Determine if F(x,y)=⟨y2,2xy⟩\mathbf{F}(x,y) = \langle y^2, 2xy \rangleF(x,y)=⟨y2,2xy⟩ is conservative.
Yes, because Py=QxP_y = Q_xPy=Qx.
No, because Py≠QxP_y \neq Q_xPy=Qx.
Yes, because it is irrotational.
No, because it is rotational.