 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/ia/paper-3/9c/i/solution

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

F_{i} = ω_{i} ω_{j} x_{j} - ∣ ω ∣^{2} x_{i} .

Since

ω

is constant, the divergence is

\partial_{i} F_{i} = ω_{i} ω_{i} - 3∣ ω ∣^{2} = - 2∣ ω ∣^{2} .

2^{3} = 8

. With the outward unit normal, the divergence theorem therefore gives

\int_{S} F \cdot d S = - 16∣ ω ∣^{2} .

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