buoyancy formula by volume...

P also factors in somehow

the buoyancy force depends on your depth and the liquid pressure there

Yes, buoyancy force is generally equal to surface(aka area) integral of pressure, and it is very general thing, _any_ force caused by non-viscous liquid is equal to surface integral of pressure. For example, weight of water in bottle is equal to surface integral of pressure of water :-) .

Informal proof:
Now, let we have object placed in water and want to find buoyancy force:

If we have water in place of object, water must remain still.
Imagine rigid bag with shape of object, with infinitely thin walls, filled with water.
Weight of that bag(with water) is equal to surface integral of pressure from inside.
And when that bag is placed into water, at every surface point, internal pressure is equal to external. And therefore weight of bag with water is equal to buoyancy force (but buoyancy force have opposite direction to weight), and equal to that integral.

(I can prove that more formally if you like, but that proof is formal enough.)

EDIT:
I kind of rambled on down below, it's useful information but I didn't directly answer the question:

Quote:

Now the question is: does this volume formula generalize to more complex models than just simple symmetric cubes and alike? I can't come up with a proof or any kind of intuitions on the subject, but I believe the words of experts =)

The problem with that equation is finding the volume of the fluid displaced, especially when the object is not entirely submerged. If you were to implement this in a computer program, the steps would like like this:

1) Determine the plane of water with respect to your object (the water
essentially 'slides' through your object)
2) Compute the volume of the water displaced. This would be no easy task, even with the simplest objects, because objects partially in, say, water are typically oriented in a weird manner.
3) Compute the weight of the water displaced
4) Apply this force at the geometric center of the submerged part of the object.

Step 2 is especially difficult, and lends itself to a numerical solution. In physics for game developers, it gives you methods for computing volumes of meshes using triple vector products, although I'm not sure which method best balances speed and accuracy that you would want.

Now for the rest of my ramble:

Saying the buoyant force is equal to the integral of pressure over the surface is correct, but it boils down to what the OP said:

Quote:

Then it shows, constructively, that for a rectilinear object, this boils down to a buoyancy force formula that depends only on the volume of the object (so that integration through the surface isn't needed).

It's incorrect to say 'integration through the surace isn't needed', because in the strictest sense this is what is really happening. However, in general, it's this equation which is easiest to remember the buoyant force on an object:

Quote:

(that is, F = fluid_density * gravitational_constant * volume)

That equation basically just says that the buoyant force is equal to the weight of the fluid displaced, and it acts at the geometric center of the part of the object submerged in the fluid. For example, bourg's book (I assume you mean physics for game developers) shows how to predict the behavior of ship hulls and submarine hulls based on the object's center of mass versus center of volume submerged in the water.

This works for all fluids, i.e, you currently have a buoyant force acting on you because of atmosphere pressure, you are displacing air, and therefore it is pushing you back up. This is how parachutes work, because they displace a lot of air, about equal to the weight of you and your equipment load.

The reason that there is a net force upward is because of pascal's law which says the pressure at a point is zero. How this works, I don't know to be honest, but it's true and it has been proven. The implication for this, however, is what dmitry mentioned, that the vertical pressure gets greater as you travel downward in the fluid (when you go down in water, your eardrums hurt because of the increase in pressure). When you put a box into water, you've got a pressure at the top of the box pushing down, and you've got a greater pressure at the bottom of the box pushing up (because the pressure increases the further you go down). The pressure at the sides cancel (otherwise a box would move funky side to side when submerged). The pressure difference between the pressure at the top and the pressure at the bottom always yeilds a yet difference equal to the weight of the fluid displaced.

Quote:

Parachutes just slow down the air moving past you during the fall.

Yeah you are right, my mistake! Thanks for pointing that out.

Here's a fun proof of the buoyancy force:
imagine you have a submerged object
now pull it down h meters (infintely slowly :)
you can measure the difference in potential energy of the system from the initial to the final condition:
is is just g*h*(WaterDensity-ObjectDensity)*ObjectVolume (g being the gravitation acceleration)

now, note that the force that counters the pulling down of the object is the buoyancy force (nothing viscous, since we move ideally slowly)

Thereforce the work done by the pulling down force is F (the pulling down force)*d
that leads to
integral(Fdx from x=0 to h)= g*h*(WaterDensity-ObjectDensity)*ObjectVolume
derive that equation and obtain:
F = g*(WaterDensity-ObjectDensity)*ObjectVolume

of couse, in this demonstration, I neglected kinetic energy that can always be kept to a minimum by reducing variations in F

Someone stated that the buoyancy force depends on the depth of the water. This is incorrect, assuming constant water density and volume. This fact changes with a bubble due to variable volume(an interesting and important problem).

Dmytry's explanation is the most elegant, although actually there's a constant that cancels that is related to the pressure, a little bit of hand waving really isn't enough to explain it.

The rigorous proof relies on some results from vector calculus that are probably a bit advanced for this thread and really it's not saying much more. Any simple fluid dynamics maths course should cover it. I think for your purposes Dmytry's explanation is nicest and complete enough.