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Working only in cylindrical coordinates, I want to show that $\triangledown$$\cdot$$r$ = 3.

We are given that $r$ = $\rho$$\hat{e}$$_\rho$+$z$$\hat{e}$$_z$.

This seems to make sense, given that in cartesian coordinates, $\triangledown$$\cdot$$r$ = 3, where $r$ = $x$$\hat{i}$+$y$$\hat{j}$+$z$$\hat{k}$. However, it doesn't seem to work for cylindrical coordinates in this case.

We know that the del operator in cylindrical coordinates is $\triangledown$ = $\frac{\partial }{\partial \rho}$$\hat{e}$$_\rho$+$\frac{1}{\rho}$$\frac{\partial }{\partial \theta}$$\hat{e}$$_\theta$+$\frac{\partial }{\partial z}$$\hat{e}$$_z$.

Then, if we calculate $\triangledown$$\cdot$$r$, we get 2, because the $\hat{e}$$_\rho$ and $\hat{e}$$_z$ terms come out to be one each, and then the $\hat{e}$$_\theta$ terms is zero.

Can anyone spot a mistake in this work? Any help would be appreciated.

UPDATE:

The problem ended up being that I wasn't using the correct divergence equation in curvelinear coordinates. What I should have used is:

$\triangledown$ $\cdot$ $A$ = $\frac{1}{h_1h_2h_3}$[$\frac{\partial}{\partial u_1}$$(A_1h_2h_3)$ + $\frac{\partial}{\partial u_2}$$(A_2h_3h_1)$ + $\frac{\partial}{\partial u_3}$$(A_3h_1h_2)$]

where $h_1,h_2,h_3$ are scale factors.

By using this equation, I end up getting $\triangledown$$\cdot$$r$ = 3, where $r$ = $\rho$$\hat{e}$$_\rho$+$z$$\hat{e}$$_z$

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If you only work in the $(\rho,z)$ plane (with fixed $\theta$), then you are effectively limiting yourself to a 2-dimensional space. This space could be parameterised by 2 Cartesian co-ordinates (say $x,y$), and you'd find $\nabla \cdot r= 2$ there too!

In order to correct this, you need to work with the full cylindrical representation:

$$r = \rho \hat{e}_{\rho} + z\hat{e}_{z}+\rho\theta \hat{e}_{\theta}$$

Now, we can see that $\nabla \cdot r = 3$ as you would expect.

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    This makes sense! What's strange is that we earlier proved that $\rho$$\hat{e}$$_\rho$+$z$$\hat{e}$$_z$ is the cylindrical equivalent of the usual $r$ = $x$$\hat{i}$+$y$$\hat{j}$+$z$$\hat{k}$...So why is this the equivalent and not the full cylindrical representation, as you put? – 2017-01-19
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    @Nathan If you consider two points on opposite sides of the cylinder, i.e. $\theta = \pi$, both points would have the same $\rho$ and the same $z$, but they are not the same point. So in general $\mathbf{r} = x\hat{\boldsymbol{\imath}} + y\hat{\boldsymbol{\jmath}} + z\hat{\boldsymbol{k}}$ cannot be represented by $\rho \hat{\boldsymbol{e}}_{\rho} + z\hat{\boldsymbol{e}}_{z}$, there *must* be a dependence on $\hat{\boldsymbol{e}}_{\theta}$. – 2017-01-20