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Math Help - area bounded by the curves...

  1. #1
    Member grgrsanjay's Avatar
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    area bounded by the curves...

    Find The Area Bounded by the following curves
    x=a(\theta - \sin \theta  )
    y=a(1-\cos \theta
    0 \leqslant \theta \leqslant 2\pi


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  2. #2
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    Well, I would first like to point out that what you have written aren't curves but just one curve defined parametrically. It's called a cycloid.
    Anyway, I guess you'd like the area between it and the x-axis.
    Just take the integral as always, but rewrite with theta:
    \int^{2a\pi}_0 y\,dx=\int^{2\pi}_0 a(1-\cos \theta)\,d(a(t-\sin \theta))=\int^{2\pi}_0 a^2(1-\cos \theta)(1-\cos \theta)\,dt
    =\int^{2\pi}_0 a^2 (1-\cos \theta)^2\,dt

    Now you can solve this any way you like.
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  3. #3
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    Quote Originally Posted by grgrsanjay View Post
    Find The Area Bounded by the following curves
    x=a(\theta - \sin \theta  )
    y=a(1-\cos \theta
    0 \leqslant \theta \leqslant 2\pi


    THE LATEX IS AWESOME IN MHF NOW!!!
    I assume you mean the area in the first quadrant

    Just integrate

    \int_{0}^{2\pi}ydx=a^2\int_{0}^{2\pi}(1-\cos(\theta)^2dt
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  4. #4
    Member grgrsanjay's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    I assume you mean the area in the first quadrant

    Just integrate

    \int_{0}^{2\pi}ydx=a^2\int_{0}^{2\pi}(1-\cos(\theta)^2dt
    i couldn't understand why are we doing it?
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