Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Concerning the Gaussion integral in polar coordinates

  1. Mar 9, 2008 #1
    I'm looking at the proof for the Gaussion integral in polar coordinates and I don´t understand why theta reaches from 0 to 2pi in the integral since you can´t get a negative value out of an exponential function (and therefor the exponential function is never in the 3rd and fourth quadrant, which spans pi to 2pi). Here's the proof

    http://upload.wikimedia.org/math/8/5/b/85bb26bab98e69735c439dcfee9807d6.png part 1

    http://upload.wikimedia.org/math/f/1/0/f10de92bc974482b9f714bdaa7fda10d.png part 2
     
  2. jcsd
  3. Mar 9, 2008 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi naggy! :smile:

    (btw, if you type alt-p, it prints π)

    It has nothing to do with the integrand (in this case, an exponential).

    It's only conerned with the limits of integration, in changing from ∫∫dxdy to ∫∫drdtheta.

    ∫∫dxdy was over the whole plane.

    So ∫∫drdtheta must be also. That means over all r and all theta.

    r can't be negative, so theta has to go all the way round, from 0 to 2π. :smile:
     
  4. Mar 9, 2008 #3
    When you change it to a double integral, you go from [tex]R^2[/tex] to [tex]R^3[/tex], and [tex]e^{-x^2}[/tex] and [tex]e^{-y^2}[/tex] are defined for all x and all y, so in effect both exponential functions are in all two dimensional quadrants.
     
  5. Mar 9, 2008 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    It is the value of the exponential function that is never never negative, not the variables.
     
  6. Mar 9, 2008 #5

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    … integrating over neverland …

    naggy, he means [tex]R^2[/tex] (… you knew that, didn't you? … :smile:)
    Do you mean never never never never negative, or never never never never never negative? :confused:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Concerning the Gaussion integral in polar coordinates
Loading...