Find the resultant force vector acting on the unit width of the wall from the water.

Force on the curved portion of the wall:

$$\begin{gathered} h=R-R\sin \theta \\ =R\left(1-\sin \theta \right) \end{gathered}$$

The force on the element

$$\begin{gathered} dF=\left(\rho gh\right)\left(Rd\theta \times 1\right) \\ dF{x}_1=dF\cos \theta \\ =\left(\rho ghRd\theta \right)\cos \theta \\ =\rho g{R}^2\left(1-\sin \theta \right)\cos \theta d\theta \\ \therefore F{x}_1=\rho g{R}^2{\int }_0^{\pi /2}\left(\cos \theta -\frac{\sin 2\theta }{2}\right)d\theta \\ =\rho g{R}^2{\left|\sin \theta +\frac{\cos 2\theta }{4}\right|}_0^{\pi /2} \\ =\rho g{R}^2\left[\left(\sin \frac{\pi }{2}+\frac{\cos \pi }{4}\right)-\left(\sin 0+\cos 0\right)\right] \\ =\rho g{R}^2\left[\left(1+\frac{1}{4}\right)-1\right] \\ =\frac{\rho g{R}^2}{4} \\ {F}_y=\rho g{R}^2{\int }_0^{\pi /2}\left(1-\sin \theta \right)\sin \theta d\theta \\ =\rho g{R}^2{\int }_0^{\pi /2}\left(\sin \theta -{\sin }^2\theta \right)d\theta \\ =\rho g{R}^2{\left|-\cos \theta +\frac{\sin 2\theta }{4}-\frac{\theta }{2}\right|}_0^{\pi /2} \\ =\rho g{R}^2\left[\left(-\cos \frac{\pi }{2}+\sin \frac{\pi }{4}-\frac{\pi }{4}\right)-\left(-\cos 0+\sin 0-0\right)\right] \\ =\rho g{R}^2\left(1-\frac{\pi }{4}\right) \\ =\rho g{R}^2\frac{\left(4-\pi \right)}{4} \end{gathered}$$

Force on the flat vertical position of the gate

$$\begin{gathered} F_{x2}=P_{av}\times A \\ =\left(\frac{\rho gR+\rho g\left(2R\right)}{2}\right)\times \left(R\times 1\right) \\ =\frac{3}{2}\rho g{R}^2 \end{gathered}$$

Thus $F_x=F_{x1}+F_{x2}$

              $$\begin{gathered} =\frac{\rho g{R}^2}{4}+\frac{3}{2}\rho g{R}^2 \\ =\frac{7}{4}\rho g{R}^2 \end{gathered}$$

Thus resultant force $\overrightarrow{F}=F_x\hat{i}+F_y\left(-\hat{j}\right)$

                                    $=\frac{7}{4}\rho g{R}^2\hat{i}-\rho g{R}^2\frac{\left(4-\pi \right)}{4}\hat{j}$

After substituting the values and simplifying, we get

$\overrightarrow{F}=\left(961.38\hat{i}-103.16\hat{j}\right)kN$