TITLE: Why is the area of a circle not $2 \pi R^2$?
QUESTION [6 upvotes]: Let $$ C=\{(x,y,z)\mid x^2+y^2=1,  0 \le z \le 1\} $$
Counter-calculus, my intuition says that the area of a circle with radius $r$ should be $2 \pi \cdot R^2$, because I think that if all radii segments are to be raised perpendicularly to the perimeter of the unit circle (at plane $xy$), then the area is preserved. 
I know I am wrong but I don't (qualitatively) understand why (note I never learnt measure theory).

REPLY [21 votes]: I have created an animation that shows what I think the OP means by "all radii segments are to be raised perpendicularly to the perimeter of the unit circle":

If this is indeed the intended interpretation, then let me make a few observations:

The radii segments, when "raised up perpendicularly to the unit circle", do indeed seem to form a cylinder (with no top or bottom); the lateral surface area of that cylinder is, as the OP says, $2\pi R^2$.
The problem with this visualization can be best understood by imagining each of the radii segments, when in the "upright" position, as being thickened into a rectangular strip.  In the upright position, those strips form a cylindrical surface.  But as they fold down, the strips will overlap one another.  See animation below:



How to prevent the rectangular strips from overlapping one another as they fold down?  Cut each one into a triangular shape with its base on the perimeter of the circle, and then they will fit together nicely as they fold down.  (See animation below.)



Of course, if you cut the rectangular strips into triangles then they will no longer form the cylindrical surface --- each one has had half of its area removed, so the area that remains is just $\pi R^2$.

So at an intuitive level, that's where the factor of $1/2$ comes from:  the need to prevent the infinitesimal rectangular strips from overlapping each other as they fold downwards.