# Green's Formula, polar co-ordinates

Printable View

• Nov 4th 2008, 10:56 AM
pkr
Green's Formula, polar co-ordinates
http://i97.photobucket.com/albums/l2...on77/math1.jpg

where
C is the ellipse x^2/a^2 + y^2/b^2 = 1
a,b>0

So basically I get upto II(-2)dxdy, with the elipse as limits, then converting to polar co-ordinates i'm unsure on the limits, but I have a vague idea it is sqrt(ab), can anyone confirm this? Would be appreciated, also can anyone tell me the best way be able to post all the different mathematical symbols
• Nov 4th 2008, 01:25 PM
ThePerfectHacker
Let $\displaystyle \bold{F}(x,y) = (M(x,y),N(x,y))$ where $\displaystyle M(x,y) = x+y$ and $\displaystyle N(x,y) = x - y$.

Green's theorem says $\displaystyle \oint_C \bold{F} = \iint_D \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = \iint_D (1 - 1) = 0$

However, if you had $\displaystyle \bold{F}(x,y) = (N(x,y),M(x,y))$ then:
$\displaystyle \oint_C F = \iint_D 2 = 2\text{area}(D) = 2\pi ab$

Because the area of the ellipse $\displaystyle D$ is given by $\displaystyle \pi ab$.
• Nov 4th 2008, 01:29 PM
pkr
Where you have N(x,y)= x-y, should it not be = y-x

as there was a "-" in the original formula?
• Nov 4th 2008, 01:47 PM
ThePerfectHacker
Quote:

Originally Posted by pkr
Where you have N(x,y)= x-y, should it not be = y-x

as there was a "-" in the original formula?

Yes, I copied down your problem in correctly.
However, all the steps necessary to solve it are still there.