= Solution
Substitute the rotating <velocity field> into the <Euler equations for an inviscid fluid>. Its acceleration is $(-\dot\Omega y-\Omega^2x,\dot\Omega x-\Omega^2y,0)$, so
$$
\frac{p_x}{\rho}=(\alpha+\Omega^2)x+(\beta+\dot\Omega)y,\qquad
\frac{p_y}{\rho}=(\gamma-\dot\Omega)x+(\delta+\Omega^2)y,\qquad p_z=0.
$$
The preceding angular acceleration makes both mixed coefficients $(\beta+\gamma)/2$, ensuring integrability. Thus
$$
\boxed{p(x,y,t)=p_0(t)+\frac\rho2\left[(\alpha+\Omega(t)^2)x^2+(\beta+\gamma)xy+(\delta+\Omega(t)^2)y^2\right].}
$$
Pressure is determined up to the spatially uniform function $p_0(t)$. The specified force has no vertical component; if gravity were additionally included in the body force, the usual hydrostatic term $-\rho gz$ would also appear. It is not implied by the displayed force here.
Back to article page
