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Math Help - limacon area

  1. #1
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    limacon area

    find the area inside the larger loop and outside the smaller loop

     r = \frac{\sqrt{3}}{2} + cos(\theta)

    I have found the zeros to be at  \frac{5\pi}{6} and \frac{7\pi}{6}

    and have set up my integral as such to exclude the area between those two points

     \frac{1}{2} \int_0^{\frac{5\pi}{6}} (\frac{\sqrt{3}}{2} + cos(\theta))^2 d\theta + \frac{1}{2} \int_{\frac{7\pi}{6}}^{2\pi} (\frac{\sqrt{3}}{2} + cos(\theta))^2 d\theta

    Resulting in 3.9220114003

    This is incorrect, where am I going wrong?

    EDIT:

    I tried integrating in a different manor

     2(\frac{1}{2} \int_0^{\frac{5\pi}{6}} (\frac{\sqrt{3}}{2} + cos(\theta))^2 d\theta - \frac{1}{2} \int_{\frac{5\pi}{6}}^{\pi} (\frac{\sqrt{3}}{2} + cos(\theta))^2 d\theta)

    This was correct, but I do not understand why the first solution didnt work
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  2. #2
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    Quote Originally Posted by silencecloak View Post
    find the area inside the larger loop and outside the smaller loop

     r = \frac{\sqrt{3}}{2} + cos(\theta)

    I have found the zeros to be at  \frac{5\pi}{6} and \frac{7\pi}{6}

    and have set up my integral as such to exclude the area between those two points

     \frac{1}{2} \int_0^{\frac{5\pi}{6}} (\frac{\sqrt{3}}{2} + cos(\theta))^2 d\theta + \frac{1}{2} \int_{\frac{7\pi}{6}}^{2\pi} (\frac{\sqrt{3}}{2} + cos(\theta))^2 d\theta

    Resulting in 3.9220114003

    This is incorrect, where am I going wrong?

    EDIT:

    I tried integrating in a different manor

     2(\frac{1}{2} \int_0^{\frac{5\pi}{6}} (\frac{\sqrt{3}}{2} + cos(\theta))^2 d\theta - \frac{1}{2} \int_{\frac{5\pi}{6}}^{\pi} (\frac{\sqrt{3}}{2} + cos(\theta))^2 d\theta)

    This was correct, but I do not understand why the first solution didnt work
    when you integrate from 0 to \frac{5\pi}{6} , the upper half of the inner loop gets included because of the overlap.

    same idea for integrating from \frac{7\pi}{6} to 2\pi ... the lower half of the inner loop gets included because of the overlap.
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