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C4 integrating parametric equations question help please! watch

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    The diagram shows (sorry cant actually get the diagram, youll have to use an internet plotter or something :P) a curve with parametric equations x = t - sin(t), y = 1-cos(t)

    Find the area bounded by the x axis and one loop of the curve.

    At x axis, y = 0

    0 = 1- cos(t)
    cos(t) = 1
    t = 0

    When t = 0, x = 0 - sin 0 = 0 =/

    I know Ive got to sort out the limits of x and y (0.0 is on there but its not a loop obviously)

    Also how would you integrate that? (need some advice please)

    Thanks
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    Subscribing. I need help with integrating parametric equations too.
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    Think ive done it can someone please check my working?

     x = t-sin(t), y = 1-cos(t)

     I = \int ydx \int y\frac{dx}{dt}dt

     \frac{dx}{dt} = 1-cos(t)

     \rRghtarrow \int ydx = \int y(1-cos(t))dt

     = \int (1-cos(t))(1-cos(t))dt

     = /int (1-2cos(t) + cos^2(t))dt

     = \int (1-2cos(t) + 1-sin^2(t))dt

     1-2sin^2t = cos2t

     sin^2t = \frac{1-cos2t}{2} \rightarrow -sin^2t = \frac{-1+cos2t}{2}

     I = \int (1-2cos(t) - \frac{1+cos2t}{2})dt

     = \int dt - 2 \int cos(t)dt - \frac{1}{2} \int (1+cos2t)dt

     = t - 2sin(t) - \frac{1}{2}(t + \frac{1}{2}sin2t)

     = t - 2sin(t) - \frac{t}{2} - \frac{1}{4}sin2t + C

    Limits for area:

     y = 0 \rightarrow 0 = 1-cos(t) \rightarrow cos(t) = 1 \rightarrow t = 0

     t = 2\pi (from t = 2pi-t)

    Limits are now t = 0 and t = 2\pi (dont know how to put limits on integration
     A = \int(1-cos(t))^2dt 

= (t-2sin(t) - \frac{t}{2}  - \frac{1}{4} sin2t) from t = 0 to 2\pi

     = (2\pi - 2sin(2\pi) - \pi - \frac{1}{4} sin(4\pi)) - (0)

     = \pi

    Thanks (sorry about the dodgy latex near the bottom)
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    (Original post by Chemhistorian)
    ...
    I'll assume everything else in error is a typo.

    See attached.
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