 Solution

Direct differentiation of the velocity field gives $(\mathbf{u}\cdot \nabla )\mathbf{u}=(-{\Omega }^{2}x,-{\Omega }^{2}y,0)$. 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 $p_r=\rho {\Omega }^{2}r$ and $p_z=-\rho g$, hence $p=\rho ({\Omega }^2{r}^2/2-gz)+C$. Constant atmospheric pressure on the free surface gives $z_s(r)=H+{\Omega }^2{r}^2/(2g)$. Conservation of fluid volume determines $H$:ThereforeThe lowest point is the axis, where $z_s(0)=h-{\Omega }^2{a}^2/(4g)$. The free surface first meets the bottom whenThis 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.