Determine if the vector field F=⟨2xy,x2+z,y⟩\mathbf{F} = \langle 2xy, x^2 + z, y \rangleF=⟨2xy,x2+z,y⟩ is conservative.
Yes, it is conservative.
No, it is not conservative because curl F≠0\mathbf{F} \neq 0F=0.
It is conservative because divergence F=0\mathbf{F} = 0F=0.
It is conservative because it has a potential function f(x,y,z)=x2y+yzf(x, y, z) = x^2 y + yzf(x,y,z)=x2y+yz.