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Stokes's theorem in spherical coordinates

  1. Sep 12, 2013 #1
    Problem:

    Say we have a vector function ##\vec{F} (\vec{r})=\hat{\phi}##.

    a. Calculate ##\oint_C \vec{F} \cdot d\vec{\ell}##, where C is the circle of radius R in the xy plane centered at the origin

    b. Calculate ##\int_H \nabla \times \vec{F} \cdot d\vec{a}##, where H is the hemisphere above the xy plane with boundary curve C.

    c. Calculate ##\int_D \nabla \times \vec{F} \cdot d\vec{a}##, where D is the disk in the xy plane bounded by C.

    (Verify Stokes's theorem in both cases.)

    Solution:

    First off, isn't it unnecessary to say verify Stokes's theorem in both cases? If we calculate a line integral, and then a surface integral bounded by the same curve, Stokes's theorem states that they're the same, yes?

    Moving on...

    a. We have [tex]\vec{F} \cdot d\vec{\ell}=(0,0,1)(dr, d\theta, rsin\theta \ d\phi)=rsin\theta \ d\phi[/tex]

    So [tex]\oint_C \vec{F} \cdot d\vec{\ell} = Rsin\theta \int_0^{2 \pi} d\phi= 2 \pi R sin \theta [/tex]

    Is this correct?

    b. We have ##F_r=F_{\theta}=0## and ##F_{\phi}=1##, so for the curl I find [tex]\nabla \times \vec{F}=(\frac{1}{r}tan\theta, 0, -\frac{1}{r})[/tex]

    I think this is right so far. But in calculating ##\int_H \nabla \times \vec{F} \cdot d\vec{a}##, I'm not sure what to use as ##d\vec{a}##. Any help? Thanks in advance.
     
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  3. Sep 12, 2013 #2

    pasmith

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    "Verify Stoke's theorem" means "observe that the line integral is indeed equal to the surface integrals"

    No. The circle in question is [itex]r = R[/itex], [itex]\theta = \pi/2[/itex] and [itex]\phi \in [0, 2\pi)[/itex]. Thus the element of line length is [itex]\mathrm{d}\vec{l} = R\hat\phi \mathrm{d}\phi[/itex], and the integral is
    [tex]
    \oint_C \vec F \cdot \mathrm{d}\vec{l} = \int_0^{2\pi} R\mathrm{d}\phi
    [/tex]

    I get [itex]\nabla \times \vec F = r^{-1}\tan\theta \hat r - r^{-1}\hat\theta[/itex].


    [itex]d\vec a[/itex] is the normal to the surface in question times the element of area. Given the choice of parametrization of the line integral, it's the normal with the non-negative [itex]\hat z[/itex] component.
     
  4. Sep 12, 2013 #3
    Ah, I see. So it's simply the perimeter of a circle. Makes sense. [tex]\int_0^{2\pi} R\mathrm{d}\phi = 2\pi R[/tex]

    Whoops, yeah that's what I meant.

    Well in class my prof used ##r^2sin \theta d\theta d \hat{\phi}##. Is that correct? If so, I would like to understand why that's the case.

    Thanks, pasmith.
     
    Last edited: Sep 12, 2013
  5. Sep 12, 2013 #4

    vela

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    No, that's not correct. ##d\vec{a}## points in the direction normal to the surface. For the hemisphere, this would be radially outward, so ##d\vec{a}## points in the ##\hat{r}## direction. The magnitude of the area element is ##R^2\sin\theta\,d\theta\,d\phi##.
     
  6. Sep 12, 2013 #5
    That makes perfect sense. So it's ##d\vec{a}=R^2sin \ \theta \ d\theta \ d\phi \ \hat{r}##, right?
     
  7. Sep 12, 2013 #6

    vela

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    Yup.
     
  8. Sep 12, 2013 #7
    So when I calculate ## \int_H \nabla \times \vec{F} \cdot d\vec{a} ## does ##r \rightarrow R## in ##\nabla \times \vec{F}=(\frac{1}{r}tan \ \theta, -\frac{1}{r}, 0)##? (ie. is it ##\nabla \times \vec{F}=(\frac{1}{R}tan \ \theta, -\frac{1}{R}, 0)##?)
     
  9. Sep 12, 2013 #8

    vela

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    Yes. You're evaluating the curl of ##\vec{F}## on the surface of the sphere, so you can set r=R.
     
  10. Sep 12, 2013 #9
    Okay. I have to integrate ##sin \ \theta \ tan \ \theta##. This is going to be a mess...
     
  11. Sep 12, 2013 #10
    Uh oh, on wolfram alpha it's saying this integral does not converge. Something has to be wrong.
     
  12. Sep 12, 2013 #11
    ## \int_H \nabla \times \vec{F} \cdot d\vec{a} = \int (\frac{1}{R}tan \ \theta, -\frac{1}{R},0) \cdot (R^2 sin \ \theta \ d\theta \ d\phi, 0, 0) = \int_0^{2 \pi} \int_0^{\pi} R sin \ \theta \ tan \theta \ d\theta \ d\phi ##. Yes?
     
  13. Sep 12, 2013 #12

    vela

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    Mathematica gives
    $$\vec{\nabla} \times \vec{F} = \frac{\cot \theta}{r}\,\hat{r} - \frac{1}{r}\,\hat{\theta}.$$ That should fix things for you.
     
  14. Sep 12, 2013 #13
    Yeah I get ##\int_0^{2 \pi} \int_0^{\pi} R cos \ \theta \ d\theta \ d\phi = 2 \pi R##.

    Hopefully I can get part c. with no difficulty. I'll keep you posted.
     
  15. Sep 12, 2013 #14
    So, for the disk we have ## d\vec{A}= - R^2 sin \ \theta \ d\theta \ d \phi \ \hat{\theta}##?
     
  16. Sep 12, 2013 #15

    vela

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    That integral as written should evaluate to ##4\pi R.##

    Nope. For one thing, r isn't constant over the disk. What should the magnitude of the area element be? Which way should it point?
     
  17. Sep 12, 2013 #16
    Ah, ##\theta## should range from 0 to pi/2. Then the integral is correct.

    I'm not sure about the magnitude, but it should point in the +z direction.
     
  18. Sep 13, 2013 #17
    ##d\vec{A}=-rdrd\phi \hat{\theta}##. Using this in the integral (with r:0 to R and Φ:0 to 2π), I get ##2 \pi R##, as expected. Is this correct?
     
    Last edited: Sep 13, 2013
  19. Sep 13, 2013 #18

    vela

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    Yup.
     
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