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The natural domain $D$ of the differential form $\omega$ is $D=D_1 \cup D_2$ where $D_1=B\bigl((0,0),2\bigr)$ is the open ball of center in the origin and radius $r=2$ and $D_2=\mathbb{R}^2-\bar{D_1}$. Since $\omega$ is closed on the simply connected $D_1$, we have that $\omega$ is exact on $D_1$. How can i prove if $\omega$ is exact on $D_2$ too? Thanks

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up vote 2 down vote accepted

$\omega(x,y) = \mathrm{d} \log((x^2+y^2-4)^2)/2$, and is closed, being an exact form for all $(x,y)$ such that $x^2+y^2 \not=4$, i.e. on $D = D_1 \cup D_2$.

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