Stoke’s theorem is a generalization of Green’s theorem which relates the integral over a surface to the integral around the boundary of the surface. These terms are a little different from what occurs in ℝ^{2}. To describe this, consider a sock. The surface is the sock and its boundary will be the edge of the opening of the sock in which you place your foot. Another way to think of this is to imagine a region in ℝ^{2} of the sort discussed above for Green’s theorem. Suppose it is on a sheet of rubber and the sheet of rubber is stretched in three dimensions. The boundary of the resulting surface is the result of the stretching applied to the boundary of the original region in ℝ^{2}. Here is a picture describing the situation.
Recall the following definition of the curl of a vector field.
Definition 18.3.1 Let

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

This is also called curl
The following lemma gives the fundamental identity which will be used in the proof of Stoke’s theorem.
Lemma 18.3.2 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 .
 (18.3) 
Proof: Start with the left side and let x_{i} = R_{i}
The proof of Stoke’s theorem given next follows [7]. First, it is convenient to give a definition.
Definition 18.3.3 A vector valued function R : U ⊆ ℝ^{m} → ℝ^{n} is said to be in C^{k}
Theorem 18.3.4 (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}

where n is the normal to S defined by

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.