- #26

Regla

thanks for showing me how I was supposed to form the integral, my only question is what do you mean that ρgL factors out?@Nidum is trying to lead you past the integration step so that you do not have to evaluate that integral.

The key is that you are being asked to evaluate ##\int_0^h \rho g L \ x \ dx##

The ##\rho g L## factors out and all that's left is evaluating ##\rho g L\ \int_0^h x \ dx##

That integral should be easy to evaluate. It should give the same result that @Nidum is jumping to: The force on a side is the average pressure on a side multiplied by the [wetted] area of that side.