# A force field is described by <F_x,F_y,F_z> = < xy +z , xy-x, 2y -zx > . Is this force field conservative?

Jul 14, 2016

NO

#### Explanation:

if the field $\vec{F}$ is conservative, then there exists a potential function, $f$, such that $\vec{F} = - \nabla f$

and as $\nabla \times \nabla f = 0$, curl of gradient, it follows that if $f$ exists then $\nabla \times \vec{F} = 0$ also

so we can test the curl of the vector field to see if it is indeed zero, as follows

$\nabla \times \vec{F} = \det \left[\begin{matrix}\hat{x} & \hat{y} & \hat{z} \\ {\partial}_{x} & {\partial}_{y} & {\partial}_{z} \\ x y + z & x y - x & 2 y - z x\end{matrix}\right]$

$= \hat{x} \left(2 - 0\right) - \hat{y} \left(- z - 1\right) + \hat{z} \left(y - x\right)$

$= \left[\begin{matrix}2 \\ z + 1 \\ y - x\end{matrix}\right] \ne 0$

This is not conservative. It is necessary that the $\nabla \times \vec{F} = 0$ for $\vec{F}$ to be conservative