Using Green's Theorem when the function is not defined at the origin.

Question

I need to use Green's theorem to calculate the following integral over the curve $C = \frac{x^2}{16} + \frac{y^2}{9} = 1$, orientated with the hands of a clock: $$ \int_{C}\frac{y^3dx-xy^2dy}{(x^2+y^2)^2}. $$ The problem is that the vectorfield is not defined in the point $(0,0)$ but the hint says that I should be able to use some form of Green's theorem. I thought that I should just exclude that point from the calculation by changing up the curve a little bit as to make it like a keyhole, but that doesnt seem right. Could anyone help me out?

Thanks for reading,

K.

Answer 1

Use Green's Theorem on the region $\{ (x,y) \mid \epsilon^2 \leq x^2+y^2 \, \wedge \, \frac{x^2}{16} + \frac{y^2}{9} \leq 1 \}.$ Then calculate the path integral over the inner curve $x^2+y^2 = \epsilon^2.$ Use these two results to calculate the path integral over the outer curve $\frac{x^2}{16} + \frac{y^2}{9} = 1.$

Answer 2

By setting $(x,y)=(4\cos\theta,3\sin\theta)$ we get $dx=-4\sin\theta\,d\theta$, $dy=3\cos\theta\,d\theta$
and the given integral equals

$$ \int_{0}^{2\pi}\frac{-108\sin^4\theta+144\cos^4\theta}{(16\sin^2\theta+9\cos^2\theta)^2}\,d\theta = \color{red}{\frac{1669 \pi }{2352}}$$ The last identity can be derived through the residue theorem (that can be seen as a form of Green's theorem) or through symmetry and the substitution $\theta=\arctan t$, reducing the original problem to the computation of $\int_{0}^{+\infty}g(t)\,dt$ with $g(t)$ being a rational function.