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Math Help - Area under a curve with parametric equations? help plz.

  1. #1
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    Exclamation Area under a curve with parametric equations? help plz.

    Hello mates, there's this wild calculus problem in my text that I can't figure out.

    I have to find the area under this parametric curve with
    x = 6 ( T - Sin(T) )
    y = 6 ( 1 - Cos(T) )

    and 0 \le T \le 2\pi

    I tried solving both for T but I don't think that was a good first step.
    Thanks for any help, I really appreciate it.

    I GOT HERE and said UGH: I solved y =6 ( 1 - Cos(T) ) for T and pluged T into x = 6 ( T - Sin(T) )
    \theta = arccos( 1 - \frac{y}{6})
     x = 6 ( arccos( 1 - \frac{y}{6}) - sin(arccos( 1 - \frac{y}{6})))
    Yeah, Kinda lost. Any help would be great, thanks a lot for looking guys.
    Last edited by Intsecxtanx; November 19th 2009 at 05:43 AM.
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  2. #2
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    Hello, Intsecxtanx!

    Find the area under this parametric curve with:

    . . \begin{array}{ccc}x &=& 6 (\theta - \sin\theta) \\ y &=& 6 (1 - \cos\theta)\end{array}\quad 0\le \theta \le 2\pi

    Recall the ancient formula: . A \;=\;\int^b_a y\,dx

    We have: . \begin{array}{ccc}y &=& 6(1-\cos\theta) \\ dx &=& 6(1 - \cos\theta)\,dx \end{array}

    Substitute: . A \;=\;\int^{2\pi}_0 \underbrace{6(1-\cos\theta)}_y\cdot \underbrace{6(1-\cos\theta)\,d\theta}_{dx} \;=\;36\int^{2\pi}_0(1-\cos\theta)^2d\theta

    Got it?

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  3. #3
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    Thumbs up

    ugh it makes perfect sense now, it's so obvious and i was staring at it for so long, thanks math god!!!!
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