We now make one of two virtually equivalent choices

• Suppose that the integration is over the entire depth of the ocean, the term $[w]_B^T$ vanishes given a rigid lid and a flat bottom. If the stress at the top of the ocean is given by the wind stress, and at the bottom of the ocean it is parameterized by a linear drag $$ \beta \overbrace{\underbrace{\bar{v}}_{\partial \psi / \partial x}}^{\text{vertical integral}} = F_\tau(x,y) - r \overbrace{\underbrace{\bar{\zeta}}_{\nabla^2 \psi}}^{\text{vertical integral}} \Rightarrow r \nabla^2 \psi + \beta \frac{\partial \psi}{\partial x} = F_\tau(x,y) $$
• Suppose that the integration is between two thin Ekman layers at the top and bottom of the ocean. The stress is zero at the interior edge of these layers, but the vertical velocity is not. At the base of the upper Ekman layer, at $z = -\overbrace{\delta_T}^{\text{upper Ekman thickness}}$, the vertical velocity is $$ w(x,y,-\delta_T) = \text{curl}_z (\boldsymbol{\tau}_T/f_0) $$ where the top of the ocean is at $z=0$
  At the top of the lower Ekman layer, the vertical velocity is $$ w(x,y,-H+\underbrace{\delta_B}_{\text{bottom Ekman thickness}}) = \delta_B \zeta $$ where $z=-H$ at the ocean bottom
  Neglecting the advective derivative, and integrating over the ocean between the two Ekman layers, \(\boxed{\frac{D\zeta}{Dt} + \beta v = f_0 \frac{\partial w}{\partial z} + \text{curl}_z \frac{\partial \boldsymbol{\tau}}{\partial z}}\) becomes $$ \beta \bar{v} = \text{curl}_z \boldsymbol{\tau}_T - f_0 \delta_B \zeta = \text{curl}_z \boldsymbol{\tau}_T - \underbrace{f_0 \delta_B \bar{\zeta}/H}_{\text{assume bottom drag using vertically integrated vorticity } \bar{\zeta}} $$ Defining the drag coefficient $r$ by $r=f_0\delta_B/H$, and introducing a streamfunction gives $$ r \nabla^2 \psi + \beta \frac{\partial \psi}{\partial x} = \text{curl}_z \boldsymbol{\tau}_T $$