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.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 333 / 2 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-333/2/b/solution