calculus

**tone999** · August 11th 2009, 01:18 PM

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

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

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

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

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

= $10^{-6}1/3 \int_{y=0}^3\int_{z=0}^3[9/2yz+9y^{2}z^{2}]dydz$

**Matt Westwood** · August 11th 2009, 01:24 PM

From evaluating the integral with respect to x between the limits 0 and 3. $3^2 = 9$ so having evaluated that, you can now go on to do the integrals in y and z.

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