Yes, it can be done, though I don't have a working TI-83 to check on. I don't think I've done this in polar form, but it should work.
There should be an example on Piecewise Functions in the manual (under Activities); basically, you multiply the function by inequalities, which evaluate to 0 when they are invalid. In this case, it should be r = ... (θ>π)(θ<2π). That will include r=0 in the plot, since the function returns 0 outside the domain; if you don't want that, I think you can use r = ... /(θ>π)/(θ<2π), which is undefined outside the domain.