Proof of moment of inertia

Considering a volume element in spherical coordinate
dV = r² sinφ dr dφ dθ ,

then the infinitesimal mass is
dm = M / (4πR³/3) dV = [3M/(4πR³)] r² sinφ dr dφ dθ

The distance of the infinitesimal mass to the rotation axis is r sinφ.

Therefore the moment of inertia is
I = ∫ (r sinφ)² dm

= ∫∫∫ (r sinφ)² [3M/(4πR³)] r² sinφ dr dφ dθ (integral for r:0->R, φ:0->π, θ:0->2π)

= [3M/(4πR³)]∫ r4 dr ∫ sin³φ dφ ∫dθ

= [3M/(4πR³)] [ r5 /5] [cos³φ/3 - cosφ] [θ] (with r:0->R, φ:0->π, θ:0->2π)

= [3M/(4πR³)] [ R5 /5] [4/3][2π]

= (2/5) M R²


note:
∫ sin³φ dφ = -∫ sin²φ d cosφ = ∫ (cos²φ - 1) d cosφ = cos³φ/3 - cosφ

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