Notation in Vector calculus, Stokes' theorem I have a question regarding Stokes' theorem:
$$\oint_c \vec{F} \, d\vec{r} = \iint_S \nabla \times \vec F({r}(u,v)) \cdot d\vec S = \iint_S \nabla \times \vec F({r}(u,v)) \cdot (r_u \times r_v)\, dS$$
In this case I know that the subscript $c$ means a contour region but how about the $S$ and $dS$, what do they mean? 
Is $S$ the region in the $x$-$y$ plane such that it contains the contour region $c$ as its 'border'?
Moreover, can I just think of $dS$ as $dA$?
Thank you.
 A: First, Stokes' Theorem is 
$$\bbox[5px,border:2px solid #C0A000]{\oint_C \vec F\cdot d\vec r=\iint_S \nabla \times \vec F\cdot \hat ndS}$$
where $C$ is a contour that is the boundary of the surface $S$.  Note that $S$ need not be a planar surface.  And note that the contour $C$ actually bounds an infinite number of surfaces.  
Heuristically, think of the surface of a opened "bag" where the opening is the contour $C$.  Yet, the bag could be of a variety of shapes and sizes even though its opening stays fixed.


ASIDE:
  To see the non-uniqueness of $S$, we use the Divergence Theorem
$$\oint_S \vec A\cdot \hat n dS=\int_V \nabla \cdot \vec A dV$$
  and apply this to $\vec A =\nabla \times \vec F$.  Since $\nabla \cdot \nabla \vec F=0$, then 
  $$\oint_S \nabla \times \vec F \cdot \hat n dS=0$$
  This shows that the closed surface integral of $\nabla \times \vec F$ is identically zero and therefore, we see that in Stokes' Theorem, the contour $C$ bounds an infinite number of open surfaces.


The unit vector $\hat n$ points normal to $S$ and its orientation (outward or inward) aligns with the "direction" of the contour's orientation and conforms with the right-hand rule.  
The term $dS$ is simply a differential element of surface such that the surface area of $S$ is given by $\int_S dS$. If we wish to project that surface element onto one of the coordinate planes, say the $x-y$ plane, then the $dS=\frac{dx\,dy}{\cos \gamma}$, where $\cos \gamma=\hat n\cdot \hat z$.  
If we parameterize the surface such that $\vec r(u,v)$ locates a point on $S$, then by using the Jacobian for coordinate transformation we have 
$$\hat n dS=\frac{\partial \vec r}{\partial u}\times \frac{\partial \vec r}{\partial v}du\,dv$$
and therefore
$$\bbox[5px,border:2px solid #C0A000]{\oint_C \vec F\cdot d\vec r=\iint_{S_{u,v}} \left.\left(\nabla \times \vec F\right)\right|_{\vec r=\vec r(u,v)}\cdot \left(\frac{\partial \vec r}{\partial u}\times \frac{\partial \vec r}{\partial v}\right)\,du\,dv}$$
A: Yes, S is the two dimensional surface that c bounds.  dS is the "differential of surface area" on S.  For example, if S is that portion of the paraboloid, $z= 4- x^2+ y^2$ above the xy-plane, its boundary is the circle $x^2+ y^2= 4$.  Writing this as the level surface $F(x,y,z)= z- 4+ x^2+ y^2$, the gradient of 4, $\nabla z- 4+ x^2+ y^2= 2x\vec{i}+ 2y\vec{j}+ \vec{k}$ gives the "vector differential of surface area", $d\vec{S}= (2x\vec{i}+ 2y\vec{j}+ \vec{k})dxdy$.  The length of that vector, $|d\vec{S}|= \sqrt{1+ 4x^2+ 4y^2}dxdy$ is the scalar differential of surface area.
(Notice that, while a bounded two dimensional surface has a unique boundary, a given boundary, c, can be the boundary of an infinite number of different surfaces.  This says that the integral of all those surfaces is the same.)
