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Help w/ Seaborn's Mathematics for the Physical Sciences

  1. Feb 15, 2009 #1
    1. The problem statement, all variables and given/known data
    In spherical polar coordinates, a vector F is given by F=(r*costheta*cosphi)rhat + (r*costheta*sinphi)thetahat + (r*sintheta*cosphi)phihat

    Check the validity of the divergence theorem for this vector and the volume that is one octant of a sphere radius b.

    2. Relevant equations
    [tex]\oint[/tex]F [tex]\cdot[/tex] dA = [tex]\int[/tex]Vgrad[tex]\cdot[/tex]Fd3r

    3. The attempt at a solutionThe solution for this problem is in my book, and I am having difficulty following it, I think because it is in polar coordinates, and I am still trying to get used to them.

    The author proceeds to divide the surface into four area elements:

    spherical surface : dA = rhat(b^2)*sintheta*dtheta*dphi

    before I go on to any of the other elements, could someone please help me understand how the author derived this expression? Thanks.
  2. jcsd
  3. Feb 15, 2009 #2


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    Hi bcjochim07! :smile:

    Mark a "rectangle" on the surface, with change in latitude dθ and change in longitude dφ.

    Then its width is bsinθ dθ, and its height is bdφ, so its area is b2sinθdθdφ. :wink:
  4. Feb 15, 2009 #3

    So I think I understand where the bdφ comes from; isn't that an arc length? Then I would think that the other dimension would be bdθ, so where does that sinθ come from?

  5. Feb 15, 2009 #4


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    Just look at a globe of the Earth …

    a circle of latitude has radius bsinθ

    (while a circle of longitude has radius b). :smile:
  6. Feb 20, 2009 #5
    Thanks. Great explanation. Now I am trying to figure out how the book derives the other surfaces. For example, the area element in the xz plane is in the negative phi hat direction with dA = r dr dθ. Considering the area formula for a circular sector, I thought it should be (1/2)rdrdθ.
  7. Feb 20, 2009 #6
    Ok... I think I see it now. I just need to divide the area up into little "rectangles" just like I did for the spherical surface. Each of these rectangles has dimensions dr by rdθ.
  8. Feb 22, 2009 #7


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    Yup! … it's all simple geometry …

    just draw the right diagram, and it becomes obvious! :biggrin:
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