Dot product in cylidrical coordinates

Question

I'm given the vector:

$$\vec{V}{(r,θ,z)}=\frac{1}{r}\hat{e_r} + (r\cosθ)\hat{e_θ}+\frac{z^2}{r^2}\hat{e_z}$$

I want the scalar product ${\vec{\nabla}}\cdot{\vec{V}}$

We know that in cylindrical coordinates : $$\vec{\nabla}=\left<\frac{\partial}{\partial r},\frac{1}{r}\frac{\partial}{\partial θ},\frac{\partial}{\partial z} \right>$$

So , the product should be $${\vec{\nabla}}\cdot{\vec{V}} =\frac{\partial}{\partial r}\left(\frac{1}{r}\right) + \frac{1}{r}\frac{\partial}{\partial θ}(r\cosθ)+\frac{\partial}{\partial z}\left(\frac{z^2}{r^2}\right) = -\frac{1}{r^2}-\sinθ +\frac{2z}{r^2}$$

However , in the answers , the answer given is this : $${\vec{\nabla}}\cdot{\vec{V}}=\frac{1}{r}\Big\{\frac{\partial}{\partial r}(1)+\frac{\partial}{\partial θ}(r\cosθ)+\frac{1}{r}\frac{\partial}{\partial z}(z^2)\Big\}=-\sinθ+\frac{2z}{r^2}$$

I don't understand why $\frac{1}{r}$ was factored out and how is that possible. I understand you can factor it out for the partial derivative with respect to $θ$ and $z$ but in the first one, which is with respect to $r$, it shouldn't be factored out, it should be differentiated. Any thoughts? Am I missing something or is there a typo in the answers?

Answer 1

The divergence operator in cylindrical coordinates is actually different from what you believe it to be: $$ \nabla\cdot\mathbf A=\frac{1}{r}\frac{\partial}{\partial r}\left(r A_r\right)+\frac{1}{r}\,\frac{\partial A_\theta}{\partial\theta}+\frac{\partial A_z}{\partial z} $$ You seem to be confusing it with the gradient operator, which as the form you specify: $$ \nabla f=\frac{\partial f}{\partial r}\hat{r}+\frac{1}{r}\,\frac{\partial f}{\partial \theta}\hat{\theta}+\frac{\partial f}{\partial z}\hat{z} $$ (though obviously you're ignoring the unit vectors).