Green's theorem.

1. Feb 9, 2005

semidevil

I'm giving a presentation on Green's theorm for class, and someone gave me this article that pretty much tells you how green's theorem work. It starts with 2 squares, and then you combine to square to form a rectangle, and then when you add the double integeral, line integeral, and the paths cancel, leaving the integeral around the rectangle.

it has a lot of descriptions, such as if I draw a half circle on top of the square, and I add the integeral, it will cancel, and again, give the perimeter.

so basically it concludes that if greens theorem holds for individual shapes(square, triangel, triangle w/ curvy hypotnuse), then it will hold for entire region.

then it goes to prove for the shapes.

I can't find a site that actually does all the proof in a "proof like" manner. and for my presentation, I would also like to draw the diagrams, but I dont want to mess up. I would rather quote from someone rather then do it myself....does anyone have any sites or know any sites that have a better visual proof of this for green's theorem?

2. Feb 9, 2005

Sirus

Not sure if this is what you're looking for, but http://tutorial.math.lamar.edu/AllBrowsers/2415/GreensTheorem.asp [Broken] is a site with a couple problems including specific shapes. Be sure to consult google, as well, if you already haven't.

Last edited by a moderator: May 1, 2017
3. Feb 9, 2005

mathwonk

i myself have been perfectly happy with greens theorem after doing the proof for a rectangle. that is the main case.

Of course I also know how to change variables in integration, so it follows that if greens theorem is true for a rectangle it is also true for any smooth image of a rectangle, which is pretty much anything. that is a lot more efficient than chopping up figures.

the point is that if H :R-->S is a reasonable map from a rectangle R to a region S, and if Q is a one form in S, and if ∂R is the boundary of R, then ∂(H(R)) = H(∂R), and

d(H*(Q)) = H*(dQ), and then the integral of dQ over H(R) equals the integral of

H*(dQ) over R, which equals the integral of d(H*(Q)) over R, which by green in the

rectangle R, equals the integral of H*(Q) over ∂R, which equals the integral of Q over H(∂R) = ∂(H(R)).

I.e. finally we get that the integral of Q over ∂(H(R)) = the integral of dQ over H(R).

thats green for the image set H(R).

QED.

4. Feb 13, 2005

xorbie

I think it's actually easy enough to prove the theorem for any general, nice convex set (a convex set is just a set in which the line between any two points is also in the set... so this holds for any nice geometrical shape like an n-gon, a circle, etc). This is because you can state that the set is bound by x in [a,b] and y in [f(a), f(b)] for some f. Then the math works out pretty nice, I think.

It's also clear enough if you just draw a picture that if you add two convex sets to make a set that isn't convex (say making a heart by adding the two left and right halfs), Green's theorem still works because the line integrals cancel out on the line that joins the two sets (in this case, the vertical line down the middle of the heart).

I'm not sure how much this helps.

5. Feb 13, 2005

mathwonk

this was his original approach. the point of my post was that actually the extra generality obtained by these tricks is illusory, as should be clear by the factr that the arguments are so trivial.

The rectangle case is indeed the general case, if you understand how integrals change variables under transformations.

this is an example by the way of the contrast and yet the intimate relation between the books of courant and of spivak. Courant explains this idea clearly but somewhat intuitively in his second volume. Many years later, Spivak gives a very precise and technically correct proof using "chains" in his little "calculus on manifolds", which is however exactly the same proof in essence.