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Integration of Gaussian equation

  1. Aug 24, 2004 #1

    Need help desperately!

    I am trying to figure out the area under a gaussian cone by finding the integral of

    2PIArEXP[-(r^2)/(2sigma^2)] dr

    My supervisor thought it is

    2PI [A sigma^2 EXP(-(r^2)/(2 sigma^2)] Between 0 and infinity

    and he came up with the answer

    2 PI A sigma^2

    I hope you can understand that!

    I don't think that his integration is right to get to the second step there. But I'm not sure why. I don't think sigma will come down like that. Please help me. I haven't done maths like this for a few years and I'm very rusty at it!


  2. jcsd
  3. Aug 24, 2004 #2
    I think he has taken
    2PIArEXP[-(r^2)/(2sigma^2)] dr

    -- AI
  4. Aug 24, 2004 #3


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    To find [itex]2\pi {A} \int_0^{\infinity}{r e^{\frac{-r^2}{2\sigma^2}}dr}[/itex], let [itex] u= \frac{r^2}{2\sigma^2}[/itex]. Then [itex]du= \frac{r}{\sigma^2}[/itex] so [itex]rdr= \sigma^2du[/itex]. When r=0, u= 0, when r= infinity, u= infinity so the integral becomes [itex]2\pi {A} \sigma\int_{o}^{\infinity}{e^{-u} du}[/itex].
    That IS [itex]-2\pi {A} \sigma e^{-u}[/itex] evaluated between u=0 and u= infinity:
    [itex]2\pi {A} \sigma[/itex].
    Last edited: Aug 24, 2004
  5. Aug 24, 2004 #4
    rdr = sigma^2 du
    the final answer is 2*pi*A*sigma^2

    -- AI
  6. Aug 24, 2004 #5
    I see what he has done now!

    Thanks so much. You guys are lifesavers!

  7. Aug 24, 2004 #6
    Thanks for your help. I am still a little confused. As I said I haven't done integration like this for a long time!

    I can't see how you substituted [itex]rdr= \sigma^2du[/itex] back in to get [itex]2\pi {A} \sigma\int_{o}^{\infinity}{e^{-u} du}[/itex]. I can see that when r=0, u= 0, when r= infinity, u= infinity. That is fine but where does the sigma come from and where does the r go?

  8. Aug 24, 2004 #7
    Look at the substitution hurkyl makes ....
    u = r^2/(2*sigma^2)
    find du/dr

    -- AI
  9. Aug 24, 2004 #8
    I understand the substitution that is made and I can find

    du/dr = r/sigma^2

    That can then be rearranged to

    rdr = sigma^2du

    How then do I get to this!

    [itex]2\pi {A} \sigma\int_{o}^{\infinity}{e^{-u} du}[/itex]

    Forgive me I think it is part of the substiution method that I don't understand. I haven't done it for many years.
  10. Aug 24, 2004 #9
    Thanks for all your hlep. I have been playing around with it and I understand it now!

    It was not remembering how to use the substitution method that was causing me problems!

    Thansk again for all your help!

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