Analysis
Calculating the force on a curved surface on an incline directly via integration is complex. Instead, we consider the equilibrium of the water contained within the hemispherical dent. The dent surface exerts a reaction force on the water, which is equal and opposite to the pressure force we need to find.
Vector Derivation
The water inside the hemisphere is in equilibrium under the action of three forces:
- Weight ($W$): The weight of the water in the hemisphere. $W = mg = \frac{2}{3}\pi r^3 \rho g$. This acts vertically downwards.
- Force from above ($F_p$): The force due to the pressure of the liquid column above the flat circular face of the hemisphere. The area is $A = \pi r^2$. The pressure at the center of this face (approx) is $\rho g (h+r)$. So $F_p \approx \pi r^2 \rho g (h+r)$. This acts perpendicular to the inclined base (at angle $\theta$ to the vertical).
- Reaction Force ($R$): The force from the dent surface on the water.
The force of the liquid on the dent is equal to the magnitude of the resultant of $F_p$ and $W$. Using the law of cosines for vector addition: