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I am looking for a counter example and am explanation for why the following statement is false:

If $F$ is a conservative field in region $A$ and $B$, so $F$ is conservative in $A \cup B$.

And what about $A \cap B$ (is it true in that case)?

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The usual example is $$ v = \left(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2},0\right) $$ defined on the punctured plane $\mathbb R^2\setminus\{0\}$. This is not a conservative field, even though we have $\text{curl}(v(x,y)) = 0$ for all $(x,y)\in\mathbb R\setminus \{0\}$.

However, we may write $\mathbb R^2\setminus\{0\}$ as the union of two simply-connected regions. $v$ is necessarily conservative on each of these regions, since any vector field on a simply-connected set whose curl is $0$ is conservative.

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For the union, think of some non-conservative, curl-free vector field. Split its domain into two parts so that it's conservative on each part, and you have your counterexample.

As for the intersection, if any integral of the field over a closed loop in $A$ gives zero, then clearly any integral over a closed loop in $A\cap B$ must give zero.

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