Bernoulli equation for pressure (with the vortex) An incompressible inviscid fluid of constant density ρ moves subject to a gravitational acceleration −gk.
The vortex caused by water flowing down a plughole is modelled by the steady velocity ￼￼￼field
when $x^2+y^2\le a^2$
$$u=\left(\frac{Γy}{a^2},-\frac{Γx}{a^2},0 \right)  $$ 
when  $x^2+y^2\gt a^2$
$$\left(\frac{Γy}{x^2+y^2},-\frac{Γx}{x^2+y^2},0 \right)   $$
 Evaluate the pressure in such a flow.
need to show that when  $x^2+y^2\le a^2$
a free surface at a constant pressure takes the form of 
$$C − \frac{Γ^2(2a^2-x^2-y^2)}{2ga^4} $$
and when $x^2+y^2\gt a^2$
a free surface takes the form of 
 $$C − \frac{Γ^2}{2g(x^2+y^2)} $$
I did work out the  $x^2+y^2\gt a^2$ one by using the Bernoulli's equation for irrotational flow but don't know the other one with vorticity since for steady irrotational flow $p+\frac{1}{2}\rho|\mathbf u|^2+\rho gz=constant$ 
But how to get the rotational part,should I also use the Bernoulli's equation? But it doesn't have the constant part then how should I approach it?
Thank you so much!
 A: For inviscid, rotational flow a weaker form of the Bernoulli equation $p + \frac{1}{2} \rho | \mathbf{u}|^2 + \rho g z  = C(\psi)$ is valid, where $C(\psi)$ is constant along a streamline. You can work with this or solve for the pressure field directly from the governing equations for inviscid flow, which reveals the Bernoulli relationship as well.
Using  cylindrical coordinates we have $x = r \cos \theta, \, y = r \sin \theta,$ and the basis vectors
$$\mathbf{e}_r = \cos \theta \, \mathbf{e}_x + \sin \theta \, \mathbf{e}_y, \\\mathbf{e}_\theta = -\sin \theta \, \mathbf{e}_x + \cos \theta \, \mathbf{e}_y $$
The velocity field is,  for  $r^2 = x^2 + y^2 \leqslant a^2 $,
$$u_r = 0, \,\, u_\theta = - \frac{\Gamma r}{a^2}, \,\, u_z = 0,$$
and for $x^2 + y^2 > a^2$,
$$u_r = 0, \,\, u_\theta = - \frac{\Gamma }{r}, \,\, u_z = 0$$
The components of the Euler equation for steady, inviscid flow, $\rho\mathbf{u} \cdot \nabla \mathbf{u} \,\, = - \nabla p + \rho \mathbf{g}$, reduce to
$$\frac{\partial p}{\partial r} = \frac{\rho u_\theta^2}{r} = \begin{cases}\frac{\rho \Gamma^2r}{a^4}, \,\, x^2+y^2 \leqslant a^2 \\  \frac{\rho \Gamma^2}{r^3}, \,\,\,\, x^2+y^2 > a^2\end{cases}$$
and  
$$\frac{\partial p}{\partial z}= -\rho g$$
Integrating, we get for $x^2 + y^2 \leqslant a^2$,
$$p = \alpha - \rho g z + \frac{\rho \Gamma^2 r^2}{2a^4}  = \alpha - \rho g z + \frac{\rho \Gamma^2 (x^2 + y^2)}{2a^4},$$
and, for $x^2 + y^2 > a^2$, 
$$p = \beta - \rho g z - \frac{\rho \Gamma^2}{2r^2} =  \beta - \rho g z - \frac{\rho \Gamma^2}{2(x^2 + y^2)} $$
The integration constant $\alpha$  can be evaluated in terms of $\beta \,$ by matching the inner pressure field with the outer pressure field at $x^2 + y^2 = a$, where we have
$$\alpha - \rho g z + \frac{\rho \Gamma^2 (a^2)}{2a^4} = \beta - \rho g z - \frac{\rho \Gamma^2}{2a^2} $$
Solving for $\alpha$ we get
$$\alpha = \beta - \frac{\rho \Gamma^2}{a^2}, $$
and the inner pressure field is
$$p = \beta - \frac{\rho \Gamma^2}{a^2} - \rho g z + \frac{\rho \Gamma^2 (x^2 + y^2)}{2a^4}$$
The free-surface where $p = C$ is given by 
$$z = \frac{\beta-C}{\rho g} - \frac{\Gamma^2}{g a^2} + \frac{ \Gamma^2 (x^2 + y^2)}{2ga^4}$$
