= Solution
Let $u_i=\langle v_i\rangle$, $\sigma_{ij}=\langle(v_i-u_i)(v_j-u_j)\rangle$, and $P_{ij}=\rho\sigma_{ij}$. The <Jeans equation> obtained by taking the first <velocity> moment of the <Collisionless Boltzmann equation> is
$$
\partial_t(\rho u_j)+\partial_k(\rho u_ju_k+P_{jk})=-\rho\partial_j\Phi.
$$
For an equilibrium population multiply by $x_i$ and integrate. <Integration by parts> gives
$$
\oint x_i(\rho u_ju_k+P_{jk})n_k\,dS-\int(\rho u_iu_j+P_{ji})\,d^3x=-\int\rho x_i\partial_j\Phi\,d^3x.
$$
Assume the integrals exist and the boundary flux vanishes, as for an isolated finite system with adequate decay. With the <stellar kinetic-energy tensor>, <stellar pressure-energy tensor> and <stellar potential-energy tensor> defined by
$$
T_{ij}=\frac12\int\rho u_iu_j\,d^3x,\qquad \Pi_{ij}=\int\rho\sigma_{ij}\,d^3x,\qquad W_{ij}=-\int\rho x_i\partial_j\Phi\,d^3x,
$$
we obtain the equilibrium <tensor virial theorem>
$$
\boxed{2T_{ij}+\Pi_{ij}+W_{ij}=0.}
$$
The <stellar kinetic-energy tensor> describes ordered motion, whereas the <stellar pressure-energy tensor> describes random motion. In particular, $T_{ij}=0$ does not mean the <stars> have no <kinetic energy>. If a surface flux is retained, the right side of the displayed equilibrium identity is that surface tensor; it cannot be discarded for an arbitrary finite aperture. For an isolated self-gravitating population, $W_{ij}$ is symmetric by pairwise interchange in the Newtonian interaction integral, and the time-dependent identity is $\tfrac12\ddot I_{ij}=2T_{ij}+\Pi_{ij}+W_{ij}$ with $I_{ij}=\int\rho x_ix_jd^3x$.
For the spherical <Newtonian gravitational potential>, $\partial_j\Phi=\Phi'(r)x_j/r$. In the pressure-supported case the <tensor virial theorem> gives $\Pi_{ii}=\int\rho\Phi'(r)x_i^2/r\,d^3x$. The notation $\Pi_{RR}$ used in this question is the <planar virial trace> $\Pi_{xx}+\Pi_{yy}$, not just the integral of the local cylindrical radial <velocity dispersion>. In <cylindrical coordinates> this trace includes both $v_R^2$ and $v_\phi^2$. Summing the two Cartesian components therefore gives
$$
\boxed{\frac{\Pi_{RR}}{\Pi_{zz}}=\frac{\int\rho\Phi'(r)R^2/r\,d^3x}{\int\rho\Phi'(r)z^2/r\,d^3x}.}
$$
The distinction explains the factor of two in the spherical limit.
Use ordinary <spherical coordinates>, $R=r\sin\theta$, $z=r\cos\theta$. The <mass density> separates as $\rho_0r^{-\gamma}h(\theta)$, where
$$
h(\theta)=(\sin^2\theta+q^{-2}\cos^2\theta)^{-\gamma/2}.
$$
Both potential-energy integrals have the same radial factor $2\pi\rho_0\int_0^\infty r^{3-\gamma}\Phi'(r)\,dr$. When this factor is finite and positive it cancels, leaving the <scale-free tracer virial ratio>
$$
\boxed{\frac{\Pi_{RR}}{\Pi_{zz}}=\frac{\int_0^\pi\sin^3\theta\,(\sin^2\theta+q^{-2}\cos^2\theta)^{-\gamma/2}\,d\theta}{\int_0^\pi\sin\theta\cos^2\theta\,(\sin^2\theta+q^{-2}\cos^2\theta)^{-\gamma/2}\,d\theta}.}
$$
\b[The angular factor in the printed intermediate formula has its sine and cosine interchanged.] Keeping its sine/cosine weights while using that printed factor would instead produce the opposite sign in the final small-flattening correction. Merely measuring $\theta$ from the equatorial plane cannot fix this: the measure and both moment weights would have to change as well.
The angular ratio is finite and positive for every fixed $q>0$ and finite real $\gamma$, because $h$ is positive and bounded above and below. However, the claim that the unrestricted global virial integrals are always well-defined needs a hypothesis. In a point-mass <Newtonian gravitational potential>, the common radial factor is proportional to $\int_0^\infty r^{1-\gamma}dr$: convergence at zero requires $\gamma<2$, while convergence at infinity requires $\gamma>2$. There is no such $\gamma$. If the global energies diverge, the boxed angular ratio is still the ratio of the two potential-energy integrals with identical spherical inner and outer cutoffs, independent of the cutoffs. It is not a quotient of two finite global <stellar pressure-energy tensors>; applying a finite-aperture <tensor virial theorem> additionally requires its surface terms. These convergence and boundary qualifications are separate from the printed angular typo.
Set $\epsilon=q^{-2}-1$. For modest flattening, the <Taylor expansion> is $h=1-(\gamma/2)\epsilon\cos^2\theta+O(\epsilon^2)$. The required elementary angular <integrals> are
$$
\int_0^\pi\sin^3\theta\,d\theta=\frac43,\quad \int_0^\pi\sin^3\theta\cos^2\theta\,d\theta=\frac4{15},\quad \int_0^\pi\sin\theta\cos^2\theta\,d\theta=\frac23,\quad \int_0^\pi\sin\theta\cos^4\theta\,d\theta=\frac25.
$$
These follow directly by $t=\cos\theta$, or from the supplied <gamma function> identity. Hence the numerator is $(4/3)[1-\gamma\epsilon/10+O(\epsilon^2)]$ and the denominator is $(2/3)[1-3\gamma\epsilon/10+O(\epsilon^2)]$. The <flattening–anisotropy virial relation> becomes
$$
\boxed{\frac{\Pi_{RR}}{2\Pi_{zz}}=1+\frac\gamma5(q^{-2}-1)+O((q^{-2}-1)^2).}
$$
For a decreasing tracer <mass density> ($\gamma>0$), oblate flattening ($q<1$) requires more random kinetic support per in-plane direction than vertically. In a spherical <Newtonian gravitational field>, directional orbital anisotropy supplies the flattening rather than an anisotropic force law. A steeper radial <mass density> amplifies the required <velocity dispersion> anisotropy. A prolate tracer reverses the leading inequality. The relation constrains integrated <velocity dispersions>, not a unique local <galactic distribution function>.
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