An Extra Credit $\pi$ day $\pi$-cnic in $\pi$-land

Suppose the island of $\pi$-land, as described above, has a radius of $1$ mile. That is, Diametric Beach has a length of $2$ miles. Again, you are picking a random point on the island for a picnic. On average, what will be the expected shortest distance to shore?

Here we want to calculate the average, minimum distance to shore $$E = \frac{1}{A} \int_{-R}^R \int_0^{\sqrt{R^2 - x^2}} \min \{ d_D (x,y), d_S (x,y) \} \,dy \,dx,$$ where $d_D$ and $d_S$ were defined as above in the Classic Fiddler answer. As we saw in the Classic Fiddler answer, the region where $d_D \leq d_S$ is given by $\Omega,$ so we can break the integral above into two sections $$E = \frac{1}{A} \int_{-R}^R \int_0^{\frac{R^2 - x^2}{2R}} y \,dy\,dx + \frac{1}{A} \int_{-R}^R \int_{\frac{R^2-x^2}{2R}}^{\sqrt{R^2 - x^2}} \left(R - \sqrt{x^2+y^2}\right) \,dy \,dx := E_1 + E_2.$$ The first integral, $E_1,$ is relatively straightforward to do, \begin{align*}E_1 = \frac{1}{A} \int_{-R}^R \int_0^{\frac{R^2 - x^2}{2R}} y \, dy \,dx &= \frac{4}{\pi R^2} \int_0^R \left( \frac{y^2}{2} \right)_{y=0}^{y= \frac{R^2 - x^2}{2R}} \,dx\\ &= \frac{4}{\pi R^2} \int_0^R \frac{1}{2} \left( \frac{R^2 - x^2 }{2R} \right)^2 \,dx \\ &= \frac{4}{\pi R^2} \int_0^R \frac{1}{8R^2} \left( x^4 - 2R^2 x^2 + R^4 \right) \,dx \\ &= \frac{1}{2\pi R^4} \left[ \frac{x^5}{5} - \frac{2R^2x^3}{3} + R^4 x \right]_{x=0}^{x=R} \\ &= \frac{1}{2\pi R^4} \left( \frac{R^5}{5} - \frac{2R^5}{3} + R^5 \right) \\ &= \frac{R}{2\pi} \left( \frac{1}{5} - \frac{2}{3} + 1 \right) \\ &= \frac{R}{2\pi} \frac{3 - 10 + 15}{15} = \frac{4R}{15 \pi} \end{align*}

