1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Change of limits when integrating with polar coordinates

  1. Oct 22, 2009 #1
    1. The problem statement, all variables and given/known data


    ∫ e^([tex]\pi[/tex]x^2) dx, with limits -∞ to ∞


    2. Relevant equations

    ∫∫ dxdy = ∫∫ rdrdθ



    3. The attempt at a solution

    Hi, here's what I've done so far:

    Introduce a dummy variable y to get

    ∫∫ e^[tex]\pi[/tex](x^2 + y^2) dxdy, with limits -∞ to ∞ for both dx and dy

    Introduce polar coordinates: x^2 + y^2 = r^2

    The equation becomes:

    ∫∫ e^([tex]\pi[/tex]r^2) rdrdθ

    But I don't know how to change the limits. Am I right in that the r limits stay the same and the θ limits change to [tex]\pi[/tex]/2 and -[tex]\pi[/tex]/2?

    If this is right, when I integrate the first part, I end up with 0. Is this correct?
    Thanks for any help.

    P.S. I don't know why the pi is higher than the other figures, but it's meant to be at the same level!
     
  2. jcsd
  3. Oct 22, 2009 #2
  4. Oct 23, 2009 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Well, first of all, the integral you give will not converge. [itex]e^{\pi x^2}[/itex] go to infinity to fast at each end. I am going to assume you meant [itex]e^{-\pi x^2}[/itex]
    Since your integration includes the entire plane, you need for r to go from 0 to [itex]\infty[/itex] and [itex]\theta[/b] to go from 0 to [itex]2\pi[/itex].

    But you might find it easier to use the fact that [itex]\int_{-\infty}^\infty e^{-\pi x^2}dx= 2\int_0^\infty e^{-\pi x^2}dx[/itex] so that the double integral is restricted to the first quadrant. r still goes from 0 to [itex]\infty[/itex] but [itex]\theta[/itex] goes from 0 to [itex]\pi/2[/itex].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Change of limits when integrating with polar coordinates
Loading...