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Math Help - trying to find area in polar form

  1. #1
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    trying to find area in polar form

    I just have a quick question about setting up the problem.
    Given r=1-1sin(theta)
    how do you find the angles to use for the upper and lower limits of the integral. I have been setting r=0 but it still requires some finesse, there has to be an easier or more proper way.

    thanks in advance.
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  2. #2
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    Quote Originally Posted by calculushelptx View Post
    I just have a quick question about setting up the problem.
    Given r=1-1sin(theta)
    how do you find the angles to use for the upper and lower limits of the integral. I have been setting r=0 but it still requires some finesse, there has to be an easier or more proper way.

    thanks in advance.
    What area are you trying to find?
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  3. #3
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    Find the area of the region lying between the inner and outer loops of the limacon r=1-2sin(theta). I know that you just set up an integration but I cannot figure out how you get the upper and lower limits.
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  4. #4
    MHF Contributor Calculus26's Avatar
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    If you are trying to find the area in general you look at the interval of theta which generates the graph. here you have a cardioid so theta varies from 0 to 2pi.

    if you have 2 graphs such as the cardioid r1 1+ cos(t) and the circle r = 3/2
    and want the area inside the cardioid but outside the circle then
    you would set r1 = r to find the limits of integration (here -pi/3 to pi/3)
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  5. #5
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    Here is a better example.

    How do I find the radian angles of cos2(theta)=r to find the area of one petal.
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  6. #6
    MHF Contributor Calculus26's Avatar
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    For for r = 1-2sin(t)

    knowing the graph is the whole story

    As with any polar graph use the usual suspects 0 - pi/2-pi 3pi/2 and 2pi

    and add the zeroes here pi/6 and 5pi/6 this gives you just the maxs, mins and zeroes which is all you need


    Half the outer loop is completed as theta varies from -p1/2 to pi/6-Multiply this integral by 2

    Half The inner loop is generated as theta varies pi/6 to pi/2 multiply by 2 and subtract this from previous
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  7. #7
    MHF Contributor Calculus26's Avatar
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    know your graph

    Its all knowing what to look for

    For cos(nt) and sin(nt) consider pi/2n Again these give the zeroes maxs and mins

    here pi/4. you get one half petal as t varies from 0 to pi/4
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