(Solved): 6 The vector field F is shown in the xy-plane and looks the same in all other horizontal planes. (I ...

6 The vector field $\mathbf{F}$ is shown in the $xy$-plane and looks the same in all other horizontal planes. (In other words, $\mathbf{F}$ is independent of $z$ and its $z$-component is 0 .) (a) Is $\mathrm{div}(\mathbf{F})$ positive, negative, or zero at $P$ ? Explain. $\mathrm{div}(\mathbf{F})$ is because the vectors that start near $P$ are those that end near $P$. (b) Determine whether curl(F) $=0$. If not, in which direction does $\mathrm{curl}(\mathbf{F})$ point at $P$ ? $\mathrm{curl}(\mathbf{F})?=0$. At $P$ the $\mathrm{curl}(\mathbf{F})$ points in the direction of positive $x$. $\mathrm{curl}(\mathbf{F})?=0$. At $P$ the $\mathrm{curl}(\mathbf{F})$ points in the direction of negative $x$. $\mathrm{curl}(\mathbf{F})?=0$. At $P$ the $\mathrm{curl}(\mathbf{F})$ points in the direction of positive $y$. $\mathrm{curl}(\mathbf{F})?=0$. At $P$ the $\mathrm{curl}(\mathbf{F})$ points in the direction of negative $y$. $\mathrm{curl}(\mathbf{F})?=0$. At $P$ the $\mathrm{curl}(\mathbf{F})$ points in the direction of positive $z$. $\mathrm{curl}(\mathbf{F})?=0$. At $P$ the $\mathrm{curl}(\mathbf{F})$ points in the direction of negative $z$. $\mathrm{curl}(\mathbf{F})=0$