I agree. Also, to actually calculate the little area element ##d\vec{S}## there are a few ways. Firstly, if you are given the surface as an equation of the form ##f(x,y,z)=0## then
[tex]\nabla (f)[/tex]
evaluated on the surface, will give you a vector normal to the surface, at the point you choose. Of course, you will still need to normalise this vector, and choose which way is 'outwards'. So then you will have a unit vector ##\hat{n}## and so your little area element ##d\vec{S} = \hat{n}dA##
Another nice way, is if you are given general curvilinear coordinates on the surface (call them ##u## and ##v##), and if you know ##\vec{r}(u,v)## i.e. the position in 3d space, as a function of ##u## and ##v##, for all positions which lie on the surface. Then the little area element is:
[tex]d\vec{S} = \left( \frac{\partial \vec{r}}{\partial u} \wedge \frac{\partial \vec{r}}{\partial v} \right) \ du \ dv[/tex]