It looks like you are on the right track with this one. However, you have a factor of 2 in your equations that I don't understand.

Since ρ is quoted in the SI unit (kg/m^{3}) and ultimately one would like to used g = 9.8 m/s^{2} , it is a good idea to convert the dimensions of the trough from cm to m. Thus the trough is 1 m deep , 10 m long with a top width of 1 m and a bottom width of 0.5 m.

The integrations will be along the vertical coordinate which can be taken to be y. For problems of this type, it is convenient to choose the zero of y to be at the top and to choose the direction of positive y to be down. The width as a function of y is then w = 1- 0.5h.

With this choice, the mass of a thin layer of water is dM = ρ 10 w dh = ρ 10 (1- 0.5h) dh, where dh is the infinitesimal thickness of the thin layer. The work required to pump this thin layer out is dM g h. Thus the work required to pump out all the water is

W = ∫ dM h

W = 10 g ρ ∫ (1 - 0.5 h) h dh The lower limit is zero, the upper limit is 1

This works out to (10/3) g ρ

The pressure on the trapezoidal face is given by P = ρ g (1- 0.5 y)

The Force is the integral of this expression over the area of the trapezoid.

Symbolically F = ∫ P dA

Since P depends on y but not on the width variable, one can write dA = w dy = (1 - 0.5 y) dy

F = rho g integral [ y(1- 0.5y) dy = (1/3) rho g