Vertical <hydrostatic equilibrium> gives
$$
\frac{dp}{dz}=-\rho\Omega_z^2z.
$$
Define vertically integrated pressure and the density-weighted <disk scale height> by
$$
P=\int_{-\infty}^{\infty}p\,dz,
\qquad
H^2=\frac1\Sigma\int_{-\infty}^{\infty}\rho z^2\,dz.
$$
Multiply the hydrostatic equation by $z$ and integrate. Since $zp\to0$ at both boundaries, <integration by parts> gives
$$
-P=-\Omega_z^2\Sigma H^2.
$$
Therefore
$$
\boxed{P=\Sigma H^2\Omega_z^2}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 321 / 3 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-321/3/b/solution