Acrylic Sphere Thickness How do I figure out how thick the walls of an acrylic sphere need to be to withstand water pressure at a certain depth?
 A: Further to Farcher's Answer, which is "correct", I suspect in practice there might be an elastic buckling theory foretelling "critical" implosion pressures lower than those that one would naïvely think would lead to failure grounded on yield stress figures and the like as in Farcher's answer.
The study of implosion buckling, together with original references, can be found in:
C. Farhata, K.G. Wang, A. Main, S. Kyriakides, , L.-H. Leed, K. Ravi-Chandar, T. Belytschko, "Dynamic implosion of underwater cylindrical shells: Experiments and Computations", International Journal of Solids and Structures, 50 2013
This is analogous to the stability analysis that foretells pretty accurately the catastrophic buckling of elastic columns. 
For simplicity, let's think of a 2D problem where a cylinder is being crushed by outside pressure. Analogous with buckling theory, you would use Euler-Bernoulli beam theory to describe possible equilibrium shapes of the cylinder for various pressures. You might begin with a cylinder with a vacuum within for a conservative, easy beginning, then build your theory up to include the effects of pressurized gas within.
What you would find would be that, below a critical pressure depending on the cylinder's Young's modulus and the thickness, the only solution for the cylinder's cross section would be a circle. 
At a critical pressure, a "multifurcation" happens where not only is a circle a possible solution, but other "flower-like" solutions where sinusoidal deviations from circularity show up around the circumference (similar to the "resonance" constraints envisaged for the electron orbits in the Bohr atom): the deviations would be of the form $A\,\sin(n\,\theta) +  B\,\cos(n\,\theta)$ for integral $n$, where $\theta$ is the polar angle measuring the position on the circumference.
The appearance of multiple solutions at critical pressures signals equilibria that can be upset to deform catastrophically into one another. In practice the minimum such pressure is going to be pretty near to the pressure that will implode your cylinder.
A: Imagine a "perfect" spherical shell of radius $r$ and thickness $t$.
For a simple analysis assume $r\gg t$.
Cut the shell in half then the area of the exposed flat surface of the hemispherical shell is approximately $2 \pi r t$.
The force acting on this surface area due to the excess outside pressure $P$ is $P \pi r^2$.
So the compressive stress in the flat surface is $  \dfrac {P\pi r^2}{2 \pi rt}= \dfrac {Pr}{2t}$.
A value of the compressive yield strength of acrylic is given here as 95 MPa.  
The approximation made that there is only one type of stress at work is reasonable if the radius of the shell is at least 10 times the thickness of the shell and of course the analysis assumes that the shell and the material of which it is made does not have any imperfections.
A: It would have a lot to do with the size of the ball and the ratio of it's thickness to radius and the elasticity of material.
It will gradually shrink up to a certain depth depending on how round and even thickness it is. But ultimately it will collapse explosively in a mechanism similar to a column buckling due to unstable balance! If the material is flexible enough to accept a shape like donut the shrinkage will continue till the ball is flattened into a pancake with few random bubbles of air.  
A: You can't calculate it.   A uniform spherical shell will shrink slightly under pressure.  Under great pressure, it'll just shrink MORE.   Unless there's a nonuniformity of
the acrylic, it won't fail by fracture, just get smaller...  
It's possible, for instance, that the first damage done will be by spontaneous
oxidation due to the compressed oxygen in the central void.
There's a useful submarine item, called 'syntactic foam' that uses lots of glass microspheres for reliable flotation.
