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Math Help - Area of limacon

  1. #1
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    Area of limacon

    Find area of limacon.
    <br />
r = 4 + 2 cos\Theta<br />

    Well I cant seem to get the latex to work for the integral that I set up so here it is rougly.

    Upper limit Pi
    Lower limit 0

    1/2 int (4+2cos theta)^2 dtheta

    by symmetry I multiply by 2 which gets rid of the 1/2 outside

    Then

    foiling

    int (16) + int (16 cos theta) + int (4 cos^2theta) dtheta

    16 theta - 16 sin theta + 4 int 1 + cos(2theta)/2 dtheta


    Am I going the right way with this integral? If not where am I going wrong?
    Thanks
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  2. #2
    MHF Contributor Calculus26's Avatar
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    Notice you complete the outer loop between 0 and 2pi/3 (where r=0)

    By integrating from 0 to pi you get the area inside the limacon plus the area of the innerloop
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  3. #3
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    Quote Originally Posted by Calculus26 View Post
    Notice you complete the outer loop between 0 and 2pi/3 (where r=0)

    By integrating from 0 to pi you get the area inside the limacon plus the area of the innerloop

    Well the question was find the area inside of the limacon. So isn't that what we want to be doing 0 to Pi? When I graph this limacon I do not see any loops since its an oval limacon.
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  4. #4
    MHF Contributor Calculus26's Avatar
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    I was reading 2+4cos(theta) my apologies

    The integral looks good
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  5. #5
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    Quote Originally Posted by Calculus26 View Post
    I was reading 2+4cos(theta) my apologies

    The integral looks good
    ok for some reason I am not getting the correct answer which is

    18 Pi

    so again..from above

    16 theta - 16 sin theta + 8 int(1+cos(2theta) )

    16 theta - 16 sin theta + 8 theta - 4 sin (2 theta)

    I have integrated incorrectly somewhere but I am not sure where..
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  6. #6
    MHF Contributor Calculus26's Avatar
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    4 int ((1 + cos(2theta))/2 dtheta = 2theta + sin(theta) yields 2pi

    you multiplied by 2 instead of dividing by 2

    added to your previous 16pi you get 18 pi

    In your original integral you had 4 int (1 + cos(2theta)/2 dtheta

    the 1 was divided by 2 also
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  7. #7
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    Quote Originally Posted by Calculus26 View Post
    4 int ((1 + cos(2theta))/2 dtheta = 2theta + sin(theta) yields 2pi

    you multiplied by 2 instead of dividing by 2

    added to your previous 16pi you get 18 pi

    In your original integral you had 4 int (1 + cos(2theta)/2 dtheta

    the 1 was divided by 2 also

    Yes thank you for your time!
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