Euler equations for the incompressible fluids Incompressible fluid with constant density ρ fills the three-dimensional domain below the free surface $z = η(r)$ in cylindrical polar coordinates. The flow is axisymmetric and steady, and the only non-zero velocity component is $u_θ$. Gravity acts upon the fluid. The fluid in $r\lt a$ rotates rigidly about the z-axis with angular velocity $\Omega$ and the fluid $r \ge a$ is irrotational.
Use the radial and vertical components of the Euler equations to show that the pressure $p$ in the region $z < η$, $r < a$ satisfies
$$\frac{p}{\rho} = \frac{1}{2}{Ω^2}{r^2} −gz + \text{constant}$$ and find the constant.
$$\frac{∂u_r}{∂t}+\left(u_r\frac{∂}{∂r}+u_z\frac{∂}{∂z}\right)u_r-\frac{(u_θ)^2}{r}+\frac{1}{\rho}\frac{∂p}{∂r}=0 $$
$$\frac{∂u_θ}{∂t}+\left(u_r\frac{∂}{∂r}+u_z\frac{∂}{∂z}\right)u_θ+\frac{u_θu_r}
{r}=0 $$
$$\frac{∂u_z}{∂t}+\left(u_r\frac{∂}{∂r}+u_z\frac{∂}{∂z}\right)u_z+\frac{1}{\rho}\frac{∂p}{∂z}=-g $$
I have the Euler equation in cylindrical coordinate. Then how should I figure out the $\Omega$. The next question is to show that the free surface position in $ r<a $ is 
$$ η=\frac{\Omega^2a^2}{g}(\frac{r^2}{2a^2}-1) $$ if it is helpful. Thank you so much!
 A: Sincerely, if you derived the Euler equations for cylindrical coordinates, this is a very simple exercise for you.
With the hypothesis, many terms disappear, those involving $u_r=0,\;u_z=0$ and the partials wrt time (steady flow).
Further, the motion is rigid, meaning that all fluid parts have zero relative motion. If they are moving in circles, the angular velocity for all point in the fuid $\Omega$ is the same! You don't need to figure anything, it must be known. With all this, $u_\theta=\Omega r$
$\begin{cases}
 -\dfrac{(u_θ)^2}{r}+\dfrac{1}{\rho}\dfrac{\partial p}{\partial r}=0\\
 \dfrac{1}{\rho}\dfrac{\partial p}{\partial z}=-g
\end{cases}$
Integrating the second,
$\dfrac{p}{\rho}=-gz+f(r)\tag 1$
So is $\dfrac{1}{\rho}\dfrac{\partial p}{\partial r}=f'(r)$ or $\dfrac{p}{\rho}+C=f(r)$. With the first equation,
$-\dfrac{(u_θ)^2}{r}+f'(r)=0$ or $-\dfrac{(\Omega r)^2}{r}+f'(r)=0$ or
$-\Omega^2r+f'(r)=0$. Integrating, $f(r) = \dfrac{1}{2}{Ω^2}{r^2}+h(z)$
$\dfrac{p}{\rho}= \dfrac{1}{2}{Ω^2}{r^2}+h(z)-C\tag 2$
Comparing $(1)$ and $(2)$ we have that $h(z)=-gz+C$
$\dfrac{p}{\rho} = \dfrac{1}{2}{Ω^2}{r^2} −gz +C$
I could not recover the given solution for the free surface, although I've found some relation that produces the well known paraboloid for the free surfce of the Newton's bucket. To find $C$, we can consider the heigh $z$ for the free surface ($p=0$) at some $r$. We can set $z=0$ for $r =a$.
$0=\dfrac{1}{2}{Ω^2}{a^2}+C$ or $C=-\dfrac{1}{2}{Ω^2}{a^2}$
$\dfrac{p}{\rho} = \dfrac{1}{2}{Ω^2}{r^2} −gz-\dfrac{1}{2}{Ω^2}{a^2}=\dfrac{1}{2}Ω^2(r^2-a^2)-gz$
For the free surface $z=\eta(r)$, $p=0$
$\eta=\dfrac{1}{2g}Ω^2(r^2-a^2),\;r\lt a$
Added
From other considerations, $\eta$ for $r\ge a$ is $\eta=-\dfrac{\Omega^2a^4}{2gr^2}$. As $\eta$ has to be continuous $\eta(a)=-\dfrac{\Omega^2a^2}{2g}$, for the formula we've found for $r\lt a$
$0=\dfrac{Ω^2a^2}{2}+\dfrac{\Omega^2a^2}{2}+C$ leading to $C=\Omega^2a^2$ and to $\eta=\dfrac{\Omega^2a^2}{g}(\dfrac{r^2}{2a^2}-1)$ for $r\lt a$
$$\eta=
\begin{cases}
 \dfrac{\Omega^2a^2}{g}(\dfrac{r^2}{2a^2}-1)& x\lt a\\
 -\dfrac{\Omega^2a^4}{2gr^2}&a\le x
\end{cases}$$
I've sketched the function for the free surface (setting $\dfrac{\Omega^2a^2}{2g}=1$ and $a=1$)

