= Solution
This is a <Rankine vortex>. Write $r=\sqrt{x^2+y^2}$. Inside the core the <velocity> is $u=(\Gamma y/a^2,-\Gamma x/a^2,0)$, so
$$
\boxed{\omega=\nabla\times u=-\frac{2\Gamma}{a^2}k\quad(r<a).}
$$
This is the core value obtained from the inner formula; outside, the given irrotationality makes the <vorticity> zero. For nonzero $\Gamma$, the sharp boundary $r=a$ has different one-sided velocity gradients, so a single classical <vorticity> is not defined exactly on that edge; the inner trace is the boxed value and the outer trace is zero.
Inside the core, direct differentiation gives $(u\cdot\nabla)u=-\Gamma^2(x,y,0)/a^4$. The steady <Euler equations> therefore give $p_r=\rho\Gamma^2r/a^4$ and $p_z=-\rho g$. Integrating,
$$
p_{\rm in}=A+\frac{\rho\Gamma^2r^2}{2a^4}-\rho gz.
$$
Outside, $|u|^2=\Gamma^2/r^2$, so part (a) gives
$$
p_{\rm out}=B-\frac{\rho\Gamma^2}{2r^2}-\rho gz.
$$
Matching <pressure> at the core edge $r=a$ yields $A=B-\rho\Gamma^2/a^2$. Hence the complete <pressure> field, up to one additive constant $B$, is
$$
\boxed{p(r,z)=\begin{cases}
B-\rho gz-\dfrac{\rho\Gamma^2(2a^2-r^2)}{2a^4},&r\leq a,\\
B-\rho gz-\dfrac{\rho\Gamma^2}{2r^2},&r>a.
\end{cases}}
$$
Both the <velocity> and <pressure> are continuous at $r=a$. The inside expression has been derived from the momentum equation because the core is rotational.
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