Distance from the circle's centre

Here are two different methods:

1) Integration in polar coordinates
2) Monte-Carlo simulation in Cartesian coordinates

Solution 1

Assume that the points are evenly distributed as a function of area. Divide the circle into circular rings of thickness $dr$ .

$\large{dA = 2\pi r dr}$

The average distance from the center is:

$\large{d_{av} = \frac{1}{A_{total}} \int_0^1 r dA = \frac{1}{\pi} \int_0^1 r 2\pi r dr = 2 \int_0^1 r^2 dr = \boxed{\frac{2}{3}} }$

Solution 2

Generate points randomly (uniform distribution in $(x,y)$ ) within a square circumscribing the circle. Keep two running sums:

Sum #1: The number of points falling within the square which also happen to fall within the circle
Sum #2: The sum of the distances of the points that have fallen within the circle

Run a million trials, and take the ratio of (Sum #2) to (Sum #1). The answer comes out to very nearly two thirds, as before.