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Thread: calculus

  1. #1
    Aug 2009


    Sorry if ive put this in the wrong category again, i dont know where this belongs, in uni my lecteur puts this under calculus.

    Gas is contained in a box defined so that it extends in each axis from 0 to 3.
    The density of the gas is given by the formula

    $\displaystyle p=xyz+(xyz)^2 x10^{-6}$

    $\displaystyle 10^{-6}\int_{x=0}^3\int_{y=0}^3\int_{z=0}^3xyz+x^{2}y^{ 2}z^{2}dxdydz$

    =$\displaystyle 10^{-6}1/3 \int_{y=0}^3\int_{z=0}^3[x^{2}yz/2 + x^{3}y^{2}z^{2}/3]dydz$

    Ive no idea where the two 9's come from in the next line, could anyone explain this

    =$\displaystyle 10^{-6}1/3 \int_{y=0}^3\int_{z=0}^3[9/2yz+9y^{2}z^{2}]dydz$
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  2. #2
    MHF Contributor Matt Westwood's Avatar
    Jul 2008
    Reading, UK
    From evaluating the integral with respect to x between the limits 0 and 3. $\displaystyle 3^2 = 9$ so having evaluated that, you can now go on to do the integrals in y and z.
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