Answer
${\bf{F}}$ is not conservative.
Work Step by Step
From Figure 18 we see that the vector field ${\bf{F}}$ is a function of $x$ because the length of ${\bf{F}}$ is increasing as we move toward positive $x$. However, we can vertically shift ${\bf{F}}$ without changing the magnitude and the direction. This implies that ${\bf{F}}$ is not a function of $y$. Since ${\bf{F}}$ is on the plane, we can write ${\bf{F}} = \left( {{F_1},{F_2},{F_3}} \right)$, where ${F_3} = 0$. Such function of ${\bf{F}}$ cannot satisfies the cross-partials condition
$\dfrac{{\partial {F_1}}}{{\partial y}} = \dfrac{{\partial {F_2}}}{{\partial x}}$, ${\ \ \ }$ $\dfrac{{\partial {F_2}}}{{\partial z}} = \dfrac{{\partial {F_3}}}{{\partial y}}$, ${\ \ \ }$ $\dfrac{{\partial {F_3}}}{{\partial x}} = \dfrac{{\partial {F_1}}}{{\partial z}}$
because $\dfrac{{\partial {F_1}}}{{\partial y}}$ is always zero, but $\dfrac{{\partial {F_2}}}{{\partial x}}$ is not zero. Therefore, we conclude that ${\bf{F}}$ is not conservative.
