Green Theorem

[mashadar]

Hi ,

i would like a little bit of help regarding that integration please.

line integral of x^2dy over elliptic region x^2/a^2 + y^2/b^2 = 1, first quad.

green's thm is forced.

i'm stuck when i arrive to

2 times double integral of xdA.

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The way i did it, was simply at first to set x=acost dy=bcostdt and integrate that from 0 to pi/2 giving 2a^2b/3, which is correct. But that is not green's theorem, is it ?

 

[galactus]

Yes, that is Green's Theorem.

Note \(\displaystyle A=\frac{1}{2}\oint_{C}-ydx+xdy\)

If we start with \(\displaystyle \underbrace{A=\int\int_{R}dA=\oint_{C}xdy}_{\text{set f(x,y)=0 and g(x,y)=x}} \;\ and \;\ \underbrace{A=\int\int_{R}dA=\oint_{C}(-y)dx}_{\text{set f(x,y)=-y and g(x,y)=0}}\)

Add these two together and we get a third formula. Thus, we have three formulas that express the area A of a region R in terms of a line integral over the boundary:

\(\displaystyle A=\oint_{C}xdy=-\oint_{C}ydx=\frac{1}{2}\oint_{C}-ydx+xdy\)

 

[mashadar]

Alright, just to be sure that we understand eachother, the way i solved it was the following:

line int of x^2dy

x=acost
dy=bcostdt

int from 0 to pi/2 of (a^2cos^2t)(bcost)dt = 2a^2b/3

this respects green's theorem ?

 

[galactus]

Yes

 

[mashadar]

Thanks!