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Integrating over a bounded surface

  1. Jun 12, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the average value of z for a the spherical surface of radius R that resides above the x-y plane.


    2. Relevant equations
    Equation of a sphere
    [tex]x^2+y^2+z^2 = R^2[/tex]


    3. The attempt at a solution
    I rearrange the equation above and do a double integral

    [tex]z_{total} = \int \int \sqrt{ R^2 - x^2-y^2} dx\, dy[/tex]

    To get the aver value of z, I devide the whole thing by the surface area in the x-y plane that is bounded by the sphere.

    [tex]z_{average} = \frac{1}{\pi R^2} \int \int \sqrt{ R^2 - x^2-y^2} dx\, dy[/tex]

    I tried setting the integration limits for x to 0 and [tex]\sqrt{R^2-y^2}[/tex], and y to 0 and R(and I multiplied the whole thing by 4 since the integration is only being taken over a quarter of the surface).

    [tex]z_{average} = \frac{4}{\pi R^2} \int_{0}^{R} \int_{0}^{\sqrt{R^2-y^2}} \sqrt{ R^2 - x^2-y^2} dx\, dy[/tex]

    When I try to solve this, I get some crazy number with infinity in it. I'm not sure where I'm going wrong.
    1. The problem statement, all variables and given/known data
     
  2. jcsd
  3. Jun 12, 2010 #2

    lanedance

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    depending on how you want to average I'm not sure your integral it right form the get go...

    I would think you would want to weight it by surface area, so the average should be somethng like:
    [tex] \frac{\int z.dA}{\int dA}[/tex]

    the integral you've written down is over circle, so doesn't give the same weighting, in this case dA is not equal to dxdy

    either way changing to spherical coordinates will save you a lot of time...
     
  4. Jun 12, 2010 #3

    HallsofIvy

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    Did you consider changing to "polar coordinates"?

    Then your numerator becomes [tex]\int_{r= 0}^R\int_{\theta= 0}^{2\pi}\sqrt{R^2- r^2} rd\theta dr[/tex].

    Your denominator is just the surface area of a hemi-sphere of radius R: [itex]2\pi R^2[/itex].

    (Edit: I have added an "r" to the integral. The differential of area in polar coordinates is "[itex] r d\theta dr[/itex]", not "[itex]d\theta dr[/itex]"!)
     
    Last edited: Jun 13, 2010
  5. Jun 12, 2010 #4

    lanedance

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    [tex] \int_{r= 0}^R\int_{\theta= 0}^{2\pi}\sqrt{R^2- r^2}d\theta dr = \int_{r= 0}^R\int_{\theta= 0}^{2\pi}z(r)d\theta dr[/tex]

    the point i'm making is if you perform that integral without the z(r), you do not get the correct surface area of the hemi-sphere
    [tex] \int_{r= 0}^R\int_{\theta= 0}^{2\pi}d\theta dr \neq 2\pi R^2[/tex]
     
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