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  1. MHB Master
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    #1


    OK just seeing if this is setup OK
    before I pursue all the steps
    I thot adding areas would be easier

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    #2
    I see it like so ...

    $\displaystyle \int_0^\pi \dfrac{8^2}{2} \, d\theta + \int_\pi^{2\pi} \dfrac{[8(1+\sin{\theta})]^2}{2} \, d\theta$

    ... note there are also opportunities to use symmetry.
    Last edited by skeeter; September 6th, 2017 at 23:10.

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    #3 Thread Author
    Quote Originally Posted by skeeter View Post
    I see it like so ...

    $\displaystyle \int_0^\pi \dfrac{8^2}{2} \, d\theta + \int_\pi^{2\pi} \dfrac{[8(1+\sin{\theta})]^2}{2} \, d\theta$

    ... note there are also opportunities to use symmetry.
    I'll try the symmetry .... half the limits mult by 2

    $\displaystyle A=2\left[\int_0^{\pi/2} 64\, d\theta
    + \int_{3\pi/2}^{2\pi} [8(1+\sin{\theta})]^2 \, d\theta\right]$

    $128\left[\displaystyle \left[\theta\right]_0^{\pi/2}
    +\left[\theta-2cos\theta-\dfrac{\sin\left(2x\right)-2x}{4}\right]_{3\pi/2}^{2\pi}\right]$

    sorry I just can't get this the bk ans is $16(5\pi - 8)$
    Last edited by karush; September 7th, 2017 at 14:58.

  4. Pessimist Singularitarian
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    #4
    I would use symmetry and known the area of a semicircle to write:

    $ \displaystyle A=\frac{8^2}{2}\left(\pi+2\int_{-\frac{\pi}{2}}^{0}\left(1+\sin(\theta)\right)^2\,d\theta\right)=32\left(\pi+\frac{3}{2}\pi-4\right)=16(5\pi-8)$

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