State and prove Green's theorem

Statement

Green's theorem: Let C be a piecewise-smooth, simple closed curve in the xy-plane, traversed counter-clockwise, and let R be the region it encloses. If P(x, y) and Q(x, y) have continuous first partial derivatives in R and on C, then

∮_C (P dx + Q dy) = ∬_R (∂Q/∂x − ∂P/∂y) dx dy

It converts a line integral around the boundary into a double integral over the region.

Proof (for a region simple in both directions)

Take R so that every line parallel to an axis meets C in at most two points.

   y
   |        y = f₂(x)  (upper curve C₂)
   |      ______________
   |     /              \
   |    (   region  R    )
   |     \______________/
   |        y = f₁(x)  (lower curve C₁)
   |
   +---a-----------------b------> x

Part 1: ∮ P dx = − ∬ (∂P/∂y) dx dy. Describe R as a ≤ x ≤ b, f₁(x) ≤ y ≤ f₂(x). Then

∬_R ∂P/∂y dx dy = ∫ₐᵇ [ P(x, y) ] from y=f₁(x) to y=f₂(x) dx
                = ∫ₐᵇ P(x, f₂(x)) dx − ∫ₐᵇ P(x, f₁(x)) dx

Now evaluate ∮ P dx along C counter-clockwise. The lower curve C₁ is traced from x = a to x = b; the upper curve C₂ is traced from x = b to x = a; any vertical parts have dx = 0 and contribute nothing:

∮_C P dx = ∫ₐᵇ P(x, f₁(x)) dx + ∫_b^a P(x, f₂(x)) dx
         = ∫ₐᵇ P(x, f₁(x)) dx − ∫ₐᵇ P(x, f₂(x)) dx
         = − ∬_R ∂P/∂y dx dy                              ... (1)

Part 2: ∮ Q dy = ∬ (∂Q/∂x) dx dy. Describe R also as c ≤ y ≤ d, g₁(y) ≤ x ≤ g₂(y), where x = g₁(y) is the left curve and x = g₂(y) is the right curve.

∬_R ∂Q/∂x dx dy = ∫_c^d [ Q(g₂(y), y) − Q(g₁(y), y) ] dy

Counter-clockwise, the right curve is traced upward (y from c to d) and the left curve downward; horizontal parts have dy = 0:

∮_C Q dy = ∫_c^d Q(g₂(y), y) dy + ∫_d^c Q(g₁(y), y) dy
         = ∫_c^d Q(g₂(y), y) dy − ∫_c^d Q(g₁(y), y) dy
         = ∬_R ∂Q/∂x dx dy                                ... (2)

Adding (1) and (2):

∮_C (P dx + Q dy) = ∬_R (∂Q/∂x − ∂P/∂y) dx dy      ∎

Extension

A region that is not simple in this sense is cut into finitely many simple sub-regions by lines parallel to the axes. Applying the theorem to each and adding, the line integrals along the common cuts cancel (they are traversed in opposite directions), so the theorem holds for the whole region. The same argument works for a region with holes, taking the outer boundary counter-clockwise and the inner boundaries clockwise.