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Homework Help: Integral with divergence thm

  1. Aug 14, 2010 #1
    1. The problem statement, all variables and given/known data
    Evaluate the integral
    [tex]\int\limits_{V=\infty} e^{-r} \left[ \nabla \cdot \frac {\widehat{r}} {r^2} \right] , d^3 x[/tex]

    2. Relevant equations
    Divergence theorem:
    [tex]\int\limits_{V} \left ( \nabla \cdot A \right ) \, d^3 x
    = \oint\limits_{S} A \cdot \, da}

    3. The attempt at a solution
    I know that I have to apply the div theorem somewhere, but this [tex]e^{-r}[/tex] is confusing and what does it mean if the lower limit V is infinity?
    I haven't seen the integral of [tex]\frac{1}{e^r} [/tex] before but I'm kinda guessing
    [tex] \int \frac{1}{e^r} \, dr
    = \frac{1}{e^r} \int \frac{1}{u} \frac{du}{e^r}
    = ln(e^r)
    = r
    where I used a substitution [tex]u=e^r[/tex] and [tex]du= e^r dr[/tex]
    Last edited: Aug 14, 2010
  2. jcsd
  3. Aug 15, 2010 #2


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    What is the divergence of vec(r)/r^2?

  4. Aug 17, 2010 #3
    It's defined as
    [tex]4 \pi \delta x \delta y \delta z [/tex]

    but then I don't know how to apply Stokes' (which I guess to use because of the [tex]d^3 x [/tex] and V in the integral. Could I split it into a triple integral and [tex]\delta x dx [/tex] at a time?
  5. Aug 17, 2010 #4

    Char. Limit

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    This is wrong.

    [tex]\frac{1}{e^r} = e^{-r}[/tex]

    [tex]\int e^{-r} dr[/tex]

    [tex]u=-r, du=-dr, -du=dr[/tex]

    [tex]\int -e^u du = -e^u = -e^{-r} = \frac{-1}{e^r}[/tex]

    The integral of e^-r isn't r, as that would imply that e^-r is a constant number.
  6. Aug 17, 2010 #5
    I agree, my above reasoning was useless

    Ok so I can integrate the [tex]e^-r[/tex] but I don't think that really matters when there's a delta in the integral... my main problem is how to solve a third order delta integral, probably using the Divergence theorem because of the third order and volume. So we have

    [tex]4 pi \int_{V=\infty} e^{-r} {\delta}^3 x z d^3 x[/tex]
  7. Aug 19, 2010 #6


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