suppose we consider instead a cylindrical closed can
Here's the intuitive answer. I assume everybody on this list
already thought of this, and that the intent of the original
question was to cross-check it with something more mathematical.
This is just to point out that the hand-wavy intuitive answer
can be made completely rigorous.
*) Start with the /open/ can, with water, already spinning,
already in equilibrium.
*) At some chosen depth, insert a thin, neutrally-buoyant,
rigid, porous, horizontal partition. Everything is still
in equilibrium.
*) Close the pores. Everything is still in equilibrium.
The pressure distribution is the same as before.
*) Throw away all the water above the partition.
*) Declare what remains to be a closed can. Conclude that
the pressure distribution is the same as for the corresponding
open can.
*) Not all closed cans will be the same, but each one differs
by at most an additive constant pressure (corresponding to
the depth of the partition below the center of the original
free surface).