In the screenshot below, I am trying to evaluate two closed line integrals over the regions $C_1=x^2+y^2=1$ and $C_2=4x^2+9y^2=36$. In this specific case, however, the partials of the line integral are equal to each other ($P_y=Q_x$). Thus, since this is a conservative field over a closed path, the integrals should evaluate to 0 (which means they are equal).

The part I do not understand is part B, where we are asked to actually evaluate the two line integrals. Parametrizing the path $x^2+y^2 = 1$ and evaluating it yields $2\pi$ -- which I do not get. If the vector field is conservative, and the path is closed, how does the line integral evaluate to a non-zero value?

Thanks for all the help!

enter image description here

  • $\begingroup$ A huge number of related questions can be found with Approach0. I left out a part of the vector field, because the radial part actually has a potential in the puncture plane. $\endgroup$ – Jyrki Lahtonen Aug 14 '19 at 20:53
  • $\begingroup$ Nominating this as one of the best. You are welcome to pick your favorite among those dupes. $\endgroup$ – Jyrki Lahtonen Aug 14 '19 at 20:55
  • $\begingroup$ This deals well with the potential. Observe that $$\frac1{x^2+y^2)(x,y)$$ has $\dfrac12\ln(x^2+y^2)$ as a potential, so your field needs to take that into account. $\endgroup$ – Jyrki Lahtonen Aug 14 '19 at 20:57
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    $\begingroup$ You can answer a) without computing the integrals, because Green's theorem does apply to the region bounded by $C_1$ and $C_2$. $\endgroup$ – Maxim Aug 15 '19 at 0:09

You’re making a common mistake: when the domain isn’t simply connected, being irrotational doesn’t always mean that the vector field is conservative. In this case, there’s a hole in the domain at the origin, so the integral along a closed path that surrounds this hole might not vanish.


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