= Solution
Direct differentiation of the <velocity field> gives $(\mathbf u\cdot\nabla)\mathbf u=(-\Omega^2x,-\Omega^2y,0)$. Since $u^2=\Omega^2(x^2+y^2)$, this equals $\nabla(-u^2/2)$ as required.
In cylindrical coordinates, steady rigid rotation has $u_r=u_z=0$, $u_\theta=\Omega r$. The radial and vertical <Euler equations> for constant-density fluid are $p_r=\rho\Omega^2r$ and $p_z=-\rho g$, hence $p=\rho(\Omega^2r^2/2-gz)+C$. Constant atmospheric pressure on the <free surface> gives $z_s(r)=H+\Omega^2r^2/(2g)$. Conservation of fluid volume determines $H$:
$$
\pi a^2h=2\pi\int_0^a z_s(r)r\,dr=\pi a^2H+\frac{\pi\Omega^2a^4}{4g}.
$$
Therefore
$$
\boxed{z_s(r)=h+\frac{\Omega^2}{2g}\left(r^2-\frac{a^2}{2}\right).}
$$
The lowest point is the axis, where $z_s(0)=h-\Omega^2a^2/(4g)$. The <free surface> first meets the bottom when
$$
\boxed{|\Omega|=\frac{2\sqrt{gh}}a.}
$$
This assumes no overflow and uses the full disk of wetted bottom until first contact. Once a dry central region forms, the volume constraint and geometry must be modified.
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