Average Distance in a Disk

Worked solution

The answer is 128/(45π) ≈ 0.9054. A clean route uses integral geometry. Any pair of points determines a line, and the Blaschke-Petkantschin (Crofton) change of variables says dP₁ dP₂ = |t₁ − t₂| dt₁ dt₂ dG, where t₁, t₂ are positions of the points along the line and dG = dp dφ is the standard measure on lines (p the distance from the centre, φ the direction). For a line meeting the disk in a chord of length σ, ∫∫ |t₁ − t₂| · |t₁ − t₂| dt₁ dt₂ over the chord equals σ⁴/6. A line at distance p gives σ = 2√(1 − p²). Hence ∫∫ d dP₁ dP₂ = ∫₀^π ∫_{−1}^{1} (16/6)(1 − p²)² dp dφ = π · (8/3) · (16/15) = 128π/45. Dividing by the squared area π² gives E[d] = 128/(45π), a little under the radius.