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

Tricky Cartesian to Polar Change of Variables Integral

  1. Dec 6, 2005 #1
    Hmm, I can't seem to get this double integral transformation:

    int(limits of integration are 0 to 3) int (limits of int are 0 to x) of (dy dx)/(x^2 + y^2)^(1/2)

    and i need to switch it to polar coordinates and then evaluate the polar double integral.

    i sketched the region over which i am integrating and it isn't a circular region (which is why I guess this was assigned).

    but i cant the limits of integration for the theta angle. the radius goes from 0 to 3, but i cant get the theta angle limits of integration.

    the answer is 3ln(sqrt(2) + 1))
    i can see where the 3 comes from, but i dont know where i would get the ln term.

    thank you in advance
  2. jcsd
  3. Dec 6, 2005 #2


    User Avatar
    Science Advisor
    Homework Helper

    Have you sketched the region in the xy-plane? If you do you'll find that the radius depends on the angle (since it's not a circular region like you said). For example, r goes from 0 to 3 on the x-axis (theta=0), but it goes from 0 to sqrt(18) for if theta=Pi/4.
  4. Dec 6, 2005 #3
    yeah , i had that the radius goes from 0 to sqrt(18) and that theta goes from 0 to 45*
    however, after you do the change of variables, (ie, dydx = r dr d(theta))
    im double integrating 1, and i get 3*sqrt(2)/2, but the real answer is 3 ln(sqrt(2) + 1))
    any other hints or clues?
    Last edited: Dec 6, 2005
  5. Dec 6, 2005 #4


    User Avatar
    Science Advisor
    Homework Helper

    If you let r go from 0 to sqrt(18) regardless of theta, you will get a section of a disk. r is dependent upon theta.

    You are right to let theta vary from 0 to Pi/4. But given an angle theta, between which values should r vary? Hint: look at picture.
  6. Dec 6, 2005 #5


    User Avatar
    Science Advisor

    The horizontal line y= 3 is [itex]r sin(\theta)= 3[/itex] or [itex]r= \frac{3}{cos(\theta)}[/itex]. The integral is
    [tex]\int_{\theta= 0}^{\frac{\pi}{2}} \int_{r= 0}^{\frac{3}{cos(\theta)}}drd\theta= 3\int_{\theta= 0}^{\frac{\pi}{2}}\frac{d\theta}{cos(\theta)}[/tex]

    To do that integral, multiply numerator and denominator by [itex]cos(\theta)[/itex] and let [itex]u= cos(\theta)[/itex].
    Last edited by a moderator: Dec 6, 2005
  7. Dec 6, 2005 #6

    Thanks Guys!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook