Searching for a surface integral example where I need to normalize the normal vector. I'm having a little confusion with surface integrals normal vector.  I know that $\iint_S\vec F\cdot  d\vec S=\iint_S \vec F \cdot \hat n dS$. 
Upon all of the questions I faced, if S is given by $z=f(x,y)$ I would  just $\vec n = (f_x, f_y, -1)$ (can change directions according to question) and use it like this: $\iint_D\vec F\cdot(f_x,f_y,-1)dxdy$.  
otherwise I would parametrize $S$, lets say by $\vec r(t,s)$, and then find $\vec r_t\times\vec r_s$, and use it again like this to solve $\iint_D \vec F(t,s) \cdot (\vec r_s \times \vec r_t)dtds$. 
But it's always bugging me as of why I don't need to care about $\hat n$? I've solved alot of questions and never needed to normalize a normal vector yet, and I can just by intuition assume that I might be getting the unit normal vector and $dS$ together in my normal vector. 
So I'm trying to find an example maybe where my methods won't really work, if there is, and if not and my methods are correct, I would love to hear some explanations of what's happening and if my assumption is correct. 
Any help is appreciated. Thanks in advance!
 A: Please note in surface integral $\displaystyle \iint_S \vec F \cdot \hat n \ dS$
if we parametrize the surface as $r(u, v)$
$ \displaystyle \hat n = \frac{\vec r_u \times \vec r_v}{|\vec r_u \times \vec r_v|}$
Now from your earlier question about which normal vector to use to find surface area, you know that the mapping of surface area element to the domain of $u, v$ is
$dS = |\vec r_u \times \vec r_v| \ du \ dv$
So we end up with,
$\displaystyle \iint_S \vec F \cdot \hat n \ dS =  \iint_D \vec F(u,v) \cdot \frac{\vec r_u \times \vec r_v}{|\vec r_u \times \vec r_v|} \ |\vec r_u \times \vec r_v| \ du \ dv$
$\displaystyle =  \iint_D \vec F(u,v) \cdot (\vec r_u \times \vec r_v) \ du \ dv$
Take example of a unit sphere centered at the origin and vector field $\vec F = (x, y, z)$. To find flux through the surface,
We note that the unit normal vector $\hat n = (x, y, z)$
So surface integral is $\displaystyle \iint_S (x, y, z) \cdot (x, y, z) \ dS = \iint_S  dS = 4\pi$.
(we know surface area of the unit sphere is $4\pi)$
But most surface integrals would not simplify as above. So if I parametrize the surface as $r(u, v)$ and integrate over $du \ dv$ (or $dx \ dy$), I need to map surface area element to the domain of $u, v$. So we need to multiply unit normal vector with $|r_u \times r_v|$ as per surface area formula and so we end up with $(r_u \times r_v)$.
