We’re trying to determine the average distance from a random point in a circle to the centre, for various definitions of average. The simplest, the arithmetic mean distance, is ≈ .6667.
\iint_R \ln \operatorname f x\ \delta AFirst, the geometric mean. The basic integral required to solve this is above. It’s not a very hard integral, especially for a circle. It’s equal to 2π times the integral from 0 to 1 of x2 ln x. That’s a relatively simple integration by parts, which many already know how to do–but for completion:
\int_0^1 x^2 \ln x\ \delta x \\[16pt] \int_0^1 \ln x \cdot x^2\ \delta x \implies \xcancel{\left[ {x^3\ln x\over 3}\right]}_0^1-\int_0^1 \frac{x^2}3 \delta x \\[16pt] = {\left[\frac{x^3}9\right]}_0^1 = -\frac19Substituting, the geometric mean distance is e-2/9 ≈ .8007. Note that this is -2/9 not -1/9 because we multiplied the integral by 2π.
Second: the quadratic mean. Since we’re using the identity function, this is a rather easy one. The integral of x2 times x is 1/4 x4, from 0 to 1 that’s just 1/4. We take the square root of 1/8, not 1/4, for the same reason as with the geometric mean. Thus, the quadratic mean distance is ≈ .3536.
Third: the harmonic mean. We saved the best for last: the determinant of the transformation from cartesian to polar coordinates is x. 1/x times x is 1. It’s the easiest integral of all! The harmonic mean distance is .5.
And last, the extra easy cool-down problem, that’s not even really integration at all. This one is fairly obvious: the median distance is ≈ .7071.