Using big D derivative, with del dot v = 0, assuming x is perpendicular to the inclined plane and nu=constant –

0 = g sin(theta) + nu d^2v/dx^2

v = – (g sin(theta) / 2nu) x^2 + c_1 x + c_0

Assuming the flow is stuck on the bottom (v = 0 at x = 0) and there is no shear at the air boundary (dv/dx = 0 at x = d)

v = (g sin(theta) / 2nu) ( 2xd – x^2 )

(fastest at the top)

The flow rate is integral of this over x, times width w

= (g sin(theta) / 3 nu) d^3 w

This looks right dimensionally (m^3 s^-1)

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