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Evaluating Cartesian integral in polar coordinates

  1. Nov 20, 2016 #1
    1. The problem statement, all variables and given/known data
    Transform given integral in Cartesian coordinates to one in polar coordinates and evaluate polar integral.

    ##\int_{0}^3 \int_{0}^x \frac {dydx}{\sqrt(x^2+y^2)}##

    2. Relevant equations


    3. The attempt at a solution
    I drew out the region in the ##xy## plane and I know that ##0 \leq \theta \leq \frac{\pi}{4}##.
    For ##r##, I thought that it should be ##3 \leq r \leq 3\sqrt 2##

    So my polar integral is ##\int_{0}^ \frac{\pi}{4} \int_{3}^{3\sqrt 2} \frac{1}{\sqrt(r^2)} r drd\theta##.

    The answer I get from this integral is ##\frac{3\pi}{4} (\sqrt2 -1)##

    But the answer is ##3 \ln(\sqrt2+1)##.

    I have no idea how ##ln## appears in the answer.

    Where am I wrong?

    Thanks!
     
  2. jcsd
  3. Nov 20, 2016 #2

    Mark44

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    Assuming you are working the same problem as appears in your book, I agree with you. If this is the problem --
    ##\int_{0}^3 \int_{0}^x \frac {dydx}{x^2+y^2}##
    that would explain the book's answer.
     
  4. Nov 20, 2016 #3
    Ah, thank you! I did check the question again in the book, and I did not misread it.
     
  5. Nov 20, 2016 #4

    LCKurtz

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    Your limits for ##r## are incorrect. Your text's answer is correct for the problem as stated.
     
  6. Nov 20, 2016 #5
    Are both upper and lower ones incorrect? I don't know any other way to find the limits for ##r##. Any hints?

    Or since value of ##x## is constant, are the bounds in Cartesian coordinates something like ##3 \leq \sqrt{9+y^2} \leq 3\sqrt2##?
     
  7. Nov 20, 2016 #6

    LCKurtz

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    Both are incorrect. What does the region look like? Put your pencil on the origin and move it in the ##r## direction. What does it hit?
     
  8. Nov 20, 2016 #7
    I honestly do not know. In the ##xy## plane, the region is bounded by the line ##y=x##, the ##x## axis and the line ##x=3##. What does moving it in the ##r## direction mean? If I do so, for ##\theta=0##, I get ##0 \leq r \leq 3##, but for ##\theta=\frac{\pi}{4}##, I get ##0 \leq r \leq 3 \sqrt2##.

    I'm confused.
     
  9. Nov 20, 2016 #8

    LCKurtz

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    Well, at least you are thinking about it. You have noticed that if ##\theta = 0##, ##r## goes from ##0## to ##3## and for ##\theta = \frac \pi 4##, ##r## goes from ##0## to ##3\sqrt 2##. Both observations are correct. So what you have noticed is what ##r## goes to depends on what ##\theta## is. So if you take some in between value of ##\theta## and move ##r## in that direction, don't you hit the line ##x=3##? What is the equation of that in polar coordinates? What is ##r## on that line? It will depend on ##\theta##.
     
  10. Nov 20, 2016 #9
    Ah, I see. Is it ##0 \leq r \leq \frac{3}{cos\theta}##?
     
  11. Nov 20, 2016 #10

    LCKurtz

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    No. Answer the two questions I asked before (now in red). You have about 10 minutes before I hit the sack. [Edit:] I see you corrected it so now it's correct. Write it as ##3\sec\theta## and integrate it.
     
  12. Nov 20, 2016 #11
    Thanks! if it weren't correct, it would be too much pressure for me. 10 minutes! :wink:

    Anyway, thanks so much for your time, and good night!
     
  13. Nov 20, 2016 #12

    Mark44

    Staff: Mentor

    Apologies for my incorrect conclusion. r indeed does depend on ##\theta##.
     
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