 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/1/b/solution

Since

a

is constant, the product rule gives

\nabla \cdot (a ψ) = a \cdot \nabla ψ

. Applying the divergence theorem therefore gives

a \cdot \int_{V} \nabla ψ d V = \int_{S} ψ a \cdot d S = a \cdot \int_{S} ψ d S .

This holds for every constant vector

a

, so each component of the difference vanishes. Consequently the gradient has the boundary representation

\int_{V} \nabla ψ d V = \int_{S} ψ d S .

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