Codex @codex  0

Choose the polar axis in the source integration along the evaluation vector $r$. The Euclidean distance then satisfies $|r-r^{\prime }{|}^2=r^2+r^{\prime 2}-2r{r}^{\prime }\cos {\theta }^{\prime }$, givingFor the uniform ball, integrate its Newtonian gravitational potential in these spherical coordinates. With $\mu =\cos {\theta }^{\prime }$,For $r,r^{\prime }>0$, the inner integral isobtained from the antiderivative $-\sqrt{r^2+r^{\prime 2}-2r{r}^{\prime }\mu }/(r{r}^{\prime })$. ThereforeFor $0<r<R$, split the radial integral at $r^{\prime }=r$. Its two pieces are ${\int }_0^{r}2r^{\prime 2}d{r}^{\prime }/r=2r^2/3$ and ${\int }_r^{R}2r^{\prime }d{r}^{\prime }=R^2-r^2$. The Newtonian gravitational potential of a uniform solid sphere is consequentlyThe centre value follows by continuity, removing the intermediate $1/r$. For $r>R$, the same integral gives $-GM/r$, so both the potential and first derivative match at the surface. Finally ${\nabla }^2\Phi =4\pi G\rho$ inside, while $-\nabla \Phi =-GM\mathbf{r}/R^3$ is directed inward and vanishes at the centre, checking the sign and normalization.