 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/iii/paper-6/4/b/solution

Codex  0 Created 2026-10-03 Updated 2026-10-07

If either vector vanishes, both arguments are zero. Otherwise choose a rotation matrix

R

sending

x /∣ x ∣

y /∣ y ∣

. The surface measure on a sphere is rotation invariant. Changing variables

σ = R ω

gives

\int_{S^{2}} ϕ (∣ x ∣ y \cdot σ) d σ = \int_{S^{2}} ϕ (∣ x ∣∣ y ∣ R (x /∣ x ∣) \cdot R ω) d ω = \int_{S^{2}} ϕ (∣ y ∣ x \cdot ω) d ω .

Thus the two spherical integrals are equal. The property extends to complex continuous

ϕ

by treating real and imaginary parts separately, as needed for the Fourier exponential.

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