= Solution
Use eastward $x$, northward $y$, constant layer depth $H$, constant reference <density> $\rho_0$, and a depth-independent horizontal <velocity field> $(u,v)$. Assume a flat bottom and a rigid upper boundary, giving the two-dimensional <incompressibility condition> $u_x+v_y=0$. The <traditional approximation> retains only the vertical component of planetary rotation in the horizontal equations. Take the <Coriolis parameter> $f=f(y)$, a constant <linear drag> rate $r$, and no other torques. The component <Cauchy momentum equations> are
$$
Du-fv=-\frac{p_x}{\rho_0}+\frac{\tau_s}{\rho_0H}-ru,
\qquad Dv+fu=-\frac{p_y}{\rho_0}-rv,
\qquad D=\partial_t+u\partial_x+v\partial_y.
$$
Define <relative vorticity> $\xi=v_x-u_y$. Apply $\partial_x$ to the second equation and subtract $\partial_y$ of the first. The <pressure gradient force> has zero <curl> because $\rho_0$ is constant and mixed <partial derivatives> commute. Direct expansion of the material-acceleration terms gives
$$
\partial_x(Dv)-\partial_y(Du)
=D\xi+(u_x+v_y)\xi=D\xi.
$$
The <curl> of the rotation terms is
$$
\partial_x(fu)+\partial_y(fv)
=f(u_x+v_y)+vf_y=vf_y.
$$
The imposed zonal <wind stress> contributes $-\tau_{s,y}/(\rho_0H)$, and constant <linear drag> contributes $-r\xi$. Hence
$$
D\xi+vf_y=-\frac{\tau_{s,y}}{\rho_0H}-r\xi.
$$
Since $Df=vf_y$, defining <absolute vorticity> $\zeta=\xi+f$ gives
$$
\boxed{\left(\partial_t+u\partial_x+v\partial_y\right)\zeta
=-\frac1{\rho_0H}\frac{\partial\tau_s}{\partial y}-r\xi.}
$$
Without the <incompressibility condition>, the left-hand side would contain the stretching term $\zeta\nabla_h\cdot\mathbf u$. Variable layer depth, variable <density> or spatially varying <linear drag> would introduce further terms. No small-<Rossby number> assumption was needed to obtain this constant-depth <vorticity equation>.
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