 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-3/38c/d/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

(u^{(1)}, σ^{(1)})

and

(u^{(2)}, σ^{(2)})

in the same fluid domain satisfy

\int_{\partial D} u^{(1)} \cdot σ^{(2)} n d S = \int_{\partial D} u^{(2)} \cdot σ^{(1)} n d S .

Apply it to the same rod translating with arbitrary velocities

U_{1}

and

U_{2}

. The integral at infinity vanishes, and the rod-surface integrals give

U_{1} \cdot F_{2} = U_{2} \cdot F_{1} .

Since

F_{i} = - R U_{i}

U_{1} \cdot R U_{2} = U_{2} \cdot R U_{1}

for every pair of vectors. Therefore

R = R^{T} .

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