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I got this line integral: $$ \int \frac{(x+y)dx-(x-y)dy}{x^2+y^2} $$ Curve is a circle $$ x=a\cos(t) $$ $$ y=a\sin(t) $$ $$ 0 \leq t \leq 2\pi $$

I first tried to solve it directly and got $ -2a^2\pi $ as a result, however after that I calculated partial derivatives of M and N: $$ \frac{\partial M}{\partial y}=\frac{x^2-y^2-2xy}{(x^+y^2)^2} $$ $$ \frac{\partial N}{\partial x}=\frac{x^2-y^2-2xy}{(x^+y^2)^2} $$ So, that would suggest conservative vector field, right? My question is how do I continue? Do I find potential function? $$ f(x,y)=\frac{1}{2}\ln(x^2+y^2)+\arctan(\frac{x}{y})+C$$ What next?

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What do you mean, what next? All of what you've written is true. The thing that's really going on here is that your vector field and potential function are only defined in $\mathbb{R}^2-\{0\}$, so you have something extra to think about before applying the theorems you know and love about conservative vector fields. – KReiser Mar 9 '12 at 16:53
up vote 1 down vote accepted

You have calculated that $$\int_\gamma \frac{(x+y)dx-(x-y)dy}{x^2+y^2} = -2a^2\pi.$$ (Remember to write the path over which you integrate as the "lower bound" of the integral! Also, the path you've written is not a circle unless $a = 1$. It's worth graphing for your self.) But you have also calculated that the vector field $$\mathbf{F} = \left(\frac{x+y}{x^2+y^2},\frac{-(x-y)}{x^2+y^2}\right)$$ is irrotational (it's curl is 0). If the vector field were conservative, then we ought to have $$\int_\gamma \frac{(x+y)dx-(x-y)dy}{x^2+y^2} = 0$$ without even needing to calculate a potential function. This is a basic property of conservative vector fields. So either your direct calculation of the integral is wrong or the field is irrotational but not conservative.

A vector field can be irrotational but not conservative if it is differentiable over a region which is not simply-connected. In the plane, a region is simply-connected if it has no holes. But in fact, the vector field $\mathbf{F}$ is not continuous at the point $(0,0)$ (why?), so it is certainly not differentiable. The path $\gamma$ circles the point $(0,0)$, so any region containing $\gamma$ contains $(0,0)$. We can conclude that $\mathbf{F}$ is not conservative in any region containing $\gamma$.

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So that's it? It's not a conservative vector field and the solution is $-2\pi$ with barely three written lines? If only every problem was small as this one. – aarnes Mar 9 '12 at 17:10
Well, those three written lines rely on you knowing what a path integral is. If you were assigned this problem in a multivariable calculus course, it would (likely, in the USA, etc.) be after a full year of calculus and many hundred pages of textbooks. So there's a lot of history in those three lines, and I don't think it's fair to say "that's it." As KReiser noted, the problem is meant to illustrate the difference between conservative and irrotational vector fields. – Adam Saltz Mar 9 '12 at 17:17

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