This is a very difficult theorem to do in ultimate generality. To do that will require a consideration of the Jordan curve theorem from topology. However, it is not too hard to consider a fairly general version which works well for most things of interest. Green’s theorem concerns an open bounded set in the plane, U. This theorem relates a line integral around ∂U with an area integral on U. Of course it is necessary to specify an orientation for the bounding curve which we will assume initially is a smooth curve. So how does one specify an orientation for this curve? As discussed above, this is equivalent to specifying a direction of motion around the curve and it is a highly non trivial consideration notwithstanding the simple picture we like to draw to illustrate the concept. However, one can give a good meaning to this for open sets for which the divergence theorem holds as described above. Recall that for these open sets, there is indeed a well defined outer unit normal.
Suppose U is as described in the proof of the divergence theorem above. Let n be the unit outer normal on ∂U. Let k be the normal to the plane in the direction of the positive z axis where i, j, k form a right handed system, i being in the direction of the positive x axis and j in the direction of the positive y axis as shown. Then at a point of the boundary, the direction of motion is specified in the following way. Letting the fingers of the right hand point in the direction of motion, when they are curled toward the vector k, the thumb points in the direction of n, the exterior normal which is given thanks to the divergence theorem. This defines from n a direction of motion around the curve, T. Assume the parameterization is such that T is parallel to
What is n in terms of T and k? From the geometric description of the cross product, T = k × n. From the geometric description of the cross product, this implies n = T × k and so n is parallel to the vector
Theorem 13.1.1 (Green’s Theorem) Let U be an open set in the plane whose boundary ∂U is a smooth curve and let F
 (13.1) 
Proof: As illustrated in the above picture, the unit exterior normal is a multiple of
Now let F

Note that this is completely general as long as U is one of those open sets for which the divergence theorem holds in which there were the sets Q_{i} as described earlier. Thus Green’s theorem will hold for many regions in the plane provided they are bounded by a C^{1} curve as described earlier. However, this theorem also holds for the situation where the region is bounded by a piecewise smooth curve. This follows from the generalization mentioned in Section 11.2 for the divergence theorem in which the boundary is Lipschitz and has derivatives off a suitable closed subset of the boundary. The most general version of the theorem concerns a rectifiable Jordan curve. Also, you can cut out holes and paste together regions for which Green’s theorem holds and get another region for which it holds.
Suppose the theorem is valid for U_{1},U_{2},


because if Γ = ∂U_{k} ∩∂U_{j}, then its orientation as a part of ∂U_{k} is opposite to its orientation as a part of ∂U_{j} and consequently the line integrals over Γ will cancel, points of Γ also not being in ∂U. As an illustration, consider the following picture for two such U_{k}.
Similarly, if U ⊆ V and if also ∂U ⊆ V and both U and V are open sets for which 13.1 holds, then the open set V ∖
Then


which equals

where ∂V is oriented as shown in the picture. (If you walk around the region V ∖U with the area on the left, you get the indicated orientation for this curve.)