# How to determine if a field is conservative if sometimes the integral is $0$ and sometimes not?

Given the field $P(x,y)=\frac{y}{y^2-(x-1)^2}$, $Q(x,y)=\frac{1-x}{y^2-(x-1)^2}$ determine whether the field $(P(x,y,Q(x,y))$ is conservative within its defined range.

Background:

This question follows two questions from the same problem:

1. Prove that $\int_cP\,dx+Q\,dy=0$ given curve $(x-3)^2+(y-3)^2=1$
2. Calculate $\int_cP\,dx+Q\,dy$ if the curve is $c(t)=\langle 1-\cos t, \sin t\rangle$, $0\le\theta\le 2\pi$.

First we see that $P_y=Q_x$ so it's a conservative field in the range except the point $(1,0)$ where the field is undefined. So in question 1) the integral automatically is $0$.

In the 2) question the integral is $2\pi$.

The correct answer is that the field $(P(x,y,Q(x,y))$ is not conservative because in the 2) question the integral was not equal to $0$.

This is extremely confusing to me. We clearly see that $P_y=Q_x$ which means the field is conservative! Now there's the point $(1,0)$ indeed I guess if the range includes that point then the field is not defined hence not conservative? But the answer depends on the range in that case. But the question doesn't explicitly mention the range.

• The domain $\Bbb R^2-\{(1,0)\}$ has nontrivial cohomology. – Angina Seng Aug 8 '17 at 18:32
• The error is in the following sentence: "We clearly see that $P_y=Q_x$ which means the field is conservative!" It is not true that if $P_y=Q_x$, then the field is conservative. There are topological obstructions (depending on the "shape" of the domain) that prevent the vanishing of the 2D "curl" $P_y-Q_x$ from entailing conservativity. – symplectomorphic Aug 8 '17 at 18:42