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Math Help - Volume generated by curve in one complete rotation

  1. #1
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    Volume generated by curve in one complete rotation

    Find the area bounded by the x-axis and the arc of the cycloid x=a(\theta-\sin\theta), y=a(1-\cos\theta) between \theta =0 and \theta=2\pi, rotated about the x-axis.

    My problem is that I don't have the formula for the volume for parametric equations.
    For Cartesian equations : V=\int ^a_b (\pi y^2) dx
    But I don't know how to adapt this for parametric equations.
    Thanks!
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  2. #2
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    Hello, arze!

    \text{The region bounded by the }x\text{-axis and the arc of the cycloid:}

    . . \begin{Bmatrix}x&=&a(\theta-\sin\theta) \\ <br />
y&=&a(1-\cos\theta)\end{Bmatrix}\:\text{ between }\theta =0\text{ and }\theta=2\pi

    \text{is rotated about the }x\text{-axis.}


    My problem is that I don't have the formula for the volume for parametric equations.

    For Cartesian equations: . \displaystyle V\:=\:\pi \int ^a_b  y^2\,dx

    But I don't know how to adapt this for parametric equations. . Yes, you do!

    We have: . \begin{Bmatrix}dx &=& a(1-\cos\theta)\,d\theta \\ y &=& a(1-\cos\theta) \end{Bmatrix}


    Substitute: . \displaystyle V \;=\;\pi \int^b_a y^2\,dx

    . . . . . . . . . . . \displaystyle =\;\pi \int^{2\pi}_0 \overbrace{a^2(1-\cos\theta)^2}^{y^2} \cdot \overbrace{a(1-\cos\theta)\,d\theta}^{dx}

    . . . . . . . . . . . \displaystyle =\;\pi a^3 \int^{2\pi}_0 (1-\cos\theta)^3\,d\theta

    Got it?
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