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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


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    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


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    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


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    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: Dec 6, 2005
  7. Dec 6, 2005 #6

    Thanks Guys!
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