# Finding the net force exerted by water on a vessel wall

I have a hemispherical vessel held against the wall, which holds water inside it. I had to calculate the net force required to hold this arrangement in equilibrium. The diagram is:

Whatever horizontal force $F_x$ I apply, the same will be applied by the wall on the water in it. So using calculus by taking elements of the wall of thickness ${\rm d}y$ and integrating the force due to pressure on the circular part of the wall, I got the answer $F_x = \rho g \pi R^3$, where $\rho=\,$density of water, $R=\,$radius of vessel.

However, when I checked the solution, they gave the answer as $\rho g R(\pi R^2)$, which although the same as my answer, is written in a format that seems to suggest the answer is just the product of pressure at half the total depth times the total area of the wall. Is there a general result like this that suggests the total horizontal force can be calculated just using the pressure at half the depth times the total area of the wall in contact with the fluid? How is it derived?

I divided the above in two parts as the imagining the the 3d figure the origin is the point the wall touching the line point P. From basic equation, $$F=PA$$ Here as the origin is point P,The pressure on it by the above portion is same as the lower one. $$\Rightarrow P=\rho gR$$ $$\Rightarrow A=\frac{\pi R^2}{2}$$ So this is on is for the upper porition for lower one $$P=\rho gR$$ $$A=\frac{\pi R^2}{2}$$ As the forces distributed at every point in the vessel is uniform and same below the divison. so Total force on the vessel is $$F_{u}=\frac{(\rho gR)(\pi R^2)}{2}$$ $$F_{l}=\frac{(\rho gR)(\pi R^2)}{2}$$ $$F_{total}=F_{u}+F_{l}$$ $$F_{total}=\rho gR(\pi R^2)$$.