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Math Help - Area between curves

  1. #1
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    Area between curves

    How would i find the area between the curves y=cosx, the y axis and the lines y =1/2 and y=srt3/2......i keep getting (pi(2-srt3)-6(-1+srt3))/12 but i doubt this is right can someone do it please thanks nath
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  2. #2
    TD!
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    Your answer is correct, apart from the sign mistake (you get approx -0.296 while it should be +0.296).
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  3. #3
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    Hello, Nath!

    Your answrer is negative . . . otherwise, excellent work!

    How would i find the area between the curves y = \cos x, the y-axis
    and the lines y = \frac{1}{2} and y = \frac{\sqrt{3}}{2}
    i keep getting: \frac{\pi(2 - \sqrt{3}) - 6(-1 + \sqrt{3})}{12} . . . but i doubt this is right.
    Code:
            |
            * *
        _   |     *
       √3/2 + - - - -*- -
            |::::::::::*
        1/2 + - - - - - * - -
            |
          --+------------*---
            |           π/2
    I did it "sideways" . . . with respect to y.

    We have: . y = \cos x\quad\Rightarrow\quad x = \arccos y

    The area is: . A \;= \;\int^{\frac{\sqrt{3}}{2}}_{\frac{1}{2}} \arccos y\ dy

    This integral can be done by-parts . . . or from a standard formula.

    We get: . A \;= \;y\cdot\arccos y - \sqrt{1 - y^2}\:\bigg|^{\frac{\sqrt{3}}{2}}_{\frac{1}{2}}

    We have: . \left[\frac{\sqrt{3}}{2}\cdot\arccos\frac{\sqrt{3}}{2} - \sqrt{1 - \left(\frac{\sqrt{3}}{2}\right)^2} \right]  - \left[\frac{1}{2}\cdot\arccos\frac{1}{2} - \sqrt{1 - \left(\frac{1}{2}\right)^2}\right]

    . . . = \;\left[\frac{\sqrt{3}}{2}\cdot\frac{\pi}{6} - \frac{1}{2}\right] - \left[\frac{1}{2}\cdot\frac{\pi}{3} - \frac{\sqrt{3}}{2}\right] \;= . \frac{\pi\sqrt{3}}{12} - \frac{1}{2} - \frac{\pi}{6} + \frac{\sqrt{3}}{2}

    . . . = \;\frac{\pi\sqrt{3}}{12} - \frac{6}{12} - \frac{2\pi}{12} + \frac{6\sqrt{3}}{12} \;= . \frac{\pi\sqrt{3} - 2\pi - 6 + 6\sqrt{3}}{12}

    Therefore: . A\;=\;\frac{\pi(\sqrt{3} - 2) + 6(\sqrt{3} - 1)}{12} . . . ta-DAA!

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  4. #4
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    Thanks i see my mistake for some reason i switched the terminals
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