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Math Help - Evaluating the line integral....

  1. #1
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    Evaluating the line integral....

    ...where curve C is given by the integral of (x^2)*y*sqrt(z)dz
    C: x=t^3, y=t, z=t^2, t is between 0 and 1

    I came up with the integral (from 1 to 0) (2t^12)*sqrt(1+9t^4+4t^2)

    is this integral correct? if yes, how do I integrate it?
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  2. #2
    MHF Contributor Calculus26's Avatar
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    I'm not sure what you're doing but you want the line integral

    of (x^2)*y*sqrt(z)dz on the curve x=t^3, y=t, z=t^2 ?

    F = (x^2)*y*sqrt(z) k

    Since

    x=t^3, y=t, z=t^2

    F = t^6*t*t k = t^8 k

    r = t^3 i + t j + t^2 k

    r ' = 3t^2 i + j + 2t k

    F*r ' = 2t^9

    Which is easy to integrate over the interval 0 to 1
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