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Math Help - Limits of integration with polar coordinates.

  1. #1
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    Limits of integration with polar coordinates.

    Ok, I have the following problem find the area inside the cardioid

     r = 1 -sin\theta

    and outside the circle

    r = \frac{1}{2}

    I know you check for the limits of integration by finding where the two graphs intersect which is

    \frac{\pi}{6} and \frac{5\pi}{6}

    However they show how the solution is done im a trying to re work it and they say you can just take the area of half the region and multiply it by two which gives the these limits of integration

    \frac{\pi}{6} and- \frac{\pi}{2}

    How the hell that came up? I cant see where the right half the graph starts at -pi/2
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    Quote Originally Posted by RockHard View Post
    Ok, I have the following problem find the area inside the cardioid

     r = 1 -sin\theta

    and outside the circle

    r = \frac{1}{2}

    I know you check for the limits of integration by finding where the two graphs intersect which is

    \frac{\pi}{6} and \frac{5\pi}{6}

    However they show how the solution is done im a trying to re work it and they say you can just take the area of half the region and multiply it by two which gives the these limits of integration

    \frac{\pi}{6} and- \frac{\pi}{2}

    How the hell that came up? I cant see where the right half the graph starts at -pi/2

    Do you even know how this cardioid look like? Take a peek here:

    File:CardioidsLabeled.PNG - Wikipedia, the free encyclopedia

    Your cardioid is the green one, and then you can see why the area wanted can be divided when the angle is taken from Pi/6 to -pi/2 and multipled by two (it is symmetric with respect to the y-axis)

    Tonio
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  3. #3
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    Yes I have seen the graph
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  4. #4
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    Quote Originally Posted by RockHard View Post
    Yes I have seen the graph

    Good. Then you see now why they told you to do what they did and why it works.

    Tonio
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  5. #5
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    Would doing it from the limits of integrations found by the intersection poiints not work? 5pi/6 and pi/6
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  6. #6
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    Quote Originally Posted by RockHard View Post
    Would doing it from the limits of integrations found by the intersection poiints not work? 5pi/6 and pi/6

    Yes, but you have to go from below! Looking at the picture one sees that you better integrate from \frac{5\pi}{6}\,\,\to\,\,\frac{13\pi}{6}, to force the path to "run" below the x-axis (most of the time),
    If you do as you say the path will go above the x-axis and you'll be evaluating something wrong.

    Tonio
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