Divergence and Volume Integrals

In summary: 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
  • #1
BOAS
552
19

Homework Statement



(3 i) Using [itex]\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}[/itex] calculate the divergence of the vector field [itex]\mathbf{F} = 3\mathbf{e_{\rho}} + \cos(\phi)\mathbf{e_{\phi}} + 5\rho\mathbf{e_{z}}[/itex].

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

Homework Equations

The Attempt at a Solution



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

[itex]\frac{\partial \mathbf{F_{\rho}}}{\partial \rho} = 0[/itex]

[itex]\frac{\partial \mathbf{F_{\phi}}}{\partial \phi} = -\sin(\phi)[/itex]

[itex]\frac{\partial \mathbf{F_{z}}}{\partial z} = 0[/itex]

[itex]\nabla . \mathbf{F} = \frac{3}{\rho} + \frac{1}{\rho}(-\sin(\phi) + 0 = \frac{3 - \sin(\phi)}{\rho}[/itex]

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

[itex]x = \rho \cos(\phi)[/itex]

[itex]y = \rho \sin(\phi)[/itex]

[itex]\rho^{2} \cos^{2}(\phi) + \rho^2 \sin^{2}(\phi) \leq R^{2}[/itex]

[itex]\cos^{2}(\phi) + \sin^{2}(\phi) = 1[/itex]

[itex]\rho^{2} \leq R^{2}[/itex]

[itex]\rho \leq R[/itex]

[itex]\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[/itex]

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

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

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

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

[itex]\int_{V} \nabla . \mathbf{F} dV = 12 \pi \rho H[/itex]

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!
 
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  • #2
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
Orodruin said:
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;

[itex]\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}[/itex]

Which actually gives me the same function.

[itex]\nabla . \mathbf{F} = \frac{3 - \sin(\phi)}{\rho}[/itex]

Orodruin said:
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.

Orodruin said:
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...
 
  • #4
BOAS said:
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.
 
  • #5
vela said:
It's actually not a mistake,

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

BOAS said:
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.
 
  • #6
BOAS said:
Thanks for catching that - It's a mistake on the handout.

From my notes;

[itex]\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}[/itex]

Which actually gives me the same function.

[itex]\nabla . \mathbf{F} = \frac{3 - \sin(\phi)}{\rho}[/itex]
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}##?
 

1. What is divergence?

Divergence is a mathematical operation that measures the extent to which a vector field flows away from a given point. In simpler terms, it is a measure of how much a field spreads out or converges at a particular point.

2. How is divergence related to volume integrals?

Divergence is closely related to volume integrals because it is used to calculate the flux (flow) of a vector field through a surface. This flux can then be integrated over a volume to calculate the total flow through that volume.

3. What do volume integrals represent?

Volume integrals represent the total value of a function over a given volume. In the context of divergence, they represent the total flow of a vector field through a given volume.

4. What is the significance of divergence and volume integrals in physics?

Divergence and volume integrals are important concepts in physics because they are used to describe the behavior of vector fields, such as velocity or force fields. They are particularly useful in fluid dynamics, electromagnetism, and other areas of physics where vector fields are present.

5. How is divergence and volume integrals calculated?

Divergence is typically calculated using the gradient operator, and volume integrals are calculated by integrating the function over a given volume. The specific method of calculation may vary depending on the context and the specific vector field being analyzed.

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