TITLE: Paradox about the volume of a cylinder
QUESTION [22 upvotes]: Trying to apply Cavalieri's method of indivisibles to calculate the volume of a cylinder with radius $R$ and height $h$, I get the following paradoxical argument.

A cylinder with radius $R$ and height $h$ can be seen as a solid obtained by rotating a rectangle with height $h$ and base $R$ about its height. Therefore, the volume of the cylinder can be thought as made out of an infinity of areas of such rectangles of infinitesimal thickness rotated for $360^\circ$; hence, the volume $V$ of the cylinder should the area of the rectangle $A_\text{rect} = R \cdot h$ multiplied by the circumference of the rotation circle $C_\text{circ} = 2\pi R$:
  \begin{align}
V = A_\text{rect} \cdot C_\text{circ} = 2 \pi R^2 \cdot h
\end{align}

Of course, the right volume of a cylinder with radius $R$ and height $h$ is 
\begin{align}
V = A_\text{circ} \cdot h = \pi R^2 \cdot h
\end{align}
where $A_\text{circ} = \pi R^2$ is the area of the base circle of the cylinder. 
Question: Where is the error in my previous argument based on infinitesimals?

REPLY [2 votes]: Well, first of all, the infinitesimals don't match up. You have rectangles of thickness of, say, $dw$, and to get from one to the next, you're rotating through an angle of $d\theta$. The second problem is that these volumes are not distinct. If you take a slab with height $h$ and base $R$ with thickness $dw$ and rotate it by an angle of $d \theta$, you will end up with a volume that intersects the volume that it occupied before the rotation. Because of this "double counting" (note that not all points are double-counted; but on average you're counting each point twice), you calculate twice the volume you should.
You can see similar issue when calculating the area of a circle by sweeping a rectangle with length $R$ and thickness $dw$ through the circle. If you rotate it in small enough angles that every point on the circumference is covered by a rectangle, then points in the interior will be covered by multiple rectangles.