Archimedes' Principle

Principle
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 vector field $\mathbf F$ defined over $V$ we have the divergence theorem


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

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 Reference for this eqn?.

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\ dV = \rho g \int_V\ dV = \rho g V $

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

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