Question

Find the pressure and force on the curved wall.
Given that the object is filled with liquid.

Answer 1

 see the diagram showing the cross-section (vertical) of the given tank. Width of the tank into the page is 'c'.

Pressure P at a depth h from surface DA = P = ρ g h  (in all directions)
h = R(1 - Cos θ)
Infinitesimal area on the curved surface AEB: ΔS = R dθ * c 
Since Pressure is a scalar and directionless, we can calculate the average. But we cannot total up the values at different heights to get total pressure.

Average pressure on the curved surface =
= Pressure at E (center) = ρ g R

We can find also by integration.
$P_{av}=\frac{1}{TSA} \int \limits_{\theta=0}^{\pi} {P} \, dS = \frac{1}{\pi\ R\ c} \int \limits_{\theta=0}^{\pi} {\rho g R (1-Cos\theta)} \, R d\theta\ c\\\\= \frac{\rho g R}{\pi} \int \limits_{\theta=0}^{\pi} {(1 - Cos\theta)} \, d\theta= \frac{\rho g R}{\pi} [ \theta - Sin\theta ]_{\theta=0}^{\pi}\\\\=\rho g R$

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The horizontal force on the curved surface will be same as the force on a vertical surface like CD kept jus in front of curved surface AEB. 
F_x = ρ g R * (2R c) = 2 ρg R² c.

There is a vertical component of force too, which is vertically upwards.  We find this by integration.
Fy = π/2 * ρ g R² c

Net force F on the curved surface AEB: √ [Fₓ² + Fy² ] 
F = ρ g R² c √(2² + π²/2² ). This acts through point O   (center of circle).

Angle between F and Fₓ : Φ.     Tan Φ = Fy / Fx = π/4


We can find Fy by integration as below.

Take area dS1 above OE at θ from OA, and area dS2 below OE at θ from OB. So θ varies from 0 to π/2.  Numerically dF2 is more as Pressure is more.

 dF1_y = - dF1 Cosθ = - ρ g R² c (1 - Cosθ) cos θ dθ
 dF2_y = + dF2 Cosθ = + ρ g R² c (1 + Cos θ) cos θ dθ

dFy = dF1_y + dF2_y = ρ gR² c (1 + Cos 2θ) dθ

On integration from 0 to π/2, we get Fy = π/2 * ρ g R² c   (upwards).