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:
- Prove that $\int_cP\,dx+Q\,dy=0$ given curve $(x-3)^2+(y-3)^2=1$
- 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 field is locally conservative, meaning that around any point in its domain there is a small neighborhood on which the field is indeed conservative. This is also known as curl-free, for the simple reason that the curl of the field is zero. However, once there is a hole in the domain you open the door for larger-structure non-conservativeness, just like you discovered here.