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Integral calculus: plane areas in polar coordinates

  1. Mar 6, 2012 #1
    what is the area inside the graph of r=2sinθ and outside the graph of r=sinθ+cosθ?

    so i compute for the values of 'r',... but, i only got one intersection point which is (45°, 1.41).
    there must be two intersection points right? but i've only got one. what shall i do?
    i cannot compute for the area of the said region because i've only got one limit which is ∏/4.. what shall i do?

    please help..
    thanks a lot.. :smile:
     
  2. jcsd
  3. Mar 6, 2012 #2

    tiny-tim

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    hi delapcsoncruz! :smile:

    useful tip: multiply both sides by r, then convert to cartesian coordinates …

    that (amost) immediately gives you the positions of these circles :wink:
     
  4. Mar 6, 2012 #3

    Office_Shredder

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    Alternatively not that 5pi/4 is also an intersection point
     
  5. Mar 6, 2012 #4
    can you please give me of an example using your said useful tip.. please..
     
  6. Mar 6, 2012 #5
    the value of 'r' in 5pi/4 is -1.41 , so that is also equal to pi/4 which r is 1.41
     
  7. Mar 6, 2012 #6
    can you give me an example of your useful tip.. please..
     
  8. Mar 6, 2012 #7

    tiny-tim

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    try it with r = 2sinθ …

    what do you get? :smile:
     
  9. Mar 6, 2012 #8
    r^2=2rsin(theta)
     
  10. Mar 6, 2012 #9

    tiny-tim

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    now convert to cartesian (x and y)
     
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