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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2025

Paper 3

Reflection symmetry makes the <integral> zero for $i\ne j$. Rotational symmetry makes all diagonal <integrals> equal, and their sum is $R^2A_{n-1}(R)$. Therefore
$$
\boxed{\int_{r=R}x_ix_j\,dS=\delta_{ij}\frac{R^2A_{n-1}(R)}n
=\delta_{ij}\frac{\pi^{n/2}R^{n+1}}{\Gamma(n/2+1)}.}
$$

Solution

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

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

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

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