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Flux of r through a sphere

  1. Sep 11, 2014 #1
    1. The problem statement, all variables and given/known data

    What is the flux of r through a spherical surface of radius a?

    2. Relevant equations

    I'm guessing I should use a surface integral? ∫v.da ?

    3. The attempt at a solution

    Plugging in: I would get ∫r.da ? but what is a small patch of a sphere?

    I'm kind of confused. Not really sure what r is even. Is this just any vector going through the sphere?

    Please help

    thanks
     
  2. jcsd
  3. Sep 11, 2014 #2

    CAF123

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    The da there is a vector normal to an infinitesimal surface area element on the sphere. You can find it by parametrising the surface by some appropriate coordinates, thereby finding two vectors lying in the plane of the sphere which infinitesimally span a flat plane. This constitutes the area element you speak of. The derivation is done in most books.
    Yes, it corresponds to the special case where the vector field coincides with position vector, so in Cartesian coordinates reads ##\mathbf r = \langle x,y,z \rangle##. Your problem suggests working with the equivalent expression in a different set of coordinates.
     
  4. Sep 11, 2014 #3

    vanhees71

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    Well, it might be simpler to use Gauss's integral theorem
    [tex]\int_{V} \mathrm{d}^3 r \vec{\nabla} \cdot \vec{V} = \int_{\partial V} \mathrm{d} \vec{a} \cdot \vec{V}.[/tex]
     
  5. Sep 11, 2014 #4

    HallsofIvy

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    Do you mean a sphere of radius a with center at the origin?

    Yes, you should. And to find what you call "da" (I would call it "[itex]d\vec{S}[/itex]") write the surface in parametric equations- the standard equations for spherical coordinates with "[itex]\rho[/itex]" set to "a":
    [itex]x= a cos(\theta) sin(\phi)[/itex], [itex]y= a sin(\theta) sin(\phi)[/itex], and [itex]z= a cos(\phi)[/itex].
    You can write that as a "position vector" for any point on the surface of the sphere depending on [itex]\theta[/itex] and [itex]\phi[/itex]:
    [tex]\vec{r}= a cos(\theta) sin(\phi)\vec{i}+ a sin(\theta)sin(\phi)\vec{j}+ a cos(\phi)\vec{k}[/tex].
    The derivatives with respect to [itex]\theta[/itex] and [itex]\phi[/itex] are vectors in the tangent plane at each point on the surface of the sphere:
    [tex]\vec{r}_\theta= -a sin(\theta) sin(\phi)\vec{i}+ a cos(\theta)sin(\phi)\vec{j}[/tex]
    [tex]\vec{r}_\phi= a cos(\theta)cos(\phi)\vec{i}+ a sin(\theta)cos(\phi)\vec{j}- a sin(\phi)\vec{k}[/tex].

    Finally, the cross product of those two vectors
    [tex]\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ a cos(\theta)cos(\phi) & a sin(\theta)cos(\phi) & -a sin(\phi) \\ -a sin(\theta)sin(\phi) & a cos(\theta)sin(\phi) & 0 \end{array}\right|= a^2cos(\theta)sin^2(\phi)\vec{i}+ a^2sin(\theta)sin^2(\phi)\vec{j}+ a^2sin(\phi)cos(\phi)\vec{k}[/tex]
    is perpendicular to the sphere and contains "area information"- the vector differential of surface area is [itex](a^2cos(\theta)sin^2(\phi)\vec{i}+ a^2sin(\theta)sin^2(\phi)\vec{j}+ a^2sin(\phi)cos(\phi)\vec{k})d\theta d\phi[/itex]

    [itex]\vec{r}[/itex] is the "position vector" of a point: [itex]x\vec{i}+ y\vec{j}+ z\vec{k}[/itex] in Cartesian coordinates and [itex]a cos(\theta)sin(\phi)\vec{i}+ a sin(\theta)sin(\phi)\vec{j}+ a cos(\phi)\vec{k}[/itex] in these coordinates.

    Take the dot product, [itex]\vec{r}d\vec{S}[/itex], of those two vectors and integrate with [itex]0\le\theta\le 2\pi[/itex] and [itex]0\le\phi\le \pi[/itex].

    (of course, Vanhees71 may well be right- it might be easier to use Gauss' integral theorem.)
     
    Last edited: Sep 11, 2014
  6. Sep 11, 2014 #5
    so I get an answer of :

    [tex]\frac{8\pi a^3}{3}[/tex]
     
  7. Sep 11, 2014 #6

    CAF123

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    hey, try the question using the divergence theorem as vanhees71 suggested and compare your results :)
     
  8. Sep 11, 2014 #7
    Yup got the same answer :)
     
  9. Sep 11, 2014 #8

    vanhees71

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    How did you come to this result?
     
  10. Sep 11, 2014 #9
    By plugging in the volume of the sphere (4pi a^3 /3) into the divergence theorem. My answer is two times that. I think that makes sense since the flux would indicate the outward and inward flux?
     
  11. Sep 11, 2014 #10
    Is it not correct?
     
  12. Sep 11, 2014 #11

    CAF123

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    What is ##\nabla \cdot \mathbf r##, where ##\mathbf r## is the position vector? (A calculation most easily done in Cartesian coordinates)
     
  13. Sep 11, 2014 #12
    Whoops it's 4 pi a ^3
     
  14. Sep 12, 2014 #13

    vanhees71

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    Now you've got it!:thumbs:
     
  15. Sep 12, 2014 #14

    vela

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    So did you figure out why you got a different result from the surface integral?
     
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