Question 6.3.1: Figure 1 shows an arbitrary infinitesimal triangular prism o...

Goal Prove for a liquid at rest that the pressure at an arbitrary point is the same in all directions.
Given The fluid is at rest; the liquid prism is taken from an arbitrary point; the dimensions and angle θ are as shown in Figure 1.
Assume The prism represents a particle since it is infinitesimal and the particle is in equilibrium because it is at rest. The weight of the particle can be ignored. The pressure on each of the faces is different (p_{1} ,p_{2},p_{3},p_{4}) and the pressure on any face acts perpendicular to that face.
Formulate Equations and Solve Because the particle is in equilibrium we know the forces must sum to zero. Since the pressure must be perpendicular to any surface, the only two surfaces with pressure in the x direction are 1 and 3. Based on equilibrium equation (5.3A) we write:

\sum{F_{x} } =0                                          (5.3A)
\sum{F_{x} } =0\left(\rightarrow + \right) =p_{1} dy  dz-p_{3}\left(ds \sin \theta \right)dz
p_{1}  dy = p_{3}   ds  \sin \theta                                        (1)

The only two surfaces with pressure in the y direction are 2 and 3. Based on equilibrium equation (5.3B) we write:

\sum{F_{y} } =0                                    (5.3B)
\sum{F_{y} }= 0\left(\uparrow + \right) =p_{2}  dx  dz  –  p_{3}  ds  dz\cos \theta
p_{2}  dx =p_{3}  ds \cos \theta                                                          (2)

From Figure 1 we see that

ds  \cos \theta = dx                                (3)
ds  \sin \theta = dy                                (4)

Substituting (4) into (1) and (3) into (2) we get

p_{1}  dy = p_{3}  dy
p_{2}  dx = p_{3}  dx
p_{1}= p_{3} = p_{2}

If we were to sum the forces in the z direction the analysis would show that

p_{3} = p_{4}

Combining our results we can conclude thatp_{1}= p_{2} = p_{3}=p_{4}