A journal I am reading about circular hydraulic jumps provides the equation and I quote:
$H$ Depth after hydraulic jump, $h$ depth before hydraulic jump, $V$ velocity after hydraulic jump, $v$ velocity before hydraulic jump, $R$ radius of hydraulic jump.
The condition for conservation of momentum can be written as $$\frac{dp}{dt} = 2\pi R\rho HV^2 - 2\pi R\rho hv^2 = F_1-F_2$$ where $$F_1 = 2\pi R \rho g \int_0^hx\cdot dx = \rho g\pi Rh^2$$ $$F_2 = 2\pi R \rho g \int_0^Hx\cdot dx = \rho g\pi RH^2$$
I understand that units for $F$ and $\frac{dp}{dt}$ are both $kg\ m\ s^{-2}$, and that they are equivalent.
But I don't know how, in this example, $F_1$ and $F_2$ are derived i.e. where $g$ and the integrals come from. Could somebody please explain?
Page 4 from: On the circular hydraulic jump