Surface integral of scalar function

  1. 1. The problem statement, all variables and given/known data
    Find the mass of a spherical surface S of radius R such that at each point (x, y, z) in S the mass density is equal to the distance of (x, y, z) to some fixed point (x_0, y_0, z_0) in S.

    2. Relevant equations
    Integral of a scalar function over a surface.

    3. The attempt at a solution
    I was thinking about converting this into spherical coordinates, but I see no way of doing that nicely since the distance formula would get very messy. I am also assuming they are using the euclidean distance, since this is an intro multivariable course.

    I don't need help evaluating, just with getting it set up.

    This is from Vector Calculus, 5e. by Marsden and Tromba, 7.5 #9.
  2. jcsd
  3. It doesn't matter what the fixed point [tex](x_0, y_0, z_0)[/tex] is, so you can make a choice that makes spherical coordinates less horrible.
  4. So basically, choose (x_0, y_0, z_0) to be 0, and using spherical coordinates, the distance is
    [tex]\sqrt{(2Rsin(\phi)cos(\theta))^2 + (2Rsin(\phi)sin(\theta))^2} + (2Rcos(\phi))^2}[/tex]
    = 2R.

    So, [tex]\int_0^{2\pi} \int _0^\pi 2R*R^2sin(\phi) \, d\phi d\theta ?[/tex]

    [tex] = 8\pi*R^3 [/tex]

    Edit: The answer is supposed to be (16/3)Pi*R^3, so I lost a factor of 2/3 somewhere...
    Last edited: Apr 15, 2010
  5. Well, you probably shouldn't choose [tex](x_0, y_0, z_0)[/tex] to be the origin if your sphere [tex]S[/tex] is centered at the origin. Read the question again; [tex](x_0, y_0, z_0)[/tex] is supposed to lie on [tex]S[/tex].
  6. Cyosis

    Cyosis 1,495
    Homework Helper

    Make life easy for yourself and take [itex]\theta_0=\phi_0=0[/itex].
  7. Wow, I made that way harder than it had to be. Thank you both, that was driving me nuts!
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