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Translating integration domain

  1. Jun 3, 2008 #1
    If [tex]f:\mathbb{R}^3\to\mathbb{R}[/tex] is a continuous function, [tex]x_0\in\mathbb{R}^3[/tex] a fixed point, [tex]r>0[/tex] a fixed radius, and [tex]n\in\mathbb{R}^3[/tex] a fixed vector satisfying [tex]|n|=1[/tex], then is the equation

    [tex]
    \int\limits_{B(x_0 + \alpha n, r)} d^3x\; f(x) \;=\; \int\limits_{B(x_0, r)} d^3x\; f(x) \;+\; \frac{\alpha}{r} \int\limits_{\partial B(x_0,r)} d^2x\; ((x-x_0)\cdot n) f(x) \;+\; O(\alpha^2),\quad\quad\alpha\in\mathbb{R}
    [/tex]

    true? I convinced myself of this somehow, but I'm still feeling unsure. I don't know how to deal with equations like this rigorously. There are other problems of similar nature, where the integration domain is changed a little bit, and then it is somehow possible to write the change as a functional of the restriction of the integrand onto the boundary.

    The B notation means the ball

    [tex]
    B(x_0,r) = \{x\in\mathbb{R}^3\;|\;|x-x_0|<r\},
    [/tex]

    and [tex]\partial[/tex] is the boundary,

    [tex]
    \partial B(x_0,r) = \{x\in\mathbb{R}^3\;|\;|x-x_0|=r\}.
    [/tex]
     
    Last edited: Jun 3, 2008
  2. jcsd
  3. Jun 4, 2008 #2
    Later in my calculations I encountered a situation where some factor, that was supposed to be 1, was 3/2. I cannot yet know where the mistake is, but this formula could be one potential place for it.
     
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