Buoyant Force Issue

I don't understand why the buoyant force is equal to the weight of the displaced water for objects that are not "spheres of fluid." Why does the reasoning for fluid spheres hold for other objects?

• What do you precisely mean by "fluid spheres"? Dec 2 '14 at 3:36
• I mean spheres that are assumed to be composed of the same fluid as the surrounding fluid. Dec 2 '14 at 5:14
• @user11629 Can you explain this a bit more? It isn't entirely clear what you are trying to ask. Dec 2 '14 at 5:20

To see why that statement is true, recall that to each point $\mathbf x$ in the fluid, we can associate a number $P(\mathbf x)$ which gives the pressure in the fluid at that point. This means that the force exerted on a small directed area element $d\mathbf a$ of a surface is $-P(\mathbf x)d\mathbf a$ (where $d\mathbf a$ is taken outward-pointing as is standard). The net force on any surface $S$ is obtained by breaking it up into little area elements, and summing up the forces on all of them. In other words, it's obtained by doing an integral; \begin{align} \mathbf F[S] = -\int_S P(\mathbf x)d\mathbf a. \end{align} This means that when you put an object under water, the force exerted on its surface by the water depends only on the shape of its surface, and where it is in the water, not on what it's made of. In particular, it would be the same if it were composed of water or of wood or any other material having a well-defined surface $S$ for that matter.