 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-2/6/b/i/solution

M_{ij} = \int_{V} x_{i} x_{j} d^{3} x

. The spherical shell is preserved by every rotation, so this is an isotropic tensor,

M_{ij} = A δ_{ij}

. Its trace determines the constant:

3 A = M_{ii} = \int_{V} r^{2} d^{3} x = 4 π \int_{R_{1}}^{R_{2}} r^{4} d r = \frac{4 π}{5} (R_{2}^{5} - R_{1}^{5}) .

Contracting with the components of the two unit vectors gives

\int_{V} (x \cdot n_{1}) (x \cdot n_{2}) d^{3} x = \frac{4 π}{15} (R_{2}^{5} - R_{1}^{5}) (n_{1} \cdot n_{2}) .

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