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Gauss Divergence theorem to find flux through sphere with cavity.

  1. Feb 1, 2010 #1
    1. The problem statement, all variables and given/known data
    Use the divergence theorem to find the outward flux of a vector field [tex]
    F=\sqrt{x^2+y^2+z^2}(x\hat{i}+y\hat{j}+z\hat{k})[/tex] across the boundary of the region [tex]1\leq x^2+y^2+z^2 \leq4

    2. Relevant equations
    The Gauss Divergence Theorem states [tex]\int_D dV \nabla \bullet F=\int_S F\bullet dA [/tex] where D is a 3d region and S is it's boundary.

    3. The attempt at a solution
    First, I sketched out the boundary, which I think is a sphere of radius 2 with a cavity f radius 1 at the centre. The formula requires that S is oriented outwards.
    I basically need to know if the way to do this is first to treat it first as a sphere of radius 2 without a cavity, and work out the outward flux throught this. Then treat the cavity as a sphere of radius 1 and work out the flux going into this, and add the two.
    If this is the right approach, I'm not sure how to treat an inward pointing area element. The formula seems to heavily stress that the outward normal is taken, and I don't know if taking an inward normal is allowed.
    Alternatively, I think I might be able to take the region as a whole straightaway, and then when integrating over the volume just take the limits of the radius as 2 and 1.
    This way, which is the only way can actually get an answer at the moment, gives me an answer for the flux as[tex]\int_S F\bullet dA=48\pi [/tex]
    Any help would be much appreciated.
  2. jcsd
  3. Feb 1, 2010 #2
    "Outward" means "pointing away from the interior of the region". The outward normal on the inner (radius-1) surface points toward the origin, not away from it.

    Incidentally, dot product is \cdot, not \bullet: [tex]\int_S F \cdot dA[/tex]
  4. Feb 1, 2010 #3
    Ok, does that mean the way I got my answer isn't valid? Since by doing the whole thing in one go I only treated the normal pointing away from the outer surface and not the interior normal pointing towrads the origin?
  5. Feb 1, 2010 #4
    No; if you computed [tex]\int_D (\nabla\cdot F)\,dV[/tex], then the divergence theorem says that equals the flux with the normal taken in the correct (away from the region) direction.
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