Derivation of Archimedes' principle

Question

It is my understanding that upthrust from a liquid on a body is due to pressure difference on the top of the body and the bottom of the body. How, then, is this fact used in order to derive/work out that the upthrust on a body is equal to the weight of fluid displaced? (For example, if a ball bearing is dropped through water then the upthrust is equal to $\frac 43\pi r^3\rho_{water}.g$, where $r$ is the radius of the ball bearing and $\rho_{water}$ is the density of water.)

Answer 1

Imagine your object to be made up of a lot of infinitesimally small "straws" - little cylinders.

Each cylinder has an area $dA$ and a length $\ell$. You know that the volume of such a cylinder is $\ell dA$.

Now look at the pressure difference between the top and bottom of that cylinder: at the bottom, the pressure will be greater by $\rho \ell g$ (where $\rho$ is the density of the liquid, and $g$ is the gravitational acceleration) - that's just the way water pressure works, the pressure exactly supports the weight of the column of liquid above it. The difference in pressure is proportional to the difference in depth, which is $\ell$. So with an area at top and bottom of $dA$, the force on the cylinder is $\rho \ell g dA$ - difference in pressure, times area. But if the volume is $\ell dA$, then the force can be written as

$$F = \rho g V$$

Now if an object is made up of many such cylinders, each experiencing a force equal to the weight of the liquid it displaced, then the entire object will also experience a force equal to the weight of the displaced liquid.