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Homework Help: Surface integrals and heat flow

  1. Oct 19, 2009 #1
    1. The problem statement, all variables and given/known data

    The temperature u in a star of conductivity 6 is inversely proportional to the distance from the center:

    [tex] u = \frac{3}{\sqrt{x^{2} + y^{2} + z^{2}}} [/tex]

    If the star is a sphere of radius 3, find the rate of heat flow outward across the surface of the star.

    2. Relevant equations

    3. The attempt at a solution

    I really don't know what to do here - I have not learned about "conductivity" in terms of taking surface integrals and every example problem my teacher has done has involved a force function, yet this seems to only give a scalar temperature function.

    So, I am thoroughly confused as to where to start?
  2. jcsd
  3. Oct 19, 2009 #2
    im doing the exact same one , it is due in 7 hrs
    im lost too but

    you know z^2+y^2+x^2 = r^2 so plug that into the eq. and solve for u.

    after that im lost
  4. Oct 19, 2009 #3


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    That looks to me like you need Gauss' theorem:
    [tex]\int\int\int \nabla \cdot \vec{f} dV= \int\int \vec{f}\cdot \vec{n}dS[/tex]
    where the first integral is over a bounded region and the second over the surface of that area. The right side is the flow across the surface and so the integrand is the rate of heat flow across the surface.
  5. Oct 19, 2009 #4
    you mean divergence theorem? but what is the vector field?
  6. Oct 19, 2009 #5


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    can you make a vector heat field from your scalar tempertaure field?

    any differential conduction equations that might hep?
  7. Oct 19, 2009 #6
    the gradient.

    then what?

    use DIV theorem and spherical coordinates?
  8. Oct 19, 2009 #7


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    Homework Helper

    right so you get the heat field from something like

    [tex] \vec{q} = -k \nabla u [/tex]

    [tex] u(r) = \frac{3}{r} [/tex]

    now try the left hand integral given by Halls in Guass's theorem, i think this is effectively summing up all the heat generated by the star, which in steady state will also be the amount leaving the surface of the star (by the way does the form of the field look familiar? maybe you can think of the temperature as the potential)

    knowing the surface area of the star & due to the spherical symmetry of the heat field, you should be able to work back to the integrand on the right
    Last edited: Oct 19, 2009
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