 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ia/paper-3/9a/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

\iint_{\partial V} F \cdot n d S = ∭_{V} \nabla \cdot F d V

. Here

\nabla \cdot F = 3 z + 2 yz

. The

y

term integrates to zero by symmetry, while the ellipse has area

π / ab

. Hence the flux is

\frac{π}{ab} \int_{0}^{3} 3 z d z = \frac{27 π}{2 ab} .

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