Divergence and Volume Integrals

[BOAS]

Homework Statement

(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}.

Homework Equations

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!

 

[Orodruin]

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 $\rho = 0$.

 

[BOAS]

You have the wrong expression for the divergence. There is no term which does not contain a derivative.

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}

Your integration limits are fine apart from the fact that the $\rho$ should be an $R$, $\rho$ is the coordinate which you integrate over.

Thank you, this makes sense.

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 $\rho = 0$.

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...

 

[vela]

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

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.

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...

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.

 

[Orodruin]

It's actually not a mistake,

Indeed, for some reason I was thinking gradient, my bad.

How do I deal with this?

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.

 

[Ray Vickson]

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...

Couldn't (ii) be more easily done using the divergence theorem, $\int_V \vec{\nabla} \cdot \vec{F} \, dv = \oint_{S} \vec{F} \cdot d\vec{S}$?