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Integration over a Simplex

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To illustrate the transformation formula we will integrate the function

$\displaystyle f(x,y,z) = (1-x-y)z $

over the simplex

$\displaystyle S:\quad x,y,z \geq 0\,, \quad x+y+z\leq 1 \, . $

By transforming the variables by

$\displaystyle \left( \begin{array}{r}x\\ y\\ z\end {array} \right) = g(u,v,w) =... ... v (1-u)\\ w (1-u)(1-v) \end {array} \right) \,, \qquad 0 \le u,v,w \le 1 \,, $

we can describe $ S$ as an image of an elementary cube. With

$\displaystyle g^\prime = \left( \begin{array}{rrr} 1 &0&0\\ -v&(1-u)&0 \\ -w(1-v)&-w(1-u)&(1-u)(1-v) \end {array} \right) $

one gets the amount of the jacobian determinate

$\displaystyle \vert\operatorname{det}g^\prime\vert = (1-u)^2 (1-v). $

Therefore we have
$\displaystyle \int_S f$ $\displaystyle =$ $\displaystyle \int\limits^{1}_{0}\int\limits^{1}_{0}\int\limits^{1}_{0} \underb... ...\underbrace{(1-u)^2 (1-v)}_{\vert\operatorname{det}g^\prime \vert }\,dw\ dv\ du$  
  $\displaystyle =$ $\displaystyle \frac{1}{2}\int\limits^{1}_{0}(1-u)^4\,du\, \int\limits^{1}_{0} (1-v)^3\,dv$  
  $\displaystyle =$ $\displaystyle \frac{1}{2}\ \frac{1}{5}\ \frac{1}{4} = \frac{1}{40} \,.$  

(Authors: Höllig/Much/Höfert)
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  automatisch erstellt am 20.  8. 2008