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Is this correct?

  1. Nov 11, 2015 #1
    1. The problem statement, all variables and given/known data
    set up a double integral to find the area of the region inside the cardioid r=2-2sin θ and outside the circle r=1.

    2. Relevant equations


    3. The attempt at a solution
    setting the two equations equal to each other, I got theta = pi/6 , 5pi/6

    Area= ∫-5pi/6 to pi/6 ∫1 to 2-2sinθ rdrdθ

    I am not sure if my bounds for the lower angle is correct or not.

    any help is appreciated

    here is the graph:
    Snapshot.jpg
     
  2. jcsd
  3. Nov 11, 2015 #2

    SammyS

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    The -5π/6 is not correct.

    -5π/6 + 2π = 7π/6 is at the same location as -5π/6 .

    You need a negative angle which terminates at the same place as 5π/6 .
     
  4. Nov 11, 2015 #3

    Mark44

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    Your graph shows intersections for ##\theta = \pi/6## and ##\theta = 5\pi/6##. Above you have ##-5\pi/6##.
     
  5. Nov 12, 2015 #4

    LCKurtz

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    @qq545282501: Have you figured out the correct limit yet? What is your final answer?
     
  6. Nov 12, 2015 #5
    I think @SammyS is correct, -7pi/6 should work, I think it as it goes from -7pi/6 to pi/6, counterclockwise. but i dont understand the need to add 2pi, since 7pi/6 is the actual angle if you go clockwise from 0.
    I didnt understand what @Mark44 said, or he is misunderstanding my question.

    what do you think?
     
  7. Nov 12, 2015 #6

    SammyS

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    Yes, "it goes from -7pi/6 to pi/6, counterclockwise".

    If you are referring to my adding 2π, I was merely pointing out that -5π/6 is co-terminal with 7π/6 and that is incorrect for a limit of integration.
     
  8. Nov 12, 2015 #7
    oh ! i see, anyway, thank you all ! big Love
     
  9. Nov 12, 2015 #8

    Mark44

    Staff: Mentor

    What I said was that in your drawing, you show one of the angles as being ##5\pi/6##, but in your integral (below), you are using ##-5\pi/6## as a limit of integration.
     
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