Why do hydrostatic fluid exert force on any wall normally with magnitude $ρgh$ always? I understand the derivation in middle of container, but for a container like an inverted cone the particle very near to the wall exerts pressure on wall of magnitude $\rho gh$ normally. I ask “why?” and tried to discover it, but I had some problems.
Just like in normal derivation I assumed that since very small volume of liquid is at rest, it experiences equal and opposite force from both the liquid side and wall side, but from the liquid side in the normal direction to the wall pressure is found to be $2\rho gh$ (integrating all the normal components of pressure), then the same amount of pressure (by normal force) is produced by wall, so that liquid is stationary and due to this a reaction force is developed in wall with $2\rho gh$ magnitude normally...
Where did I go wrong?
 A: 
I assumed that since very small volume of liquid is at rest, it experiences equal and opposite force from both the liquid side and wall side, but from the liquid side in the normal direction to the wall pressure is found to be 2ρgh {integrating all the normal components of pressure},

You do not need to do integration here. Let's draw a free body diagram, and lets assume that the small volume is so small that we don't need to worry about pressure changes from top to bottom but the pressure is $\rho g h$ everywhere.

Now, let's calculate $\vec f_1$, $\vec f_2$, and $\vec f_3$ using $\vec f=P \vec A$ directly since those are in the fluid and we know $P=\rho g h$. Then by Newton's 2nd law $\vec f_0=-(\vec f_1+\vec f_2+\vec f_3)$. So we have $$\vec f_1=\rho g h (0,d^2)$$ $$\vec f_2=\rho g h (d^2, 0)$$ $$\vec f_3=\rho g h \left(0,- d \left[d-d \tan(\theta) \right] \right)$$ and therefore $$\vec f_0=-(\vec f_1+\vec f_2+\vec f_3)=\left(-\rho g h d^2, -\rho g h d^2 \tan(\theta)\right) $$
So $$||\vec f_0||=f_0=\rho g h d^2 \sec(\theta)$$ $$\hat f_0 = \vec f_0/f_0=\left(-\cos(\theta),-\sin(\theta)\right)$$ And since $A_0 = d^2/\cos(\theta)$ we find that the force is directed normal to the area and has a magnitude of $\rho g h A_0$, so $f_0= P \vec A_0$ as expected, even at a slanted wall.
A: Imagine a piston of small area, A inside an empty cylinder, close to its bottom end, which is open. The cylinder is immersed in liquid, at any angle, so that the piston is at depth h and orientated at any angle. When the piston is pushed down the cylinder by a small distance $\Delta x$, let's say a force $F_\bot$ normal to the piston is needed. Thus the work done by the piston on the liquid is $F_\bot \ \Delta x$. But the displaced liquid (volume $A \Delta x$) makes the liquid level rise; effectively the displaced liquid rises by height $h$ and gains gravitational potential energy $mgh = \rho A \ \Delta x \ gh.$ If the fluid is perfectly non-viscous, or the process is very slow (quasi-static), we can equate the work done to this gain in PE. Thus: $$F_\bot \ \Delta x=\rho A \ \Delta x \ gh.$$ Dividing both sides of the equation by $A \Delta x$ and writing $p=\frac{F_\bot}{A}$ we get  $$p=h \rho g.$$
Once you've seen the power of this argument, you'll see that it works anywhere in the liquid, including at a wall of the container. Indeed you could make a small part of the wall the movable piston itself.
A: The expression ρgΔh only gives the pressure change as you go down in a (uniform) fluid.  For any small sample of fluid, the randomly bouncing molecules exert the same pressure on any surface that they encounter. Many pressurized systems (like the brake system in your car) have pressures generated by pumps.  A SCUBA tank may contain air at nearly 2000 lbs/sq.in.
A: Even though very short, this is the most intuitive answer Ive seen to this question: https://physics.stackexchange.com/a/656557/307354
It’s one thing to do the math but to see why is good too. And the author is interesting too.
