Is the circumference included in the area? Let's say we are calculating the area of a circle of radius 1. When we compute its area, are we also including the outermost circumference? Or are we computing everything inside (but not including) the circumference?
 A: The circle boundary has zero area. It is a one-dimensional curve. The area of the disk without the boundary is therefore the same as the area of the disk including the boundary.
Comments have been asking about a hypothetical situation of repeatedly removing zero area parts from the disk. Would the area ever drop?
First note that the disk without the boundary is an open set. There is no "next smaller" circle included in the set; instead you get all ratios up to but not including the original boundary. So a native way of "peeling off" one circle after the other would at that point fail to select the next circle to remove.
If you make a selection that picks some slightly smaller radius, and removes a circle if that radius, then you don't end up with a smaller disk. You end up with a combination of a smaller disk and an annulus that makes up for the difference. The total area will remain the same. If inserted you treat the smaller disk as all that remains, your effectively removed an annulus of perhaps small but definitely non-zero area, reducing the area over all.
You might have a mental picture that says the disk is just the union of infinitely many circles. If all the circles have zero area, how can the disk have non-zero area? Well, intuitively speaking the result of infinity times zero is poorly defined. The moment you get infinitely many point sets of area zero you might get non-zero area as a result.
If you split a set of points and then try to reason about the measure (area, volume etc.) of the parts and their combinations, things can get surprisingly complicated. The Banach-Tarski paradox is a prominent example for this, but way more complicated than the topics discussed so far.
