∴ {\displaystyle \therefore }
In this case the integral ∫ ρ 0 x ( F x ) 0 d τ 0 {\displaystyle \int \rho _{0}x\left(F_{x}\right)_{0}d\tau _{0}} becomes ∫ σ 0 x ( F x ) 0 d S 0 {\displaystyle \int \sigma _{0}x\left(F_{x}\right)_{0}dS_{0}} , over the surface of the sphere of radius a, and F x = 2 π σ 0 cos θ {\displaystyle F_{x}=2\pi \sigma _{0}\cos \theta } .
Thus
and