With
$$
\mathcal P(q)=\frac4\pi\sqrt{1-q^2},
$$
the factor $\sqrt{1-q^2}$ cancels from the <Abel transform>:
$$
\mathcal P(Q)
=\frac{4Q}{\pi}\int_0^Q\frac{dq}{\sqrt{Q^2-q^2}}
=\frac{4Q}{\pi}\cdot\frac\pi2.
$$
Hence
$$
\boxed{\mathcal P(Q)=2Q,
\qquad 0\leq Q\leq1}.
$$
This <probability density function> is normalized and again favors rounder projections.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 320 / 1 / f / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-320/1/f/solution