When considering the surface $S: x^2+y^2+z^2 = R^2$ we know that the surface integral
$$ \iint_S dS = 4\pi R^2$$ Since this is the area of a sphere, but while using surface integral I know that the $dS$ is the norm of the normal vector of the surface $n = (2x,2y,2z)$, $||n|| = \sqrt{4(x^2+y^2+z^2)} = 2R$, over the area of the circle it is on top of, on which it follows
$$\iint_SdS = \iint\limits_{x^2+y^2\leq R} 2RdA = 2R\cdot(\pi R^2) = 2\pi R^3$$
And I dont know why the result is diferent on an $R$ factor. The $\times 2$ might be cause the sphere is not a function and has to be split on 2 equal parts, but I dont know how I ended up with an extra $R$ term.
Best Answer
The formula you are referring to, where you integrate the magnitude of the normal vector, is dependent on the parameterization of the surface that you choose. You didn't specify a parameterization, which is maybe the source of your confusion.
In general, if a surface is given by $\vec{r}(u,v) = (x(u,v),y(u,v),z(u,v))$, with partial derivative vectors $\vec{r}_u = (x_u,y_u,z_u)$ and $\vec{r}_v=(x_v,y_v,z_v)$, then the normal vector you want to use is the cross product $\vec{n} = \vec{r}_u \times \vec{r}_v$. Notice that any re-scaling of a normal vector is also a normal vector, so it matters which particular normal vector you use.
There are many ways of parameterizing the sphere. One way is using "graph coordinates", where the parameterization is $\vec{r}(x,y) = (x,y,\sqrt{R^2-x^2-y^2})$. In this case, the normal vector from the formula above has length $$ \| \vec{n} \| = \frac{R}{\sqrt{R^2-x^2-y^2}} $$ Integrate this over the disc of radius $R$ (use polar coordinates!) to get $2\pi R^2$. This only gives half the sphere, so double it to get the correct answer.
Another convenient way is to parameterize using spherical coordinates. The parameterization would be $\vec{r}(\theta,\phi) = (R\sin(\phi)\cos(\theta), \, R\sin(\phi)\sin(\theta), \, R\cos(\phi))$. In this case, the magnitude of the normal vector would be $$ \| n \| = R^2 \sin(\phi) $$
Integrate this over the domain $0 \leq \theta \leq 2\pi$ and $0 \leq \phi \leq \pi$ to cover the whole sphere, and it gives the desired answer.