# Divergence and Volume Integrals

1. Mar 28, 2015

### BOAS

1. The problem statement, all variables and given/known data

(3 i) Using $\nabla . \mathbf{F} = \frac{\partial \mathbf{F_{\rho}}}{\partial \rho} + \frac{\mathbf{F_{\rho}}}{\rho} + \frac{1}{\rho} \frac{\partial \mathbf{F_{\phi}}}{\partial \phi} + \frac{\partial \mathbf{F_{z}}}{\partial z}$ calculate the divergence of the vector field $\mathbf{F} = 3\mathbf{e_{\rho}} + \cos(\phi)\mathbf{e_{\phi}} + 5\rho\mathbf{e_{z}}$.

(ii) Evaluate the volume integral of $\nabla . \mathbf{F}$ over the cylinder defined by $-H \leq z \leq H, x^{2} + y^{2} \leq R^{2}$.

2. Relevant equations

3. The attempt at a solution

I'm not sure if i'm doing the volume integral properly.

$\frac{\partial \mathbf{F_{\rho}}}{\partial \rho} = 0$

$\frac{\partial \mathbf{F_{\phi}}}{\partial \phi} = -\sin(\phi)$

$\frac{\partial \mathbf{F_{z}}}{\partial z} = 0$

$\nabla . \mathbf{F} = \frac{3}{\rho} + \frac{1}{\rho}(-\sin(\phi) + 0 = \frac{3 - \sin(\phi)}{\rho}$

So that's part one done - Moving on to converting this cylinder to cylindrical polar coordinates.

$x = \rho \cos(\phi)$

$y = \rho \sin(\phi)$

$\rho^{2} \cos^{2}(\phi) + \rho^2 \sin^{2}(\phi) \leq R^{2}$

$\cos^{2}(\phi) + \sin^{2}(\phi) = 1$

$\rho^{2} \leq R^{2}$

$\rho \leq R$

$\int_{V} \nabla . \mathbf{F} dV = \int^{H}_{-H} \int^{2\pi}_{0} \int^{\rho}_{0} \frac{3 - \sin(\phi)}{\rho} \rho d\rho d\phi dz$

$\int_{V} \nabla . \mathbf{F} dV = \int^{H}_{-H} \int^{2\pi}_{0} \int^{\rho}_{0} 3 - \sin(\phi) d\rho d\phi dz$

$\int_{V} \nabla . \mathbf{F} dV = \int^{H}_{-H} \int^{\rho}_{0} [3 - \sin(\phi)]^{2\pi}_{0} d\rho dz$

$\int_{V} \nabla . \mathbf{F} dV = 6\pi \int^{H}_{-H} \int^{\rho}_{0} d\rho dz$

$\int_{V} \nabla . \mathbf{F} dV = 6\pi \rho \int^{H}_{-H} dz$

$\int_{V} \nabla . \mathbf{F} dV = 12 \pi \rho H$

I'm not that confident in my treatment of the limits, so it would be good to get some feedback/know if i'm doing this wrong.

Thanks!

2. Mar 28, 2015

### Orodruin

Staff Emeritus
You have the wrong expression for the divergence. There is no term which does not contain a derivative. Your integration limits are fine apart from the fact that the $\rho$ should be an $R$, $\rho$ is the coordinate which you integrate over. Also note that the result you get is not really correct anyway, the vector field has a singularity at $\rho = 0$, which is not captured by your expression for the divergence, which only holds outside of $\rho = 0$.

3. Mar 28, 2015

### BOAS

Thanks for catching that - It's a mistake on the handout.

From my notes;

$\nabla . \mathbf{F} = \frac{1}{\rho} \frac{\partial \rho \mathbf{F_{\rho}}}{\partial \rho} + \frac{1}{\rho} \frac{\partial \mathbf{F_{\phi}}}{\partial \phi} + \frac{\partial \mathbf{F_{z}}}{\partial z}$

Which actually gives me the same function.

$\nabla . \mathbf{F} = \frac{3 - \sin(\phi)}{\rho}$

Thank you, this makes sense.

How do I deal with this? Define the interval of rho to be not inclusive of zero? That would give me a cylinder with an infinitely small hole through the center I think...

4. Mar 28, 2015

### vela

Staff Emeritus
It's actually not a mistake, so it's no surprised you ended up with the same expression for the divergence. In your original post, the product rule had already been applied to the $\partial_\rho(\rho F_\rho)$ term.

You could set the lower limit of the $\rho$ integral to $\varepsilon$ and take the limit as $\varepsilon \to 0^+$. It wouldn't change the result though.

5. Mar 28, 2015

### Orodruin

Staff Emeritus
In this case it is not really a problem, the divergence of this vector field goes to infinity at rho equals zero, but this is compensated by the rho in the volume element. Depending on the vector field, you may run into trouble or not. Try the same thing with the vector field $\vec e_\rho/\rho$ for example. The "safe" bet is applying the divergence theorem.
Couldn't (ii) be more easily done using the divergence theorem, $\int_V \vec{\nabla} \cdot \vec{F} \, dv = \oint_{S} \vec{F} \cdot d\vec{S}$?