 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/ia/paper-3/11a/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

Put

x = r ω

, with

∣ ω ∣ = 1

d S = r^{2} d ω = r^{2} sin θ d θ d φ

, so

f (r) = \frac{1}{4 π} \int_{S^{2}} ϕ (r ω) d ω .

Differentiation under this fixed-domain integral yields

f^{'} (r) = \frac{1}{4 π} \int_{S^{2}} \nabla ϕ (r ω) \cdot ω d ω = \frac{1}{4 π r ^{2}} \int_{\partial V_{r}} \frac{\partial ϕ}{\partial n} d S = \frac{1}{4 π r ^{2}} \int_{V_{r}} \nabla^{2} ϕ d V = 0.

The last step is the divergence theorem and the Laplace equation. Hence

f

is constant on radii for which the closed ball lies inside

V

. Continuity at the origin gives

lim_{r \to 0} f (r) = ϕ (0)

f (r) = ϕ (0) .

 New to topics? Read the docs here!