= Solution
Choose the polar axis in the source integration along the evaluation vector $r$. The <Euclidean distance> then satisfies $|r-r'|^2=r^2+r'^2-2rr'\cos\theta'$, giving
$$
H(r,r')=-\frac1{4\pi\sqrt{r^2+r'^2-2rr'\cos\theta'}}.
$$
For the uniform ball, integrate its <Newtonian gravitational potential> in these <spherical coordinates>. With $\mu=\cos\theta'$,
$$
\Phi(r)=-2\pi G\rho\int_0^R r'^2dr'\int_{-1}^1\frac{d\mu}{\sqrt{r^2+r'^2-2rr'\mu}}.
$$
For $r,r'>0$, the inner integral is
$$
\frac{r+r'-|r-r'|}{rr'},
$$
obtained from the antiderivative $-\sqrt{r^2+r'^2-2rr'\mu}/(rr')$. Therefore
$$
\Phi(r)=-2\pi G\rho\int_0^R\frac{r'}r[r+r'-|r-r'|]dr'.
$$
For $0<r<R$, split the radial integral at $r'=r$. Its two pieces are $\int_0^r2r'^2dr'/r=2r^2/3$ and $\int_r^R2r'dr'=R^2-r^2$. The <Newtonian gravitational potential of a uniform solid sphere> is consequently
$$
\boxed{\Phi(r)=-2\pi G\rho\left(R^2-\frac{r^2}{3}\right)=-\frac{GM}{2R^3}(3R^2-r^2),\quad 0\leq r<R,\qquad M=\frac{4\pi\rho R^3}{3}.}
$$
The 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.
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