As I saw I probably won't be getting an answer I took a more naive path. I sampled the blue square by 100 points (10x10) and just approximated the overlapping areas using the number of points contained by the other squares. It proved to be precise enough, and yielded pretty good results for my task.
I know it's not a real mathematical solution to the problem, but maybe it will help other people facing the same task.
The volume of any prismatic shape (“stacks” of uniform shape and size) is very simply the sum of the volumes of those little shapes. This works for any (integer, Euclidean) dimension, too. We all know the area of a rectangle: but why is it true? Well, if you subdivide the sides of the rectangle into units (say centimetres) then you can very easily see that the rectangle consists of a certain number of unit centimetre squares, and it all fits, and you can count how many such squares there are by multiplying the side lengths. Then you have $x$ centimetres square. You can’t really do any more proving than that; this is a definition of area - how many unit areas (in this case, unit centimetre squares) cover the shape?
Back to the cylinder now. It really is exactly the same principle. A fundamental definition of volume (in this normal, Euclidean space) would just be: how many unit volumes fill up my shape? And a unit volume can be bootstrapped from the unit area: if you have a square, and add a perpendicular height of a certain length, you can do the subdivision again and see clearly that we have $a\times b\times c$ unit cubes, where $a,b,c$ are side lengths- and then we just define the volume this way. The cylinder consists of a large stack of discs. How thin these discs are and how many discs there are depends on your interpretation, but it is intuitively good to subdivide the cylinder into stacks of unit height. Then the height of the whole cylinder is how many discs there are, and you can say that the volume of the cylinder is height unit discs. But “unit disc” is a clumsy measure of volume - so what’s the volume of a unit disc? Sort of by definition of “unit”, it is just the base area times $1$ units cubed, and again you can’t really push the issue more than that: this is a very natural definition of volume.
Altogether now, we have volume equals $h$ unit discs, and a unit disc has volume $b$, $b$ for base area, $h$ for height, so... $V=b\times h$. The only thing to be wary of is to make sure all your units agree in the same measurement system!
About your issue with the base being circular: this is not actually a problem, so long as you accept the area of a circle. A disc of a unit height is just pulling the base circle upwards, a unit distance, and has volume $1\times b$. If you want an intuition or proof of why the area of a circle is what it is, and why it is even measurable, I’m afraid you probably will have to do some calculus or calculus-type ideas - I believe that’s how the Greeks did it!
Best Answer
Given a $4$m $\times$ $4$m square, sure, I can determine its area by counting unit squares.
However, a 16m$^2$ circle's area is not measured by counting little squares, nor does its derivation pertain to little squares.
The fact that area always has dimension length$^2$ (m$^2$) does not imply that area is generally determined by counting little squares (or even rectangles).
Area formulae are not pre-tuned to count squares (or even rectangles).
Similarly, distance has length (m) as dimension, yet (1) it is valid to measure the length of a curved road by connecting short pieces of non-straight rope; (2) I can use a watch (time) to determine that I've driven 100 miles.