For the second integral, it will be easier to switch to polar coordinates, then do some trig-substitutions to get a handle on the resulting integral. The curve $y = \frac{R^2 - x^2}{2R}$ is equivalent to the equation $x^2 + 2R y - R^2 = 0$ which is given by $r^2 \cos^2 \theta + 2R r \sin \theta - R^2 = 0$ in polar coordinates. Solving the quadratic in terms of $r$ and choosing the positive solution when $\sin \theta \geq 0$ since $\theta \in [0,\pi]$, we get \begin{align*}r &= \frac{ -2R \sin \theta + \sqrt{ 4R^2 \sin^2 \theta + 4R^2 \cos^2 \theta} }{2 \cos^2 \theta} \\ &= \frac{-2R \sin \theta + 2R}{2 \cos^2 \theta} \\ &= \frac{ R \left( 1 - \sin \theta \right) }{ \cos^2 \theta } \\ &= \frac{R}{1 + \sin \theta},\end{align*} since $\cos^2 \theta = 1 - \sin^2 \theta = (1 - \sin \theta) (1 + \sin \theta).$ Therefore, we have \begin{align*}E_2 &= \frac{1}{A} \int_{-R}^R \int_{\frac{R^2 - x^2}{2R}}^{\sqrt{R^2 - x^2}} \left( R - \sqrt{x^2 + y^2} \right) \,dy \,dx \\ &= \frac{2}{\pi R^2} \int_0^\pi \int_{\frac{R}{1 + \sin \theta}}^R (R - r) r \,dr \, d\theta \\ &= \frac{2}{\pi R^2} \int_0^\pi \left[ \frac{Rr^2}{2} - \frac{r^3}{3} \right]^{r=R}_{r = \frac{R}{1 + \sin \theta}} \, d\theta \\ &= \frac{2}{\pi R^2} \int_0^\pi \left( \frac{R^3}{2} - \frac{R^3}{3} \right) - \left( \frac{R^3}{2(1 + \sin \theta)^2} - \frac{R^3}{3(1 + \sin \theta)^3} \right) \,d\theta \\ &= \frac{R}{3} - \frac{R}{3\pi} \int_0^\pi \frac{3 \sin \theta + 1}{(1 + \sin \theta)^3} \,d\theta = \frac{R}{3} - I.\end{align*} In order to calculate the last integral $I$, we need to do a trigonometric half-angle substitution with $u = \tan \frac{\theta}{2}.$ Here we see that $$du = \frac{1}{2} \sec^2 \frac{\theta}{2} \,d\theta = \frac{1}{2} \left(1 + \tan^2 \frac{\theta}{2} \right) \,d\theta = \frac{1 + u^2}{2} \,d\theta$$ and further that $$\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} = 2 \tan \frac{\theta}{2} \cos^2 \frac{\theta}{2} = \frac{2 \tan \frac{\theta}{2}}{1 + \tan^2 \frac{\theta}{2}} = \frac{2u}{1+u^2}.$$ We additionally see that when $\theta = 0,$ then $u = 0,$ whereas when $\theta = \pi,$ then $u = \tan \frac{\pi}{2} = \infty.$ Therefore we get \begin{align*} I &= \frac{R}{3\pi} \int_0^\pi \frac{3 \sin \theta + 1}{(1 + \sin \theta)^3} \,d\theta \\ &= \frac{2R}{3\pi} \int_0^\infty \frac{ 3 \frac{2u}{1+u^2} + 1 }{\left(1 + \frac{2u}{1 + u^2}\right)^3} \frac{2 \,du}{1 + u^2} \\ &= \frac{2R}{3\pi} \int_0^\infty \frac{ (1 + 6u + u^2) ( 1 + u^2 ) }{ (1 + u)^6 } \, du.\end{align*} Making a further substitution of $v = 1 + u,$ we then get \begin{align*} I = \frac{R}{3\pi} \int_0^\infty \frac{ (1 + 6u + u^2) ( 1 + u^2 ) }{ (1 + u)^6 } \, du &= \frac{2R}{3\pi} \int_1^\infty \frac{ (1 + 6(v-1) + (v-1)^2) ( 1 + (v-1)^2 ) }{v^6} \, dv \\ &= \frac{2R}{3\pi} \int_1^\infty \frac{ (v^2 + 4v - 4)(v^2 - 2v + 2) }{v^6} \,dv \\ &= \frac{2R}{3\pi} \int_1^\infty v^{-2} + 2v^{-3} - 10v^{-4} + 16v^{-5} -8v^{-6} \,dv \\ &= \frac{2R}{3\pi} \left[ -\frac{1}{v} -\frac{1}{v^2} + \frac{10}{3v^3} -\frac{4}{v^4} + \frac{8}{5v^5} \right]_{v=1}^{v=\infty} \\ &= -\frac{2R}{3\pi} \left( -1 -1 + \frac{10}{3} -4 + \frac{8}{5} \right) = \frac{32R}{45\pi} \end{align*} Putting this altogether we get $$E_2 = \frac{R}{3} - \frac{32R}{45\pi} = \frac{R(15\pi - 32)}{45 \pi}.$$

Therefore, the average minimal distance to the beach on $\pi$-land if the picnic spot is uniformly randomly chosen over the area of $\pi$-land is $$E = E_1 + E_2 = \frac{4R}{15\pi} + \frac{R(15 \pi - 32)}{45 \pi} = \frac{R(15\pi -20)}{45 \pi} = \frac{R(3\pi - 4)}{9\pi}.$$ So in particular, if $R = 1,$ then we get an average distance to the beach of $$E = \frac{3 \pi - 4}{9\pi} \approx 0.191862272807\dots.$$