Surface area of a sphere with integration of disks Why it is not correct to say that the surface area of a sphere is:
$$
2 \int_{0}^{R} 2\pi r \text{ } dr
$$
In my mind we are summing up the perimeters of disks from $r=0$ to $r=R$, so by 1 integration, we would have $\frac{1}{2}$ of the surface area of the sphere.
I know that it's not correct because that will give us $2\pi R^2$ that it's different from $4\pi R^2$, but why???
Thanks!

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.

