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Recently, I had an exam and in that I was asked to evaluate the line integral of the function $$F=\frac{-y}{x^2+y^2}i+\frac{x}{x^2+y^2}j$$ alongside the unit circle, $0 \le t \le 2\pi $ . Moreover, it was asked if this integral could be carried out with using Green's Theorem or not and why?

For the first, I did the following: $$\oint_C F\cdot dr=\int_0^{2\pi}F\big (\cos(t),\sin(t)\big)\cdot\big(-\sin(t),\cos(t)\big)dt$$ which is $2\pi$. But always I have problem with above theorem and I do know I didn't pass this part of question correctly. May I ask to help me? Thank you.

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up vote 2 down vote accepted

Your calculation of the line integral is correct. No, Green's Theorem cannot be applied directly because the vector field has a singularity at the origin.

Some years ago I taught a multivariable calculus course, and this very vector field was one of the stars of the whole show. You can see for instance $\S 5$ of these notes for an extended discussion of the fact that this vector field is irrotational -- i.e., has zero curl -- but is not conservative and in particular does not have a scalar potential function. Then, if you are feeling especially curious, you can go on to read the next section, which segues from this example to a discussion of De Rham cohomology and the problem that the main character in the film A Beautiful Mind writes on the blackboard in the multivariable calculus class attended by his future wife.

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@Oh Thanks Pete. I saw that film. Thanks for the refrences. I will follow what you noted. – Nancy Rutkowskie Aug 6 '12 at 14:00
I hope you enjoy the notes. Let me say that the one I linked to is "handout eight"; if you replace the "eight" by a different number you'll get other sections of the notes, and I believe that five through seven also have things to say about this vector field. (Or go to and scroll down to Multivariable Calculus to see the full list.) – Pete L. Clark Aug 6 '12 at 14:05
@PeteL.Clark, one question/aclaration: when you say "this vector field is irrotational yet not conservative...", you mean "in any domain containing the origin", otherwise $$\arctan\frac{y}{x}\,$$ is a potential. – DonAntonio Aug 6 '12 at 16:01
@DonAntonio: agreed. – Pete L. Clark Aug 6 '12 at 16:50

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