 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ia/paper-3/11a/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

i \neq = j

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 )} .

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