I understand how to take the divergence of a vector field defined by (x,y,z) but how can one take the divergence for the following:
$$\bar F = \bar a \times \bar r$$where $\bar a$ is a constant vector and $\bar r$ is $(bcos(\theta) , bsin(\theta) , c\theta)$, with b and c being positive constants and $\theta$ taking values between 0 to $\frac{5\pi}{2}$.
I have worked out the cross product to be $$ \begin{pmatrix} a_2c\theta - a_3b\sin(\theta)\\ a_3bcos(\theta) - a_1c\theta\\ a_1bsin(\theta) - a_2bcos(\theta)\\ \end{pmatrix} $$
if we let $\bar a = (a_1,a_2,a_3)$
However, I am not sure how to proceed after this to work out the divergence, since there is no x, y, or z (apparently the answer turns out to be 0).
Thank you for your help.
$\nabla \cdot (\overrightarrow{a}\times\overrightarrow{r})=\overrightarrow{r}\cdot (\nabla\times\overrightarrow{a})-\overrightarrow{a}\cdot(\nabla\times\overrightarrow{r})=-\overrightarrow{a}\cdot(\nabla\times\overrightarrow{r})$
for $\overrightarrow{r}$, $x=b\cos\theta, y=b\sin\theta, z=c\theta$, can be reparameterized as :
$x=b\cos(\cfrac{z}{c})$
$y=b\sin(\cfrac{z}{c})$
$z=z$
$\nabla \times\overrightarrow{r}=-\cfrac{b}{c}\Big(\cos(\cfrac{z}{c})\mathbf{i}+\sin(\cfrac{z}{c})\mathbf{j}\Big)$