected perpendicular distance from a uniform random point in the disc to the horizontal diameter. It also equals the projected mean absolute deviation along any fixed direction, by the rotational symmetry of the disc.

There is a clean slicing identity behind this. For a convex body $K$ of area $A(K)$ in the plane, a unit direction $\mathbf{n}$, and a reference level $m$, let $\ell_K(t)$ be the length of the slice $\{P \in K : \mathbf{n} \cdot P = t\}$. Then the mean absolute deviation of the coordinate $\mathbf{n} \cdot P$ about $m$ is $$\mathbb{E}_{P \sim K}\bigl[\,|\mathbf{n} \cdot P - m|\,\bigr] = \frac{1}{A(K)} \int |t - m|\, \ell_K(t)\, dt.$$ For the unit disc with $m = 0$, $A = \pi$, and $\ell(t) = 2\sqrt{1 - t^2}$, this evaluates to $\tfrac1\pi \int_{-1}^{1} |t|\, 2\sqrt{1 - t^2}\, dt = 4/(3\pi)$, recovering the answer above. (Note the absolute value: the signed deviation about the mean averages to $0$. For this symmetric disc the mean and median both equal $0$, but a median does not generally have this property. The quantity here is the mean absolute deviation.)

import numpy as np
r = np.sqrt(np.random.rand(10**7))
theta = np.random.rand(10**7) * 2*np.pi
y = r * np.sin(theta)
print(f"sim: {np.abs(y).mean():.5f}   exact: {4/(3*np.pi):.5f}")
# sim: 0.42438   exact: 0.42441