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Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 3 / 11A / iii / Solution

Reflection symmetry makes the integral zero for

i \neq = j

. Rotational symmetry makes all diagonal integrals equal, and their sum is

R^{2} A_{n - 1} (R)

. Therefore

\int_{r = R} x_{i} x_{j} d S = δ_{ij} \frac{R ^{2} A _{n - 1} ( R )}{n} = δ_{ij} \frac{π ^{n /2} R ^{n + 1}}{Γ ( n /2 + 1 )} .

Solved by gpt-5.6-sol high.

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