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Homework Help: Simple Surface Integral - Heat Flow on Surface of Star

  1. Feb 9, 2016 #1
    1. The problem statement, all variables and given/known data
    I have this problem in an online assignment. Someone told me the answer, so I already got it right, but I don't know why my logic leads me to the wrong answer. The problem:
    The temperature [itex]u[/itex] of a star of conductivity 1 is defined by [itex] u = \frac{1}{sqrt(x^2+y^2+z^2)}[/itex]. If the star is a sphere of radius 5, find the rate of heat flow outward across the surface of the star.

    2. Relevant equations
    Surface integral: [itex]\int \int_S f(x,y,z) * dS [/itex]
    Surface area of sphere = [itex] 4\pi r^2[/itex]
    The correct answer is [itex]4\pi[/itex]

    3. The attempt at a solution
    I tried this two ways and got the same answer both ways, unfortunately this answer isn't correct.
    First method was flat out doing the integral. Firstly, it is clear that [itex] u = \frac{1}{| \vec{r} |}[/itex]. However, the radius on the surface of the sphere is always 5, so [itex] u = \frac{1}{5} [/itex]. If we use the parametrization [tex] \vec{r} = (\rho \cos(\theta)\sin(\phi), \rho \sin(\theta) \sin(\phi), \rho \cos(\phi))[/tex] And dS is [itex]\rho^2 \sin(\phi) d\phi d\theta[/itex]. Once again, [itex] \rho = 5 [/itex] for every point on the surface of the star. Now we can fill out the surface integral:
    [tex] \int_0^{\pi} \int_0^{2\pi} \frac{1}{5} * 25\sin(\phi)d\theta d\phi \\
    = 5 \int_0^{\pi} \int_0^{2\pi} \sin(\phi) d\theta d\phi \\
    = 10\pi \int_0^{\pi} \sin(\phi) d\phi \\
    = 10\pi (-\cos(\phi))|_0^{\pi} = 20\pi [/tex]
    I knew that wasn't correct, so I thought of an alternative: At each point on the surface, the value of the function is just [itex] \frac{1}{r} = \frac{1}{5} [/itex]. So shouldn't the surface integral evaluate to the surface area of the sphere multiplied by the function u, and thus [itex] \frac{4\pi 5^2}{5} = 20\pi [/itex]?
    It seems like the value doesn't depend on the radius.
    Can someone explain this to me?
  2. jcsd
  3. Feb 10, 2016 #2

    Ray Vickson

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    Your ##u = u(x,y,z)## is the temperature at the point ##(x,y,z)## in the star. How is the rate of heat flow across the star's surface related to the function ##u##? In other words, how is your function ##f(x,y,z)## obtained from the function ##u(x,y,z)##?
    Last edited: Feb 10, 2016
  4. Feb 10, 2016 #3


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    In general:

    $$\iint_S f(x, y, z) \space dS = \iint_D f(\vec r(u, v)) |\vec r_u \times \vec r_v| \space dA$$

    For this particular problem we can adapt the notation:

    $$\iint_S u(x, y, z) \space dS = \iint_D u(\vec r(\theta, \phi)) |\vec r_{\theta} \times \vec r_{\phi}| \space dA$$

    You need to find a suitable ##\vec r(\theta, \phi)##, and project the sphere onto the x-y plane to obtain the region ##D##. Take ##\rho = 5## for the parameterization.
  5. Feb 10, 2016 #4

    Ray Vickson

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    Do we know that ##f = u##? I asked the OP that, but have received no response.
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