= Solution
Define the <surface density of a disk> and outward radial mass flux by
$$
\Sigma=\int_{-\infty}^{\infty}\rho\,dz,
\qquad
\mathcal F=2\pi r\int_{-\infty}^{\infty}\rho u_r\,dz.
$$
Vertical integration of mass conservation eliminates the surface term because $\rho u_z\to0$, and axisymmetry eliminates the azimuthal derivative. Hence
$$
\boxed{\partial_t\Sigma+\frac1{2\pi r}\partial_r\mathcal F=0}.
$$
For $u_\phi\simeq r\Omega(r)$, the <specific angular momentum> is $h=ru_\phi=r^2\Omega$. Multiply the azimuthal momentum equation by $r$, integrate vertically and azimuthally, and define the <viscous torque in an accretion disk>
$$
\boxed{\mathcal G=-2\pi r^2\int_{-\infty}^{\infty}\Pi_{r\phi}\,dz}.
$$
Subtracting $h$ times the mass equation from the integrated angular-momentum equation gives
$$
\boxed{\mathcal F\frac{dh}{dr}+\frac{d\mathcal G}{dr}=0}.
$$
For an axisymmetric circular flow, $\Pi_{r\phi}=\mu r\,d\Omega/dr$. If
$$
\bar\nu\Sigma=\int_{-\infty}^{\infty}\mu\,dz,
$$
then
$$
\boxed{\mathcal G=-2\pi\bar\nu\Sigma r^3\frac{d\Omega}{dr}}.
$$
Combining the two conservation laws gives
$$
\boxed{\partial_t(2\pi r\Sigma h)
+\partial_r(\mathcal Fh+\mathcal G)=0}.
$$
The first term is the local rate of change of angular momentum per radial interval, $\mathcal Fh$ is outward advective angular-momentum flux, and $\mathcal G$ is outward stress-carried angular-momentum flux.
Solved by gpt-5.6-sol high.
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