A balloon under the ocean So everybody is familiar with how buoyancy works in theory. However, if I sink a balloon filled with air underwater, the pressure of the water will compress the air inside it, reducing the volume of the balloon and the volume of displaced water, therefore reducing the buoyant force with increased depth.
From this I would understand that using a flexible container, buoyant force is not in fact linear with depth.
Am I correct?
 A: For a balloon that is held just under the surface, the pressure is approximately atmospheric, the weight of the displaced water is equal to the density of water multiplied by the volume of the balloon, and the buoyant force is equal to the weight of the displaced water.  If that balloon is pushed to a depth of approximately 10 meters, the ambient absolute pressure is 2 atmospheres, the volume of the balloon is half of what it was at the surface, and the buoyant force is one half of what it was originally.  Continuing, it is seen that at a depth of 90 meters, the absolute ambient pressure is 10 atmospheres, the balloon volume is 1/10 of its original volume, and the buoyant force is 1/10 of its original value.  This means that the buoyant force on the balloon is inversely proportional to depth.
A: Yes, you are correct. The buoyant force is given by
$B = \rho_f V_{\text{disp}}g$
where $\rho_f$ is the density of the fluid, $V_{\text{disp}}$ is the volume displaced, and $g$ is the gravitational constant (if you are for some reason dealing with a huge change in altitude, this might not be constant, of course). Note that we are using the density of the fluid here, which may change with depth. In general, we want to talk about things like the compressibility, the way the (relative) volume of a fluid or solid changes in response to pressure. This is expressed as
$\beta = -\frac{1}{V}\frac{\partial V}{\partial p}$
For water, this is typically around $4.6\times 10^{-10} \text{Pa}^{-1}$, which generally is on the same order of magnitude as rocks and mercury (water is mostly incompressible)
On the other hand, when talking about gasses, we can generally use the ideal gas law, $PV=nRT$ to see that the change in volume is inversely proportional to the change in pressure. There is, of course, a compressibility factor associated with certain gasses which tells us how much they deviate from this rule, but for room temperature, air actually behaves very close to an ideal gas up to about $250 \text{bar}$, and even up to $500 \text{bar}$ the approximation isn't bad.
I realize this is more detail than the simple "yes" your question required, but yes, the balloon filled with air will certainly compress more than the water around it, so the volume displaced will change more than pressure. To first order, this is a linear relationship, since the water density will not really change much at all while the ideal gas law tells us the balloon volume will change with $V\propto \frac{1}{P}$.
A: You are right, here is some background.
A submarine made of steel contains air at atmospheric pressure. As it sinks deeper, the water pressure outside compresses the hull, decreasing its volume and thereby decreasing its buoyancy- which causes it to sink faster, which compresses the hull more, etc., etc. and it sinks still faster- until it reaches its crush depth, at which it implodes violently. As such, a sub trimmed for neutral buoyancy is in an unstable equilibrium- which requires the submariners to be constantly vigilant in monitoring their depth and adjusting their buoyancy accordingly.
The first bathyscaphe to reach the bottom of the Challenger Deep used a "balloon" filled with gasoline to generate buoyancy, since gasoline is less dense than water. Since it is incompressible as water is, it did not progressively lose buoyancy as it sank, and the bathyscaphe could readily return to the surface by dropping ballast.
