Surface area of a sphere:

Let us consider a sphere of radius r with centre at O.

Let, PQRS be an elementary surface of area \( d\vec{A} \) on the surface of the sphere.

For each point on this area element, r is constant but \( \theta \) and \( \phi \) are variable. So the area element is given by,

\( {d\vec{A}}_r=r^2\ \sin\theta\ d\theta\ d\phi\ \hat{r} \),

the direction of this area is along the direction of \( \hat{r} \).

[To know the proof (CLICK HERE)]

The magnitude of this area element is

\( dA=|{d\vec{A}}_r|= r^2\ \sin\theta\ d\theta\ d\phi \).

In order to calculate the total area of the surface of the sphere, we have to integrate the elementary area \( dA \) with respect to \( \theta \) from 0 to \( \pi \) and with respect to \( \phi \) from 0 to \( 2\pi \).  

\( A=\displaystyle{r^2\int_{0}^{\pi}\sin\theta\ d\theta\ \int_{0}^{2\pi}d\phi} \)

\( or,\ A=\displaystyle{r^2[-\cos\theta] _{0}^{\pi}[\phi] _{0}^{2\pi}} \)

\( or,\ A=r^2[1+1][2\pi]=4\pi{r}^2 \)