Surface area of a sphere with integration of disks

In order to give a short explanation of the mistake :

In writing $\quad 2 \int_{0}^{R} 2\pi r \text{ } dr \quad$ you sum elementary areas $\quad dA=2\pi\:r\:dr\quad$

that is to say, a strip length$=2\pi r$ and width$=dr$.

The hitch is that the width is not $dr$ but is $ds$ (see the figure).

With a few calculus, you can find $\quad ds=\frac{R}{\sqrt{R^2-r^2}}dr \quad\to\quad dA=2\pi\:r\frac{R}{\sqrt{R^2-r^2}}dr $

$$\quad 2 \int_{0}^{R} 2\pi\:r\frac{R}{\sqrt{R^2-r^2}}dr =4\pi R^2$$

This is a method to find the area of the sphere. Of course, a simpler method consists in doing the job in spherical coordinates instead of Cartesian coordinates.