 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/ia/paper-3/12a/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

F = ϕ \nabla ψ - ψ \nabla ϕ

\nabla \cdot F = \nabla ϕ \cdot \nabla ψ + ϕ \nabla^{2} ψ - \nabla ψ \cdot \nabla ϕ - ψ \nabla^{2} ϕ = ϕ \nabla^{2} ψ - ψ \nabla^{2} ϕ .

The two dot products cancel. Apply the divergence theorem with outward unit normal vector

n

\int_{V} (ϕ \nabla^{2} ψ - ψ \nabla^{2} ϕ) d V = \int_{S} (ϕ \nabla ψ - ψ \nabla ϕ) \cdot n d S .

Smoothness of the functions and boundary supplies the regularity needed for the divergence theorem.

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