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Math Help - Area of region bounded by polar equation

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    Area of region bounded by polar equation

    Find the area of the region that is outside the graph of the first equation and inside the graph of the second equation.
    1. r=2
    2. r= 4cosx


    (x = theta)
    Last edited by clev88; July 29th 2009 at 11:05 PM.
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  2. #2
    Senior Member DeMath's Avatar
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    Quote Originally Posted by clev88 View Post
    Find the area of the region that is outside the graph of the first equation and inside the graph of the second equation.
    1. r=2
    2. r= 4cosx

    (x = theta)
    So {r_1} = 2,{\text{ }}{r_2} = 4\cos \theta ,{\text{ }}{S_{{r_2}}}\backslash \left\{ {{S_{{r_1}}} \cap {S_{{r_2}}}} \right\} = ?

    Let S = {S_{{r_2}}}\backslash \left\{ {{S_{{r_1}}} \cap {S_{{r_2}}}} \right\}.

    Find the intersection points of polar curves

    {r_1} = {r_2} \Leftrightarrow 2 = 4\cos \theta  \Leftrightarrow \cos \theta  = \frac{1}{2} \Leftrightarrow \theta  =  \pm \frac{\pi }{3}{\text{ }}\left( {k = 0} \right).

    Then, taking into account the symmetry, we have

    S = 2\left[ {\frac{1}{2}\int\limits_0^{\pi /3} {r_2^2d\theta }  - \frac{1}{2}\int\limits_0^{\pi /3} {r_1^2d\theta } } \right] = \int\limits_0^{\pi /3} {\left( {r_2^2 - r_1^2} \right)d\theta }  = \int\limits_0^{\pi /3} {\left( {{{\left( {4\cos \theta } \right)}^2} - {2^2}} \right)d\theta }  =

    = \int\limits_0^{\pi /3} {\left( {16{{\cos }^2}\theta  - 4} \right)d\theta }  = \int\limits_0^{\pi /3} {\left( {8\left( {1 + \cos 2\theta } \right) - 4} \right)d\theta }  = 4\int\limits_0^{\pi /3} {\left( {1 + 2\cos 2\theta } \right)d\theta }  =

    = 4\left. {\left( {\theta  + \sin 2\theta } \right)} \right|_0^{\pi /3} = 4\left( {\frac{\pi }{3} + \frac{{\sqrt 3 }}{2}} \right) = \frac{{4\pi }}{3} + 2\sqrt 3 .


    See picture

    Last edited by DeMath; July 30th 2009 at 01:57 AM.
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    Thumbs up thanks

    Thank you this helps!
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