= Solution
The eastern wall has no normal flow, so the <quasi-geostrophic streamfunction> is constant there. Choose $\psi(L,y)=0$. Integrating the Sverdrup relation gives
$$
\boxed{
\psi_{\rm int}(x,y)
=\frac{f_0\tau'(y)}{\beta\rho H_0^2}(L-x)}.
$$
Since <geostrophic balance> gives $\psi=g\eta/f_0$,
$$
\boxed{
\eta_{\rm int}(x,y)
=\frac{f_0^2\tau'(y)}
{g\beta\rho H_0^2}(L-x)}.
$$
If this interior solution were extended to both walls, the west-minus-east height difference would be
$$
\boxed{
\eta_{\rm int}(0,y)-\eta_{\rm int}(L,y)
=\frac{f_0^2L}{g\beta\rho H_0^2}\tau'(y)}.
$$
The northward volume transport per unit meridional distance is
$$
T=H_0\int_0^L v\,dx
=H_0\int_0^L\psi_x\,dx
=\frac{gH_0}{f_0}\,[\eta(L,y)-\eta(0,y)].
$$
Thus
$$
\boxed{
T=-\frac{f_0L}{\beta\rho H_0}\tau'(y)}.
$$
This is the basin-integrated form of <Sverdrup balance>.
Solved by gpt-5.6-sol high.
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