[Hunt]

Since I have posted this elsewhere, I thought I would give it a try here too ... perhaps someone would enlighten me. Green's theorem  is widely known and considered as much as important as the fundamental theorem of calculus. The theorem simply converts line integrals to double integrals.

We can obtain Green's theorem area formula by going backwards :

A = ? ? dydx = ? ?  ( 1/2 + 1/2 ) dydx = 1/2 ? ( xdy - ydx ) , this is given as a definition in most cases.

Ofcourse this holds if the domain is closed.( a closed ring or path or region or whatever ... )

But it seems sometimes we can compute the area directly from A = ?  xdy

Example:

Determine the area between y=x² and y = x  across C⁺

A = 1/2 ( ? xdy - ydx )

I divided the line integral into  arc OA and a straight line OA where A (1,1), the point of intersection of the parabola and the straight line,  and O is the origin ofcourse. Plotting the parabola and the straight line in an xy-plane would make this a lot easier.

I chose the parametirc equations,

For [OA] : y = x , x=x      0 ? x ? 1
For arc OA : x=y² , y=y     1 ? y ? 0

I₁ = 1/2 ? (x-x)dx = 0 ----> for [OA]

I₂ = 1/2 ? (y² - 2y²)dy = 1/6 ---> for arc OA

A = I₁ + I₂ = 1/6

Now notice that the area could have been calculated directly from:

A = ? xdy ( line integral across C⁺ ) = ? xdy ( across [OA] ) + ? xdy ( across arc OA )

Using the same variables, A =  ? xdx ( 0 ? x ? 1 ) + ? y²dy ( 1 ? y ? 0 ) = 1/6

The interesting part is that A = ?  xdy gives the same answer for 1/2 ? ( xdy - ydx ) in many cases , but sometimes it doesn't. I can't seem to figure out when it does not apply ...

[Yggdrasil]

The usefulness of Green's theorem comes because it will work for functions which are not one-to-one.  For example, if you have a curve which cannot be expressed in explicit form as y=f(x), Green's theorem will be useful for calculating the area enclosed by the curve.

[Hunt]

The usefulness of Green's theorem comes because it will work for functions which are not one-to-one.  For example, if you have a curve which cannot be expressed in explicit form as y=f(x), Green's theorem will be useful for calculating the area enclosed by the curve

Indeed, but do you have any idea why the area formula for green's theorem can be simplified as above? I'm sure it's not by coincidence.

[FeLiXe]

the only thing to consider is that you have to go in a continuous curve and that the area is always on your left. (if the area is always on your right you get the same result with a minus in front)

besides that, the rule generally applies. I don't know how to prove it but I don't think my math professor lied to me 

[Hunt]

Yeah I gave up too ... perhaps it has something to do with advanced calculus.

[Donaldson Tan]

Green's theorem assumes the area is transversed.

[Hunt]

The formula can be expressed not only as A = ∫xdy , but also as A = -∫ydx , both cases would get the same result for a specific area. I read it today in my calculus book in some problem, and ironically, the question required a proof for this.