= Solution
Treat the cylinder as infinitely long, with a vacuum exterior and no <surface tension>. Let $\Phi_0$ be its <gravitational potential>. Integrating the gravitational <Poisson equation> over a coaxial cylinder of radius $r<R$ and length $L$, then applying the <divergence theorem>, gives
$$
2\pi rL\frac{d\Phi_0}{dr}=4\pi G(\pi r^2L\rho),\qquad
\frac{d\Phi_0}{dr}=2\pi G\rho r.
$$
Thus the inward <gravitational acceleration> is $-2\pi G\rho r\widehat{\mathbf r}$. <Hydrostatic equilibrium> requires
$$
\frac{dp_0}{dr}=-\rho\frac{d\Phi_0}{dr}=-2\pi G\rho^2r.
$$
Impose zero <pressure> on the <free surface>, $p_0(R)=0$, and integrate inward:
$$
\boxed{p_0(r)=\pi G\rho^2(R^2-r^2),\qquad 0\le r<R.}
$$
This is the <hydrostatic self-gravitating incompressible cylinder>. Its central <pressure> is $\pi G\rho^2R^2$, and the units of the expression are $\mathrm{Pa}$. A nonzero external <pressure> would add that constant to $p_0(r)$.
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