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Double integral for loop of rose r=cos2θ

  1. Apr 25, 2013 #1
    1. The problem statement, all variables and given/known data

    attachment.php?attachmentid=58209&stc=1&d=1366915312.jpg

    2. Relevant equations



    3. The attempt at a solution
    Solution is given.

    I don't understand how +-∏/4 is found as a range for θ
    Also why is 0 <= r <= cos2θ

    r is always r which is defined as cos2θ
     

    Attached Files:

  2. jcsd
  3. Apr 25, 2013 #2

    LCKurtz

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    Where? I don't see it.

    Did you draw the graph or have one to look at?

    That is r on the curve. If you want the area enclosed, r goes from r=0 to r on the curve, no?
     
    Last edited by a moderator: May 6, 2017
  4. Apr 25, 2013 #3

    LCKurtz

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    I see the graph has now appeared. Plot a few points by hand, say for ##\theta = -\frac {\pi}4,\,-\frac {\pi}6,\,0,\,\frac {\pi}6,\,\frac {\pi}4## and see if you don't get that right hand loop.
     
  5. Apr 27, 2013 #4
    Okay I got it. I haven't done much work with polar co-ordinates so I didn't really process that x^2+y^2=r and the circles radius is now going to vary according to cos....

    Thanks :)
     
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