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Math Help - Center of Mass

  1. #1
    Junior Member
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    Center of Mass

    Here is the question:

    A solid is formed by rotating the region bounded by the curve and the -axis between and , around the -axis. The volume of this solid is . Assuming the solid has constant density , find and .

    To find the moment, I did:

    \int_0^1 x dx = \frac {1} {2}

    As density was constant, I divided the moment by the volume and found to be 1/2.

    And I am not sure that is right and so I haven't even attempted to find . All help is much appreciated. Thanks.
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  2. #2
    MHF Contributor

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    The volume, when y= y(x) is rotated around the x- axis is (by the disk method) \pi \int_{x_0}^{x_1} y^2(x)dx.

    In this case, that would be \pi \int_0^1 e^{-7x} dx which gives the volume you quote.

    But the moment is NOT just that volume times \int x dx. It is, rather, that same integral with another x in the integral: \pi \int_0^1 xe^{-7x}dx. Use "integration by parts".

    Since this is rotated around the x-axis, it is symmetric with respect to both y and z and it immediately follows that \overline{y}= 0 and \overline{z}= 0.
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