Suppose that $\nabla \cdot \mathbf{F}=0$ and $\nabla \cdot \mathbf{G}=0$ . Does $\mathbf{F} \times \mathbf{G}$ necessarily have zero divergence?

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The statement is False.

We will give a counter example as follows.

Let

\mathbf{F}=\left<x,-y,0\right>\text{ and }\mathbf{G}=\left<0,-y,z\right>\qquad{(1)}

\begin{align*} \nabla\cdot\mathbf{F}=&\left<\dfrac{\partial\ }{\partial x},\dfrac{\partial\ }{\partial y},\dfrac{\partial }{\partial z}\right>\cdot\left<x,-y,0\right>\tag{from (1)}\\ =&\left(\dfrac{\partial\ }{\partial x}\ x+\dfrac{\partial\ }{\partial y}\ (-y)+\dfrac{\partial }{\partial z}\ 0\right)\\ =&1+(-1)+0=0\tag{general power rule}\\ \end{align*}

\begin{align*} \nabla\cdot\mathbf{G}=&\left<\dfrac{\partial\ }{\partial x},\dfrac{\partial\ }{\partial y},\dfrac{\partial }{\partial z}\right>\cdot\left<0,-y,z\right>\tag{from (1)}\\ =&\left(\dfrac{\partial\ }{\partial x}\ 0+\dfrac{\partial\ }{\partial y}\ (-y)+\dfrac{\partial }{\partial z}\ z\right)\\ =&0+(-1)+1=0\tag{general power rule}\\ \end{align*}

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