 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-2/6/d/solution

Let

J_{ij} = \int_{B} x_{i} x_{j} d^{3} x

J

J_{ij} = C δ_{ij}

. Contracting the indices and integrating in spherical coordinates gives

3 C = \int_{B} r^{2} d^{3} x = 4 π \int_{0}^{1} r^{4} d r = \frac{4 π}{5} .

I = M_{ij} J_{ij} = \frac{4 π}{15} tr M .

Only the scalar part of

M

contributes; its antisymmetric tensor and symmetric-traceless parts give zero.

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