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Math Help - area between polar curve

  1. #1
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    area between polar curve

    Find the area of the region that lies inside the curve r=1 + cosΘ and outside of the curve r= 3cosΘ

    I set them equal to eachother and solved for Θ = pi/3, 5pi/3
    I then set up the integral from pi/3 to pi of (0.5)(1+cosΘ)^2 and got
    pi - [8+rad(3)]/8

    can anyone help me solve this problem

    thanks
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  2. #2
    Newbie Mirado's Avatar
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    Solved

    Ok, so what you need to do is to take the integral of the  r=1+cos\Theta minus  r=3cos\Theta

    So, it'll look like this:
     2\int_{\pi/3}^{\pi} (1+cos\Theta - 3cos\Theta) d\Theta
    2\int_{\pi/3}^{\pi} (1-2cos\Theta) d\Theta
    2 \times [(\int_{\pi/3}^{\pi} 1 d\Theta) - 2 (\int_{\pi/3}^{\pi} cos\Theta d\Theta)]
    2 \times [\Theta |_{\pi/3}^{\pi} - 2sin\Theta |_{\pi/3}^{\pi}]
    2[\pi - \pi/3 - 2(0-\sqrt{3}/2)]
    2[(2\pi)/3 + \sqrt{3}]
    (4\pi + 6\sqrt{3})/3

    there ya are, have fun
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  3. #3
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    I don't think that is the correct answer. I believe you need to use the formula
    integral of .5[f(x)^2 - g(x)^2]. Can anyone else guide me to the correct answer.
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  4. #4
    Newbie Mirado's Avatar
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    New York
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    solved for real

    You didn't like my work before? lol. Ok, so to do it correctly this time. The trick is that the function r=1+cos\theta will need to be integrated from \pi to \pi/3 whereas the other function r=3cos\theta needs to be integrated from \pi/2 to \pi/3

    This actually comes out very nicely:
    2 \times ([1/2 \times \int_{\pi/3}^{\pi} (1 + cos\theta)^2 d\theta]  - [1/2 \times \int_{\pi/3}^{\pi/2} (3cos\theta)^2 d\theta])
    \int_{\pi/3}^{\pi} (1 + 2cos\theta + cos^2\theta) d\theta]  - \int_{\pi/3}^{\pi/2} (9cos^2\theta) d\theta
    (3\theta/2 + 2sin\theta + sin(2\theta)/4) |_{\pi/3}^{\pi} - (9\theta/2 + 9sin(2\theta)/4) |_{\pi/3}^{\pi/2}
    3\pi/2 - \pi/2 - 9\sqrt{3}/8 - (9\pi/4 -6\pi/4 - 9\sqrt{3}/8)
    \pi/4

    there ya go. That's the correct formula and the correct answer.
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