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Converting volume elements to area elements

  1. Dec 13, 2012 #1
    1. The problem statement, all variables and given/known data

    I need to evaluate an integral of the form:

    [itex]h(\vec{x}) = \int d^3\vec{x}' \frac{\delta(r' - R)}{|\vec{x}-\vec{x}'|}[/itex]

    where [itex]r' = |\vec{x}'| = \sqrt{x'^2 + y'^2 + z'^2}[/itex].

    2. Relevant equations

    The above, and maybe the fact that [itex]\delta(g(x)) = \sum_i \frac{\delta(x - x_i)}{|g'(x_i)|}[/itex] where [itex]x_i[/itex] are the roots of g.

    3. The attempt at a solution

    I can solve the problem with some physical motivation as follows:

    The given integral is the same as what you would do if you wanted to calculate the electrostatic potential of a charged spherical shell: [itex] V(\vec{x}) = \frac{1}{4\pi\epsilon_0}\int d\tau' \frac{\rho(\vec{x}')}{|\vec{x}-\vec{x}'|}[/itex] where [itex]d\tau[/itex] is the volume element and [itex]\rho(\vec{x}') = \delta(r' - R)[/itex]. In this case, the delta function restricts the "charge density" rho to the surface of the shell. The same physical problem can be treated with the formula for the electrostatic potential of a surface charge distribution: [itex] V(\vec{x}) = \frac{1}{4\pi\epsilon_0}\int da' \frac{\sigma(\vec{x}')}{|\vec{x}-\vec{x}'|}[/itex] where [itex]da[/itex] is the area element and, in this case, we can take [itex]\sigma = 1[/itex]. This integral can now be computed straightforwardly in spherical coordinates, e.g. Griffiths "Introduction to Electrodynamics" example 2.7, yielding [itex]h(\vec{x}) = 4\pi R^2 / |\vec{x}|[/itex].

    However, I would like to be able to do this calculation without having to use physical intuition to interpret a volume integral with a delta function as being equivalent to a surface integral without the delta function; that is, I just want to explicitly integrate out one of x,y, or z with the delta function and then go from there. How do I do this? I've tried using the equation I gave in (2), but it just produces a giant mess from which I don't know how to proceed.

    I guess an equivalent question would be: given the expression above for the electrostatic potential of a volume distribution (which is really a surface dist.) of charge with [itex]\rho(\vec{x}) = \sigma \cdot \delta(f(\vec{x}))[/itex], how do you rigorously derive the expression for a surface distribution?
     
  2. jcsd
  3. Dec 14, 2012 #2

    HallsofIvy

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    It sounds like you need the "divergence theorem":
    [tex]\int\int_V\int \nabla\cdot \vec{f} dV= \int_S\int \vec{f}\cdot d\vec{S}[/tex]
    where S is the surface bounding volume V.
     
  4. Dec 14, 2012 #3
    Can you expand on that? I know what the divergence theorem is, but there are no divergences in this problem. My question isn't: "given an integral over a region, how do you convert it to an integral on the boundary"? It's: "given an integral over all space but over a delta function, how do you convert it to an integral without the delta function over the surface that the delta function restricts to?"
     
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