Let Green's theorem hold for two adjacent squares. Combine the two squares into one rectangle and apply Green's theorem to the new (larger) shape. The double integrals can be added together, giving an integral that covers the entire rectangle. Add the line integrals can be added as well, creating a path that loops around the right square and then the left. One path proceeds down the common wall and the other path travels up the common wall. These subpaths cancel, leaving the line integral around the rectangle.
Now place a half circle on top of the rectangle and reason as above. Add the double integrals to get an integral that covers the entire area. Add the line integrals and cancel the common border (the top of the rectangle), giving a line integral around the perimeter.
As you can see, one can always break a region up into small shapes, and if Green's theorem holds for the individual shapes it holds for the entire region. It is enough to prove Green's theorem for a few simple shapes: a square, a triangle (two triangles make a square), or a triangle whose hypotenuse has been replaced with a smooth curve. Let's describe Green's theorem and prove it for these simple shapes, whence it is true for all rectifyable simple closed curves.
Let h(x) be a continuous function from 0 to b, thus describing the "hypotenuse" of our simple shape. Assume h is strictly decreasing, falling from its apex ad h(0) down to 0 at x = b. Let c be the counterclockwise curve that runs along the x axis from 0 to b, up the curve to h(0), and back down the y axis to the origin.
Let the vector field f have continuous partials throughout an open set containing this shape. Let p() be the first component of f and let q() be the second component. Think of f as p,q.
We would like to show that the line integral of f around c equals the double integral of the partial of q with respect to x minus the partial of p with respect to y. In other words, the line integral of f around c = the double integral inside c of q∂x-p∂y.
Add two vector functions together, and the individual line integrals around c are added, as are the double integrals inside. If Green's theorem holds for two functions it holds for their sum. Thus we can prove Green's theorem for two simple functions, one where p is set to 0 and one where q is set to 0. Once this is done, add these functions together to resurrect f.
Start by setting q = 0. Fix x and let y run from 0 up to h(x). We are integrating minus the derivative of p with respect to y, and y is also the variable of integration, so invoke the fundamental theorem of calculus to get p(x,0)-p(x,h(x)). Integrate this with respect to x as x runs from 0 to b. The integral of p(x,0) is precisely the line integral of the vector function p,0 along a direct path from the origin to b. Similarly, - the integral of p(x,h(x)) gives the line integral defined by the path that climbs up h(x), our curved hypotenuse. Finally the line integral running down the y axis is 0. Therefore the double integral of -p∂y gives the line integral of p,0 around c.
The proof for the vector function 0,q is almost identical, but the order of integration is reversed. Fix y and let x run from 0 to g(y), where g and h are inverse functions. This gives q(g(y),y)-q(0,y). Integrate this with respect to y and find an expression that equals the line integral of 0,q around c.
Add p,0 + 0,q, and Green's theorem holds for f, running around an through our simple shape. Since it holds for triangles, rectangles, and triangles with curved hypotenuses, it holds for all reasonable closed curves in the plane.
As a corollary, you can find the area inside a curve by calculating a line integral around it. Set q = x/2 and p = -y/2, and the difference in their partials is 1. The double integral of 1 is area, which is the same as the line integral of -y/2,x/2. If the curve is the unit circle, x = cos(t) and y = sin(t), and the aforementioned line integral becomes have of sine squared plus cosine squared, or ½. Integrate from 0 to 2π to get π. Sure enough, π is the area of the circle.
Technical considerations follow; you may skip this if you wish.
Wait a minute! How do you know each partial square is two line segments and a curvy hypotenuse? What if the curve is ever wiggly, at the microscopic level? No matter how fine the checkerboard, there will be some squares where the curve waves in and out, in and out.
This is a good question, and it is difficult to answer with complete rigor. I said the curve had to be "reasonable". In this case reasonable means piecewise smooth. The path can have finitely many sharp corners, where each corner defines a specific nonzero angle, but the sections between the corners are continuously differentiable, and the derivative (speed) is nonzero throughout. These conditions allow us to invoke the tube theorem, which encloses the path in a tube of radius r. Of course r is small, so the walls of the tube do not collide as the path bends around and wiggles in and out. There is always such a tube. Review that theorem and then return here.
Great! You're back. Choose r so that the path is enclosed in a tube of radius r. Furthermore, make sure r is small enough so that the direction of the path doesn't change by more than 1° when traveling through a distance of r. This was discussed in the tube theorem. finally, divide r in half, just to be on the safe site.
Select a checkerboard whose squares measure r on a side. Overlay this checkerboard onto the plane. These are the squares and partial squares we will use to prove Green's theorem, with a few modifications, as described below.
If a square has a path-corner in its interior, it cannot have two such corners, otherwise the walls of the tube would collide. In other words, the squares are smaller than the width of the tube, and one square can never have two sections of path running through it. So we have at most one path corner to eal with.
draw horizontal and vertical lines through this corner, cutting the square into four rectangles. Rectangles are just as good as squares in the above proof. Do this for every corner in the path; then we can concentrate on smooth arcs.
Watch as a continuous arc enters and leaves a square. Assume the path is moving up and to the right. For most entry angles, the path enters at the left or bottom, and leaves at the right or top. As it passes through, the direction of travel is almost constant. We chose r so that the path cannot change direction very much while traveling through a distance of r. It looks like a line as it goes through a single square. It may curve a little, but it cannot bend backwards to exit from whence it came.
If the path enters at the left side, draw a horizontal line from the entry point. If it enters from the bottom, draw a vertical line at the entry point. If it enters at the lower left corner, we don't have to do anything. Do the same for the path's exit point. We have cut the square into rectangles, and one of them has two legs and a curvy hypotenuse, as required by the above proof.
The only troublesome case occurs when the path is practically horizontal or practically vertical. Assume it is practically vertical, without loss of generality. Let c be the path's entry point and let d be the point directly above c. Let e be the paths exit point. Note that e cannot be far from d. Draw a line starting at c, up and to the left, at 45°, until it reaches the top of the neighboring square, at a point we will call b. Drop a perpendicular from b down to a, at the bottom of that square. Now abc is a triangle that satisfies Green's theorem. Next drop a perpendicular from e to the line bc, intersecting bc in f. This produces the triangles bfe and cfe. The former triangle is standard, the latter has a wavy hypotenuse. Both are compatible with Green's theorem. Well I guess that's not obvious, since our earlier triangles were always oriented along the x and y axes, and these aren't. We need to prove Green's theorem again, when the two legs of the triangle are at 45° to the axes. Sorry, I'm not going to do that here. It's pretty much done the same way.
This description has glossed over a lot of nontrivial δ ε algebra, but I think you get the idea. It's amazing how hard it is to prove theorems about arbitrary curves in the plane.