Green's theorem- integral over an ellipse

aylwin
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Homework Statement


Use Green's theorem to find the integral ∫C (y^2dx+xdy) when C is the following curve (taken counterclockwise): the ellipse x^2/a^2 + y^2/b^2 =1.

Homework Equations


Green's theorem: ∫C Mdx+Ndy = ∫∫R (∂N/∂x-∂M/∂y)dA

The Attempt at a Solution


I tried parametrizing the ellipse as r(t)=(acost,bsint), but didn't know how to go on...

I don't know how to solve the double integral over the ellipse ∫∫R (1-2y)dA.

Thanks.
 
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Begin with simpler things to get an idea: how to calculate the area of a circle of r=1?
\int_{-1}^1\int_{?}^{?} dx dy

Replace the ?, then it should be clear for the ellipse.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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