= Solution
Using the same cylindrical-shell <volume>,
$$
\boxed{L=2\pi h\int_0^a r^2v(r)\,dr
=2\pi h\int_0^a r^3\omega(r)\,dr}.
$$
For unit <density>, this is the axial <angular momentum>. To bound it, apply the integral <Cauchy-Schwarz inequality> to $f(r)=\sqrt r\,v(r)$ and $g(r)=r^{3/2}$:
$$
\left(\int_0^ar^2v(r)\,dr\right)^2
\leq\left(\int_0^arv(r)^2\,dr\right)\left(\int_0^ar^3\,dr\right)
=\frac{K}{\pi h}\frac{a^4}{4}.
$$
Multiplication by $(2\pi h)^2$ gives $L^2\leq\pi ha^4K$. Since the chamber's <volume> is $V=\pi a^2h$, this is precisely
$$
\boxed{K\geq\frac{L^2}{Va^2}}.
$$
Back to article page
