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Really easy question about boundaries.

  1. Feb 6, 2010 #1
    The ∂B(0,1) in R is the set {-1,1} so

    ∫f over ∂B(0,1) in R = f(1) - f(-1) ?

    Or would it be interpreted differently? This is a really basic question that for some reason I can't answer confidently for myself. I'm mainly dealing with averages and I'm trying to figure out what that would look like over the boundary of the ball in R1. Then, would the volume of that set just be 2?
  2. jcsd
  3. Feb 6, 2010 #2
    This is the integral of f over the subset {-1,1} of R, using the integration measure on R. The answer would therefore be zero, since the volume of {-1,1} is zero.

    If you for some reason define another integration measure on {-1,1}, e.g. by defining that Vol({-1})=Vol({1})=1, then the integral would reduce to a sum of two terms, i.e. f(-1)+f(1). But requires that you make this special definition of a new integration mesure.

    I suspect that your expression f(1)-f(-1) doesn't not make sense here. That would have been the result of integrating f'(x) over [-1,1], but this is not the case here.

  4. Feb 7, 2010 #3
    That's what I was afraid of, that the measure would be 0.
    I'm integrating an average over the boundary of a ball, so is that just not defined for R1 or am I missing something important in a definition somewhere?
  5. Feb 7, 2010 #4
    Yes, I think that in the case of R^1 you are suddenly encountering a topological difference, as compared to R^n for n>1. I did see a physicist once who derived a formula that depended on the surface area of n-spheres, and actually made special definition for the integral measure in the case of the "0-sphere" {-1,1} in order to make his result look simpler. Maybe you can do the same. Since you are doing averages, you are probably dividing by the "volume" of the domain over which you are integrating. In the case of the 0-sphere that would be 0. If you define som sort of limiting procedure to make in well-defined, you could end up with a finite and nonzero result for you integral over {-1,1}.

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