Divergence theorem

  • Thread starter bowlbase
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  • #1
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Homework Statement


Find the divergence of [tex]\vec v = \frac{\hat{v}}{r}[/tex]
Then use the divergence theorem to look for a delta function at the origin.

Homework Equations



[tex] \int ∇\cdot \vec v d\tau = \oint \vec v \cdot da [/tex]

The Attempt at a Solution



I got the divergence easy enough: [tex] \frac{1}{r^2} [/tex]

And, really I have the integrals set up as well. But I'm getting hung up on my limits:

[tex] \int \frac {1}{r^2} d\tau = \oint \frac {1}{r} da [/tex]
[tex] \int_0^?\int_0^?\int_0^R \frac {1}{r^2} (r^2 sin(\theta) dr d\theta d\phi = \oint_0^?\oint_0^? \frac {1}{R} (R^2 sin(\theta) d\theta d\phi [/tex]


These are easy integrals so my only issue is limits. To me it seems like Phi and Theta should be 2 pi. But I know that when we did integrals with hemispheres that the integration was from 0 to half pi. Though, I felt it should have been to pi there instead. Essentially, my issue is visualization of this.
 

Answers and Replies

  • #2
146
2
As soon as I posted this I realized my problem. So, imagining myself in the center of a sphere and my arm pointed out holding a paint brush as the r vector:

I spin 360 degrees painting a line all the way around. Then raising my arm and spinning again I paint a little more above the previous line. I keep doing this until my arm is 90 degrees and the top half of the sphere is painted. In other words I've painted half the sphere by increasing theta by pi over 2. To get the rest I can just flip the sphere and continue another 90 degrees. Meaning my integration is from 0 to pi for theta and 0 to 2 pi for phi.

It's kind of a weird visualization I guess but it helped me so I'm just leaving it here for anyone else stuck on such a problem.

Thanks anyway!
 
  • #3
vanhees71
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I don't understand your notation. On the left-hand side you have a vector field [itex]\vec{v}=\vec{v}(\vec{r})[/itex], right? On the right-hand side you have [itex]\hat{v}[/itex]. What is this? Usually it means the unit vector in direction of [itex]\vec{v}[/itex], but then your vector field is not clearly determined. So what's the proper definition of your vector field to start with?
 

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