Why is $d \vec S = (\frac{\partial r}{\partial u}) \times (\frac{\partial r}{\partial v}) \ du\ dv$ and not $=(\nabla F)\ du \ dv$ on $S$?

Question

I understand the motivation is to construct a vector that points out of the surface at every point, but I am under the assumption that the gradient of a function describing the surface would do this already. At least, in my calculus courses, I'm told that "the gradient is always normal to the surface" and for $d \vec S$ we desire a vector that points normal to the surface at every point. So, naturally, I would think $d\vec S$ is proportional to $\nabla F$, but is instead a cross product of two partial derivatives of sensibly chosen coordinates. This is fine with me too, but I cannot rule out still why $\nabla F$ is not a good idea here.

Answer 1

When the surface is parametrized by $(u,v)$ it is convenient use cross product when we have cartesian form we can of course use gradient vector.

Answer 2

In the first place $\nabla F\, {\rm d}(u,v)$ does not make any sense.

If a surface $S$ is given by an equation $F(x,y,z)=0$ then at any point ${\bf p}\in S$ the surface normal direction is given by $\nabla F({\bf p})$ (assumed $\ne{\bf 0}$). But the length of $\nabla F$ has no geometric meaning, and cannot be used to measure areas: If you replace your equation by $2F(x,y,z)=0$ then $\nabla F$ has doubled, but $S$ has remained the same.

The essential point of a parametric representation is that it provides you with a two-dimensional measure on $S$. If $S$ is given only in the form $F(x,y,z)=0$ you would try to solve this for $z$ and, if you are lucky, obtain $z=g(x,y)$. This then would allow you to present $S$ in the form $$S:\qquad (x,y)\mapsto {\bf f}(x,y):=\bigl(x,y, g(x,y)\bigr)\ .$$ In possession of such a representation (in terms of general parameters $u$, $v$) you then have a scalar or vectorial area measure $${\rm d}\omega=|{\bf f}_u\times {\bf f}_v|\>{\rm d}(u,v),\quad{\rm resp.},\quad {\rm d}\vec\omega=({\bf f}_u\times {\bf f}_v)\>{\rm d}(u,v)\ .$$