I thought of deriving the formula for the surface area of a sphere using integration.
So, below are my calculations:- S.A. of sphere = 2 × S.A. of a hemisphere
Now thinking of each of the hemisphere as being made up of an infinite number of circles having their radii in the range r= R to r=0 and knowing that circumference of a circle = $2 \pi r$, we have$:$ $$S.A. = 2× \int_R^0 2 \pi r \,dr$$ $= 2× ( \pi R^2 - 0) $ $= 2\pi R^2$
Why is my derivation wrong?
You need infinitesimal "ribbon" areas. Now it's ok to construct these areas as the circumfence $2 \pi r$ times their height d$h$. Note that this height is not d$r$ (then the ribbon would lie flat - that's why you get two times the area of a circle with your integration). Since the ribbon is slanted, a simple drawing of two similar triangles shows that
$$ \frac{d h }{d r}= \frac{R}{\sqrt{R^2 - r^2}} $$
So the total surface area, as you already started, is
$$S.A. = 2× \int_{h(r=R)}^{h(r=0)} 2 \pi r \,dh = 2× \int_R^0 2 \pi r \frac{R}{\sqrt{R^2 - r^2}} dr $$
Now you can integrate this, using the substitution $r = R \sin \phi$:
$$S.A. = 2× \int_0^{\pi/2} 2 \pi R^2 \sin \phi \, d \phi = 4 \pi R^2 $$