Let C be a positively oriented, piecewise smooth, simple closed curve in the plane R2, and let D be the region bounded by C. If L and M are functions of (x, y) defined on an open region containing D and have continuous partial derivatives there, then
Sometimes a small circle is placed on the integral symbol to indicate that the curve C is closed. For positive orientation, an arrow pointing in the counterclockwise direction may be drawn in this circle.
In physics, Green's theorem is mostly used to solve two-dimensional flow integrals, stating that the sum of fluid outflows at any point inside a volume is equal to the total outflow summed about an enclosing area.
The following is a proof of the theorem for the simplified area D, a type I region where C2 and C4 are vertical lines. A similar proof exists for when D is a type II region where C1 and C3 are straight lines.
If it can be shown that
are true, then Green's theorem is proven in the first case.
Define the type I region D as pictured on the right by:
where g1 and g2 are continuous functions on [a, b]. Compute the double integral in (1):
Now compute the line integral in (1). C can be rewritten as the union of four curves: C1, C2, C3, C4.
With C1, use the parametric equations: x = x, y = g1(x), a ≤ x ≤ b. Then
With C3, use the parametric equations: x = x, y = g2(x), a ≤ x ≤ b. Then
The integral over C3 is negated because it goes in the negative direction from b to a, as C is oriented positively (counterclockwise). On C2 and C4, x remains constant, meaning
Combining (3) with (4), we get (1). Similar computations give (2).
To see this, consider the unit normal in the right side of the equation. Since is a vector pointing tangential along a curve, and the curve C is the positively-oriented (i.e. counterclockwise) curve along the boundary, an outward normal would be a vector which points 90° to the right, which would be . The length of this vector is . So .
Now let the components of . Then the right hand side becomes