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If I understand your diagram, your spinning the air inside the bell. So average pressure reaches an equilibrium where it's spinning at some constant rate. We'll assume the water has negligible rotation from its interface from the air (is that a good assumption? Idk). I think the air pressure will be higher the father from the center of rotation, so you get a low point in the center. So water level will go up in the middle and down towards the sides, inverse to what you see when you stir coffee in a cup. If your bell is totally full of air, air will spill out the sides at the bottom, how much dependent on the size of the container and the rotation speed (depth might not matter but I'd have to actually write it out to see)
