= Solution
On a <beta plane>, $f=f_0+\beta y$, the preceding <vorticity equation> is
$$
\xi_t+u\xi_x+v\xi_y+\beta v
=-\frac{\tau_{s,y}}{\rho_0H}-r\xi.
$$
In the broad ocean interior, assume a statistically steady flow, weak <advection> of <relative vorticity>, and negligible <friction> compared with planetary-<vorticity> advection. The constant-depth, constant-<density> assumptions of 2(a) exclude topographic and <baroclinic vorticity generation> contributions. More precisely, require $|\xi_t|,|\mathbf u\cdot\nabla\xi|,|r\xi|\ll|\beta v|$. Then <Sverdrup balance> is
$$
\boxed{\beta v=\frac{(\nabla_h\times\boldsymbol\tau)_z}{\rho_0H}
=-\frac1{\rho_0H}\frac{d\tau_s}{dy}.}
$$
Small <Rossby number> and slow forcing help justify these conditions in the interior; they do not justify them inside an intense <western boundary current>. For fluctuations, the mean <eddy> contribution to relative-<vorticity> advection must also be negligible.
The depth-integrated northward <volume flux> per unit zonal distance is $V=Hv$. For a basin $0<x<L$ and <wind stress> independent of $x$, the northward \b[interior] <volume flux> is
$$
\boxed{T(y)=\int_0^L Hv\,dx=-\frac{L}{\beta\rho_0}\frac{d\tau_s}{dy}.}
$$
A negative value denotes southward transport. The cancellation of $H$ is why this interior <volume flux> is insensitive to the chosen constant depth. In a closed, steady basin the net <volume flux> across a full latitude section is zero: a boundary return must compensate $T$. Thus the displayed formula cannot represent the total closed-basin transport including that return.
For a concrete circulation sketch, use a northern-hemisphere subtropical basin $0<y<B$ and
$$
\tau_s(y)=-\tau_0\cos(\pi y/B),\qquad
V(y)=-\frac{\pi\tau_0}{\beta\rho_0B}\sin(\pi y/B).
$$
This represents westward <trade winds> at the southern edge and eastward westerlies at the northern edge. The negative <wind stress curl> drives southward interior flow. Define an <ocean transport streamfunction> by $Hu=-\Psi_y$, $Hv=\Psi_x$ and impose no normal flow at the eastern wall through $\Psi(L,y)=0$. Integrating <Sverdrup balance> westward gives
$$
\Psi_I=(x-L)V(y),\qquad
Hu_I=(L-x)V_y(y),\qquad Hv_I=V(y).
$$
Since $V=0$ at $y=0,B$, this also gives no normal flow at the north and south walls. Interior water moves westward in the southern part, southward across the basin, and eastward in the northern part. A northward <western boundary current> completes the clockwise <ocean gyre>.
\Image[/past-exam-of-the-geophysical-fluid-dynamics-course-of-the-university-of-oxford/2024-sverdrup-gyre.png]
{title=Northern-hemisphere subtropical Sverdrup interior and a schematic western return, with arrows indicating volume transport}
Pure <Sverdrup balance> cannot satisfy the western impermeability condition as well: $\Psi_I(0,y)=-LV(y)$ varies along that wall. Indeed imposing no normal flow on both east and west walls would require their constant <streamfunction> values to agree in this simply connected closed basin, whereas their difference here is the nonzero $T(y)$. The shaded return in the sketch represents a boundary region where the interior approximation breaks down. It is supplied by <friction> in 2(c), rather than inferred from <Sverdrup balance> alone. A southern-hemisphere subtropical <ocean gyre> has the reflected, anticlockwise circulation.
For an order-of-magnitude estimate, use $\tau_0=0.1\ \mathrm{N\,m^{-2}}$, $\rho_0=10^3\ \mathrm{kg\,m^{-3}}$, $\beta=2\times10^{-11}\ \mathrm{m^{-1}\,s^{-1}}$, $L=5\times10^6\ \mathrm m$ and a <wind stress> variation scale $\ell_\tau=10^6\ \mathrm m$. Then
$$
\boxed{|T|\sim\frac{L\tau_0}{\beta\rho_0\ell_\tau}
=2.5\times10^7\ \mathrm{m^3\,s^{-1}}=25\ \mathrm{Sv}.}
$$
Here $1\ \mathrm{Sv}=10^6\ \mathrm{m^3\,s^{-1}}$ is a <Sverdrup>. The sinusoidal example has maximum $|\tau_{s,y}|=\pi\tau_0/B$; taking $B=\pi\times10^6\ \mathrm m$ makes it agree with the chosen variation scale. Other reasonable basin widths and wind profiles change the coefficient, but retain a transport of order tens of <Sverdrups>.
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