Paradox about the volume of a cylinder

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?

Where is the error in my previous argument based on infinitesimals?

The error is here:

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°.

If you approximate the cylinder with areas of a finite thickness you can see that these "areas of such rectangles of ... thickness" are not cuboids but triangular prisms.

The volume of a triangular prism however is not $A_\text{rect}\cdot l$ but only $\frac 1 2 A_\text{rect}\cdot l$.

Therefore you have to calculate: $V=\frac 1 2 A_\text{rect}\cdot C_\text{circ}$

... which leads to the correct volume of a cylinder.

The rotation contributes to the volume by $2\pi R$ only for the side of the rectangle that is opposite to its rotation axis. The circumference covered by each point of the rectangle depends on its distance from the rotation axis.

Keeping an infinitesimal approach, think of the rectangle as made out of an infinity of vertical lines of infinitesimal thickness $dr$ at a distance $r$ (with $0 \leq r \leq R$) from the rotation axis. Every vertical line (or infinitesimal rectangle) rotates for $2\pi r$, so its contribution to the volume of the cylinder is $dV = h \cdot dr \cdot 2 \pi r$. Therefore, the volume of the cylinder is the infinite sum of such infinitesimal volumes, i.e. \begin{align} V = \int dV = 2 \pi h \int_0^R r dr = 2 \pi h \left[\frac{r^2}{2} \right]^R_0 = \pi R^2 h. \end{align}

You are assuming that all points follow a circular trajectory of circumference $2\pi R$. But this is wrong, $2\pi r$ goes from $0$ to $2\pi R$ linearly, and its average value over the rectangle is $\pi R$.