Green's theorem, reduce integral Let S be the region of the plain bounded by the graph of x^2-y^2=4 and the lines y=2 and y=-2, and F(x,y)=($\frac{-y}{x^2+y^2}$,$\frac{x}{x^2+y^2}$). Use Green's theorem to reduce the line integral $\iint_{ds}$F dx to the integral over a simpler curve, then using this technique evaluate the line integral.
So I know Green's theorem is $\iint_{ds}$F dx =$\iint_{S}$($\frac{dF_2}{dx_1}$-$\frac{dF_1}{dx_2}$)dA. But in this case $\frac{dF_2}{dx_1}$=$\frac{dF_1}{dx_2}$, so the answer is just zero? Isn't that wired since the question ask to reduce the line integral into a integral over a simpler curve...(plus there are lots of space for this question)
 A: I'm really having a hard time understanding the questions as you are writing them. Is this from a textbook? Could you copy it exactly maybe??
I think the question is getting at this: All curves that go around the singularity at (0,0) will become the "same" integral using Green's Theorem - they will become the double integral of 0 as you said. This DOES NOT mean that the answer is zero, but I guess they are saying it means that all the line integrals will be the same.
So instead of computing the line integral given, you could use a simpler curve which goes around the origin. They are probably wanting you to use the unit circle, which will make the computation very easy... I think it will be $2\pi$ as before (is this a complex variables class by any chance? Because these are tricky questions for a normal multi-variable calc class..)
Anyway, the reason that the answer is not 0 is that Green's cannot actually be used to find the answer - the singularity at (0,0) inside the region of integration violates one of the hypotheses for the theorem.
Hope that helps!
A: The equations of your curves are not coming through correctly. As a point of notation: a line integral should be demoted to a single integral in order to avoid confusion.
If the region over which you are integrating includes the origin, then you have to be careful because Green's theorem doesn't apply to regions where either the function or a first-order derivative has a singularity. However, you should be able to reduce the outer contour to a new one around, say, a small circle enclosing the origin because the region between the outer curve and a small circle is one where Green's Theorem applies, after you remove a thin cut joining the inner and outer curves.
A: It is not equality, $\frac {\partial F_1}{\partial x_1}=\frac{2xy}{(x ^ 2 + y^2)^2}$
$\frac {\partial F_2}{\partial x_2}=-\frac{2xy}{(x ^ 2 + y^2)^2}$
