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.

  • $\begingroup$ I believe you made a mistake, they are the same $\endgroup$ – user108297 Mar 24 '14 at 22:23
  • $\begingroup$ 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! $\endgroup$ – Abbas Mar 24 '14 at 22:41
  • $\begingroup$ wolframalpha.com/input/… $\endgroup$ – user108297 Mar 24 '14 at 22:50
  • $\begingroup$ 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. $\endgroup$ – Abbas Mar 25 '14 at 9:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.