Confusion regarding area element in vector surface integrals I was solving vector surface integrals, and I ran into some problems. Suppose I have some vector $\vec{F}$, and I have to integrate this over a surface $S$.
In general I know that $$I=\int\vec{F}.d\vec{A}=\int\vec{F}.\hat{n}\frac{dxdy}{\hat{n}.\hat{k}}$$
Here $\hat{n}=\frac{\vec{\nabla}S}{|\vec{\nabla} S|}$ is the unit normal to the surface, and $\frac{dxdy}{\hat{n}.\hat{k}}$ is the projection of the surface on the $xy$ plane.
However, in case of surfaces like a circle $x^2+y^2=a^2$ on a plane $z=b$, this above formula doesn't seem to work. In these cases, directly take $d\vec{A}=rdrd\theta \hat{k}$.
However, suppose, we took the surface as $S=x^2+y^2-a^2$
Then we would have : $$\hat{n}=\frac{2x\hat{i}+2y\hat{j}}{\sqrt{4a^2}}$$
However, now we can't do the integral anymore. For example, I can't find the projection on the $xy$ plane, since I would need $\hat{n}.\hat{k}$ which comes out to be $0$. I cannot take projection on any other plane, because then I'd get a $dz$ term in the integral, which is a problem, since, by definition, $z$ is a constant. In these cases, we have to find the surface area as $|r_u\times r_v|dudv \hat{n}$.
In case of spherical surfaces, or the surface of a cone and so on, both the above formulas seem to work. However, the formula seems to fail, when we try it for the surface of a circle on the plane.
Is it because, $z=2$ is the $xy$ plane, and the projection of a bounded surface on the $xy$ plane over itself is meaningless ?
Or is there any special reason why this formula doesn't seem to work in this special cases ? Any help would be highly appreciated.
 A: $ \displaystyle I=\iint_S\vec{F}.d\vec{S}=\iint_R (\vec{F}.\hat{n}) ~\frac{dx~dy}{\hat{n}.\hat{k}}$
works when the surface is given by $z = f(x, y)$ and $R$ is its projection in xy-plane. So it will not work for a cylinder $x^2 + y^2 = a^2$.
Also note $x^2 + y^2 = a^2$ in a plane itself is a curve and if you want to define a disk in plane $z = c$, it is $~x^2 + y^2 \leq a^2, z = c$.
$ \displaystyle \hat{n}=\frac{2x\hat{i}+2y\hat{j}}{\sqrt{4a^2}} ~$ is not the correct unit normal vector to this surface.
Instead as the disk lies in plane $z = c$, its unit normal vector is $(0, 0, 1)$ $[$of course it can be $(0, 0, -1)$ depending on orientation $]$. You can confirm that by parametrizing the surface as below and finding the normal vector,
$\phi(x, y) = (x, y, c), ~x^2 + y^2 \leq a^2~$ or as,
$\phi(r, \theta) = (r \cos\theta, r\sin\theta, c), 0 \leq r \leq a, 0 \leq \theta \leq 2\pi$
Alternatively I suggest that you do not need to find unit normal vector vector to the surface when doing surface integral. Rather if you are projecting surface $z = f(x, y)$ in xy-plane, use the normal vector with $\hat k$ component being $1$ or parametrize the surface as $\phi(u, v)$ and simply consider $(\phi_u \times \phi_v)$ or $(\phi_v \times \phi_u)$. For example if we have a paraboloid surface $z = x^2 + y^2, 0 \leq z \leq a^2$, considering outward normal vector we can write, $\vec n = (2x, 2y, -1)$. Now instead of funding unit normal, we can directly write the surface integral as,
$ \displaystyle I = \iint_{x^2 + y^2 \leq a^2} \vec F \cdot (2x, 2y, -1) ~ dx ~ dy$
