 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/ii/paper-4/9b/a/solution

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

Take the sheet to lie in the

x_{1} x_{2}

a, b

in those directions. Its surface density is

M / (ab)

. Reflection symmetry about the coordinate axes makes the off-diagonal entries of the inertia tensor zero. Its principal moments of inertia are

I_{1} = \frac{M}{ab} \int_{- a /2}^{a /2} \int_{- b /2}^{b /2} x_{2}^{2} d x_{2} d x_{1} = \frac{M b ^{2}}{12},

and similarly

I_{2} = M a^{2} /12

. Since

I_{3}

integrates

x_{1}^{2} + x_{2}^{2}

I_{1} = M b^{2} /12, I_{2} = M a^{2} /12, I_{3} = M (a^{2} + b^{2}) /12.

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