# Find the surface area of a sphere of radius r?

May 12, 2017

It is easier to use Spherical Coordinates, rather than Cylindrical or rectangular coordinates. This solution looks long because I have broken down every step, but it can be computed in just a few lines of calculation

With spherical coordinates, we can define a sphere of radius $r$ by all coordinate points where $0 \le \phi \le \pi$ (Where $\phi$ is the angle measured down from the positive $z$-axis), and $0 \le \theta \le 2 \pi$ (just the same as it would be polar coordinates), and $\rho = r$).

The Jacobian for Spherical Coordinates is given by $J = {\rho}^{2} \sin \phi$

And so we can calculate the surface area of a sphere of radius $r$ using a double integral:

$A = \int {\int}_{R} \setminus \setminus \mathrm{dS} \setminus \setminus \setminus$

where $R = \left\{\left(x , y , z\right) \in {\mathbb{R}}^{3} | {x}^{2} + {y}^{2} + {z}^{2} = {r}^{2}\right\}$

$\therefore A = {\int}_{0}^{\pi} \setminus {\int}_{0}^{2 \pi} \setminus {r}^{2} \sin \phi \setminus d \theta \setminus d \phi$

If we look at the inner integral we have:

${\int}_{0}^{2 \pi} \setminus {r}^{2} \sin \phi \setminus d \theta = {r}^{2} \sin \phi \setminus {\int}_{0}^{2 \pi} \setminus d \theta$
$\text{ } = {r}^{2} \sin \phi {\left[\setminus \theta \setminus\right]}_{0}^{2 \pi}$
$\text{ } = \left({r}^{2} \sin \phi\right) \left(2 \pi - 0\right)$
$\text{ } = 2 \pi {r}^{2} \sin \phi$

So our integral becomes:

$A = {\int}_{0}^{\pi} \setminus 2 \pi {r}^{2} \sin \phi \setminus d \phi$
$\setminus \setminus \setminus = - 2 \pi {r}^{2} {\left\{\cos \phi\right]}_{0}^{\pi}$
$\setminus \setminus \setminus = - 2 \pi {r}^{2} \left(\cos \pi - \cos 0\right)$
$\setminus \setminus \setminus = - 2 \pi {r}^{2} \left(- 1 - 1\right)$
$\setminus \setminus \setminus = - 2 \pi {r}^{2} \left(- 2\right)$
$\setminus \setminus \setminus = 4 \pi {r}^{2} \setminus \setminus \setminus$ QED