I must verify the divergence theorem in the following case :

$\displaystyle F(x,y,z)=x \vec i + y \vec j + z \vec k$. And $\displaystyle S$ is the surface of the sphere $\displaystyle x^2+y^2+z^2=4$.

My attempt :

$\displaystyle \iint _S F \bold n dS=\iiint _S div FdS$.

I calculate the right hand side : $\displaystyle div F=3$.

So $\displaystyle \iiint _S div FdS =$ 3 times the volume of a sphere with radius $\displaystyle 2$, hence the triple integral is worth $\displaystyle 32 \pi$.

Now comes the problems. I ain't able to calculate $\displaystyle \iint _S F \bold n dS$.

A normal unit vector to the sphere is $\displaystyle \bold n =\frac{2x \vec i+2y \vec j +2z \vec k}{2\sqrt 3}$.

So the double integral is worth $\displaystyle \frac{1}{2 \sqrt 3} \iint _S (x \vec i + y \vec j + z \vec k)(2x \vec i+2y \vec j +2z \vec k) dS=\frac{1}{2\sqrt 3} \iint _S 2x^2+2y^2+2z^2 dS$. And I'm stuck here because I have 3 variables ($\displaystyle x$, $\displaystyle y$ and $\displaystyle z$) but a double integral instead of a triple one.

I know I could solve the problem faster using spherical coordinates, but for the sake of my comprehension let's do it via Cartesian coordinates please.

Do you have any idea how could I continue, in order to reach the $\displaystyle 32 \pi$?