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Equation of Fluid Statics

The sum of all forces acting on a fluid element equals the product of its density, volume, and the acceleration due to gravity.
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The statement of the theorem

The Equation of Fluid Statics is derived from the principle of mechanical equilibrium applied to a differential volume element dV\mathrm{d}\mathcal{V} within a continuous fluid domain Ω⊂R3\Omega \subset \mathbb{R}^3 at rest. Let ρ\rho be the fluid density, PP be the thermodynamic pressure, and g⃗\vec{g} be the gravitational acceleration vector. The condition for static equilibrium requires that the net force F⃗net\vec{F}_{net} acting on dV\mathrm{d}\mathcal{V} vanishes.\n\nFormally, the governing equation is expressed as the balance of forces per unit volume:\n\n∇P+ρg⃗=0⃗\nabla P + \rho \vec{g} = \vec{0} \n\nThis vector equation implies three scalar partial differential equations (PDEs) in Cartesian coordinates (x,y,z)(x, y, z): \n\n∂P∂x+ρgx=0  ⟹  ∂P∂x=−ρgx(x-component)\frac{\partial P}{\partial x} + \rho g_x = 0 \quad \implies \quad \frac{\partial P}{\partial x} = -\rho g_x \quad \text{(x-component)}\n∂P∂y+ρgy=0  ⟹  ∂P∂y=−ρgy(y-component)\frac{\partial P}{\partial y} + \rho g_y = 0 \quad \implies \quad \frac{\partial P}{\partial y} = -\rho g_y \quad \text{(y-component)}\n∂P∂z+ρgz=0  ⟹  ∂P∂z=−ρgz(z-component)\frac{\partial P}{\partial z} + \rho g_z = 0 \quad \implies \quad \frac{\partial P}{\partial z} = -\rho g_z \quad \text{(z-component)}\n\nIntegrating these partial derivatives yields the fundamental hydrostatic pressure relation:\n\nP(z)=P0+∫z0zρgzdz′P(z) = P_0 + \int_{z_0}^{z} \rho g_z \mathrm{d}z'
Source: Wikipedia