When you have a region for which Green’s theorem holds, you can quickly get a version of Stoke’s theorem. This approach follows Davis and Snyder [36]. Thus the difficult questions concerning Stoke’s theorem on a two dimensional surface in ℝ^{3} can be reduced completely to Green’s theorem in the plane. I am assuming you have seen some discussion of Stoke’s theorem in multivariable calculus.
Recall from calculus the following definition of the curl of a vector field. Recall also how the cross product of two vectors defined as

is perpendicular to both vectors u,v and how u,v,u × v forms a right handed system. In this formula, it is assumed that the vectors i,j,k pointing respectively in the direction of the positive x axis, y axis and z axis also form a right handed system.
Also recall from calculus that when you have U a bounded region in ℝ^{2}, you can define a parametric surface S as R

R : U → ℝ^{3} has continuous partial derivatives and satisfies

Then when you write ∫ _{S}fdS, you mean

Stoke’s theorem is really just a version of Green’s theorem for a surface in three dimensions rather than a surface in the plane. It relates a line integral around the boundary of the surface to a surface integral like what was just described.
Definition 12.2.1 Let

be a C^{1} vector field defined on an open set, V in ℝ^{3}. Then

This is also called curl

where ε_{ijk} is the permutation symbol and summation is over repeated indices. This symbol is defined as

In words, when you switch two subscripts, the result is −1 times what you had. Note that the above implies that if two indices are equal, then the symbol equals 0. A fundamental identity is available. For

it follows that
 (12.3) 
Note that δ_{ij}x_{j} = x_{i} where it is always understood that summation over repeated indices is taking place.
The following lemma gives the fundamental identity which will be used in the proof of Stoke’s theorem.
Lemma 12.2.2 Let R : U → V ⊆ ℝ^{3} where U is an open subset of ℝ^{2} and V is an open subset of ℝ^{3}. Suppose R is C^{2} and let F be a C^{1} vector field defined in V.
 (12.4) 
Proof: Start with the left side and let x_{i} = R_{i}
The proof of Stoke’s theorem given next follows [36]. First, it is convenient to give a definition.
Definition 12.2.3 A vector valued function, R : U ⊆ ℝ^{m} → ℝ^{n} is said to be in C^{k}
Let U be a region in ℝ^{2} for which Green’s theorem holds. Thus Green’s theorem says that if P,Q are in C^{1}

Here the u and v axes are in the same relation as the x and y axes. That is, the following picture holds. The positive x and u axes both point to the right and the positive y and v axes point up.
Theorem 12.2.4 (Stoke’s Theorem) Let U be any region in ℝ^{2} for which the conclusion of Green’s theorem holds and let R ∈ C^{2}

and dS is the increment of surface area defined as

for dm_{2} indicating 2 dimensional Lebesgue measure discussed earlier.
Proof: Letting C be an oriented part of ∂U having parametrization,

for t ∈


By the assumption that the conclusion of Green’s theorem holds for U, this equals
Note that there is no mention made in the final result that R is C^{2}. Therefore, it is not surprising that versions of this theorem are valid in which this assumption is not present. It is possible to obtain extremely general versions of Stoke’s theorem if you use the Lebesgue integral.