calculus *astroid*

THe equation x^2/3 + y^2/3 = 4
describes an astroid. Determine the length of the astroid by finding the length of a portion of it,found in the first quadrant, being y = (4-x^2/3)^3/2 for 0<x<8 and multiplying that valor by 4.

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  1. just plug and chug

    s = 4∫[0,8] √(1+y'^2)dx

    since x^2/3 + y^2/3 = 4
    2/3 x^(-1/3) + 2/3 y^(1/3)y' = 0

    y' = -y^(1/3)/x^(1/3)

    1+y'^2 = 1 + y^(2/3)/x^(2/3)
    = (x^(2/3) + y^(2/3)]/x^(2/3)
    = 4/x^(2/3)

    So the integrand is really simple after all. You should come up with 48

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  2. Everything is so easy once you know how to work around it! Thanks again Steve.

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