A sphere is obtained when the region between $x=-r,x=r$ of the circle $x^2+y^2=r^2 $ in the first two quadrents is rotated about the $x-axis$

Consider the section by the sphere obtained by rotating the hashed area between $x=a,x=b( b >a) $

The surface area $S=2 \pi \int \limits_a^b y \sqrt {1+\bigg(\large\frac{dy}{dx}\bigg)^2}dx$

Now $x^2 +y^2 =r^2 $

$\therefore 2x+2y \large\frac{dy}{dx}$$=0=>\large\frac{dy}{dx}=\frac{-x}{y}$

$ 1+\bigg(\large\frac{dx}{dy}\bigg)^2 $$=1+\large\frac{x^2}{y^2}$

$\qquad= \large\frac{x^2+y^2}{y^2}=\frac{r^2}{r^2-x^2}$

$y \sqrt {1+\bigg(\large\frac{dy}{dx}\bigg)^2}=\sqrt {r^2-x^2}. \large\frac{r}{\sqrt {r^2-x^2}}$$=r$

$\therefore S=2 \pi \int \limits_a^b r dx=2 \pi r (x)_a^b$

$\qquad= 2 \pi r (b-a) \;sq.units $

The surface of the sphere is obtained by substituting $a=-r,b=r$

$\therefore S= 2 \pi r (2r) =4 \pi r^2 sq.units$