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Symmetry of Hydrostatic Pressure

Hydrostatic pressure acts equally in all directions at a given depth within a fluid, due to the isotropic nature of fluid stress.
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The statement of the theorem

Consider a fluid domain Ω⊂R3\Omega \subset \mathbb{R}^3 in static equilibrium under a constant gravitational field g⃗=−gk^\vec{g} = -g \hat{\mathbf{k}}. The pressure PP must satisfy the momentum balance equation (Cauchy's equation of motion for a fluid at rest): \n\n∇P=−ρg⃗\nabla P = -\rho \vec{g} \n\nSubstituting the gravitational vector yields:\n\n∇P=ρgk^\nabla P = \rho g \hat{\mathbf{k}} \n\nThis vector equation implies that the pressure PP is independent of the horizontal coordinates xx and yy. Formally, the partial derivatives must vanish:\n\n∂P∂x=0 and ∂P∂y=0\frac{\partial P}{\partial x} = 0 \text{ and } \frac{\partial P}{\partial y} = 0 \n\nConsequently, the pressure PP is a function solely of the vertical coordinate zz (or depth hh): P=P(z)P = P(z). Integrating the differential equation yields the fundamental hydrostatic relation:\n\nP(z)=P0+ρgzP(z) = P_0 + \rho g z