Gradient of surface area element $\nabla . dS$

When finding a surface integral you first must parametrise the surface into $r = r(u,v)$. So the area element is a function of $u$ and $v$. Does that mean that grad of it is zero? That is, is $$\nabla . dS = 0?$$

• We compute the gradient of functions of real functions $\mathbb{R}^n \to \mathbb{R}$, not of vector functions $\mathbb{R}^n \to \mathbb{R}^m$. – Mark Fantini Aug 13 '15 at 10:57
• @Thanks: Could you please expand the reasoning behind "is the grad of [the area element] zero [because] the area element is a function of $u$ and $v$"? As Mark Fantini notes, the "usual" gradient is defined for a real-valued function of several variables while an area element is something else (containing differentials, and perhaps vector-valued depending on context), but it seems there's something deeper in your question. – Andrew D. Hwang Aug 13 '15 at 11:20