I've taken another look at my integral I have set up. Now it seems as if it would work from a mathematical standpoint that for r<R, the separation distance (i.e. little "r" in my integral) becomes equal to R which would give me a constant for the potential and therefore an E value of 0, however I don't see a justification for setting r=R within the integral on physical grounds.
Note: I have realized that I have miswritten the "overall" equation in my second section. I apologize and will correct it now.