 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/ia/paper-3/11c/a/solution

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

\nabla \cdot (f \nabla g) = \nabla f \cdot \nabla g + f \nabla^{2} g .

n

, and use the vanishing Laplacian of the harmonic function

g

\int_{V} \nabla f \cdot \nabla g d V = \int_{S} f \nabla g \cdot n d S .

The boundary value of

f

is one. Another application of the divergence theorem therefore gives

\int_{S} f \nabla g \cdot n d S = \int_{S} \nabla g \cdot n d S = \int_{V} \nabla^{2} g d V = 0 .

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