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Math Help - Find the centroid of the plane region

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    Find the centroid of the plane region

    Find the centroid of the plane region in the second quadrant that is bounded by the curve y = 3 - (3x^2)/4



    Use Pappus's theorem (not integration) to find the volume of the solid swept out when the region in the problem above is revolved about the y-axis


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  2. #2
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    Quote Originally Posted by victorxz90044 View Post
    Find the centroid of the plane region in the second quadrant that is bounded by the curve y = 3 - (3x^2)/4

    Use Pappus's theorem (not integration) to find the volume of the solid swept out when the region in the problem above is revolved about the y-axis

    \bar{x} = \frac{\int_a^b x \cdot f(x) \, dx}{\int_a^b f(x) \, dx}

    \bar{y} = \frac{\int_a^b \frac{[f(x)]^2}{2} \, dx}{\int_a^b f(x) \, dx}


    find the cm in quad I and use symmetry ...

    \bar{x} = \frac{\int_0^2 x\left(3-\frac{3x^2}{4}\right) \, dx}{\int_0^2 3-\frac{3x^2}{4} \, dx} = \frac{3}{4}

    so, in quad II, \bar{x} = -\frac{3}{4}

    \bar{y} = \frac{\int_0^2 \frac{\left(3-\frac{3x^2}{4}\right)^2}{2} \, dx}{\int_0^2 3-\frac{3x^2}{4} \, dx} = \frac{6}{5}


    using Pappas ...

    V = A \cdot 2\pi r

    rotated about the y-axis ...

    V = \int_0^2 3-\frac{3x^2}{4} \, dx \cdot \frac{3\pi}{2} = 6\pi

    rotated about the x-axis ...

    V = \int_0^2 3-\frac{3x^2}{4} \, dx \cdot \frac{12\pi}{5} = \frac{48\pi}{5}
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    Thank you very much...

    Thank you very much....your a life saver
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