TITLE: How to check if this weird vector field is conservative
QUESTION [0 upvotes]: I need to check if this vector field is conservative: ('$\mathrm{sgn}$' is the sign function)
$$F=(\dfrac{y\cdot\mathrm{sgn}(xy)}{1+|xy|},\dfrac{x\cdot\mathrm{sgn}(xy)}{1+|xy|})$$
Do I need to check separately for $x<=0,x=0,x>0$? but what about the absolute value? Can you help me figure this out?

REPLY [1 votes]: Observe that
$$F(x,y)=\begin{cases}\left(\frac y{1+xy}\;,\;\;\frac x{1+xy}\right)\,,&xy\ge0\\{}\\\left(-\frac y{1-xy}\;,\;\;-\frac x{1-xy}\right)\,,&xy<0\end{cases}$$
and now
$$\frac{\partial}{\partial x}\left(\frac x{1+xy}\right)=\frac{1+xy-xy}{(1+xy)^2}=\frac1{(1+xy)^2}=\frac{\partial}{\partial y}\left(\frac y{1+xy}\right)$$
and by symmetry (observe both minus signs in the second row of the definition of $\;F$) , we get that $\;F\;$ is conservative wherever it is definee, i.e.: in $\;\Bbb R^2\;$
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