Exercise 8 — Active Calculus Multivariable

Let $\vF=\langle{F_1,F_2,F_3}\rangle$ and let $\begin{gathered} \vG = (\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z})\vi- (\frac{\partial F_3}{\partial x}-\frac{\partial F_1}{\partial z})\vj + (\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y})\vk \end{gathered}$ Show that $\divg(\vG)=\vec{0}$ .
Vector fields with a zero divergence everywhere in their domain are called divergence-free or incompressible vector fields. Which of the following vector fields are divergence-free?
1. $\vF=\langle{-y,z,x}\rangle$
2. $\vF=\langle{\cos(yz),3xe^{z-x},6(x+y+z)^3}\rangle$
3. $\vF=\langle{4xyz,y^2z,yz^2}\rangle$
4. $\vF=\nabla f$ where $f$ is a scalar function of $x$ , $y$ , and $z$
Let $\vF_1=\langle{3(x-z)^2,2\cos(x)+3yz+y,-(z-1)^2+e^{xy}}\rangle$ . Calculate the divergence of $\vF_1$ and give a point where $\divg(\vF_1)=0$ .
Is $\vF_1$ a divergence free vector field?

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