$F_x={\displaystyle \underset{0}{\overset{F}{}}}dF={\displaystyle \underset{0}{\overset{D}{}}}lg(D-z)Wdz=\frac{1}{2}lgW{D}^2$

Consider a cylindrical dam with diameter D$(=2$R$)($the width of the river, i$.$e$.,$ the width of the cylinder, is W$).$ At one point, the water surface of the river is equal to the height of the dam $($i$.$e$.,$ the height of the dam is equal to the diameter D of the cylinder$-$shaped dam$).$ At this time, the river is in contact with the cylinder$-$shaped dam on one hemisphere. Assuming that the density of water is $10^3$ kg$/$m$^3,$ when the total horizontal force applied by one hemisphere of the dam from the water of the river is obtained, let's call it Fx$.$ Since the water surface of the river has gone down a lot because it has not rained for a while, one day I checked that the water surface of the river has come down to half $(=$R$)$ of the cylindrical dam. So what is the total horizontal force that the dam receives from the river at this time? Answer using Fx$.$