Archimedes' Principle

Principle
Let $V$ be a compact object with a piecewise smooth boundary, submersed in an incompressible fluid.

Then the net pressure in the vertical direction effected upon the object by the fluid is equal to the weight of the fluid displaced.

This is often quoted (and probably better considered) as the informal statement:


 * A body immersed in a fluid experiences a buoyant force equal to the weight of the displaced fluid.

Proof
Let $V$ be the submerged object, and let $S = \partial V$ be its boundary.

Recall for a smooth vector field $\mathbf F$ defined over $V$ we have the divergence theorem:


 * $(1): \quad \displaystyle \oint_S \mathbf F \cdot \mathbf{dS} = \int_V \nabla \cdot \mathbf F\ dV$

provided that $\partial V$ is piecewise smooth and compact.

The pressure on the surface $S$ depends only on the depth within the fluid, so accounting for atmospheric pressure $p_0$ the force is:
 * $ p = -\rho gz + p_0$

where $\rho$ is the density of the fluid, $g=9.81\ldots$ is the gravitational acceleration and $z$ is the vertical displacement.

Letting $\mathbf F = -p\cdot \mathbf k$ (with $\mathbf k$ a unit vector in the $z$ direction) we see that the left hand side of $(1)$ becomes the buoyancy force acting on the object, for it is the sum over the surface of the $z$ component of the pressure.

Clearly $\nabla \cdot \mathbf F = \rho g$, so we have:


 * $\displaystyle \int_V \nabla \cdot \mathbf F \ \mathrm d V = \rho g \int_V \ \mathrm d V = \rho g V$

where we have let $V$ denote the scalar volume of $V$.

Note that we have assumed incompressibility and thus constant density of the fluid.

This is precisely the weight of the fluid in the volume $V$ Ref? so we are done.