# Use green's theorem to reduce the line 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 what should I do?o

Let us clarify the notation a bit. Let $x_1 = x$ and $x_2 = y$, let also $F_1 = \frac{-y}{x^2 + y^2}$ and $F_2 = \frac{x}{x^2 + y^2}$. Using the product or the quotient rules for differentiation you can verify that
$\frac{\partial F_1}{\partial y} = \frac{y^2 - x^2}{(x^2+y^2)^2},$
$\frac{\partial F_2}{\partial x} = \frac{y^2 - x^2}{(x^2+y^2)^2}.$
$\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} = 0. Let me know if you need further clarification. • I believe you made a mistake, they are the same – user108297 Mar 24 '14 at 22:23 • It is a bit confusing. However,$\frac{\partial{F_1}}{y}$and$\frac{\partial{F_1}}{y}$can not be the same, otherwise$F_1$and$F_2$remain the same if we replace the substitution$x<->y$, which is clearly not true. I hope this is more convincing! – Abbas Mar 24 '14 at 22:41 • wolframalpha.com/input/… – user108297 Mar 24 '14 at 22:50 • Yes you are right it should be zero. I have edited my post. I guess the minus sign in$F-1\$ should be removed for the question to make sense. – Abbas Mar 25 '14 at 9:21