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Consider the following integral \begin{equation} \int_{\Omega} f(y,My-z_2)\, g(z_1,z_2,y) ~ dz_1\, dz_2\, dy \end{equation}

$f(y,My-z_2) = 1$ (a constant function for each value of $y$ and $z_2$)

$g(z_1,z_2,y) = z_2$

The integration region $\Omega$ is defined as $\{(z_1, z_2, y) ~|~ z_1 = 0; 0 \leq z_2 \leq 1 - My; 0 \leq y \leq 1 \}$. I wish to calculate this integral, but I am facing some problems, due to the fact that $\Omega$ is defined in such a way that $z_1 = 0$.

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If is a triple integral, is zero because the domain $\Omega$ has measure zero.

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  • $\begingroup$ Suppose I eliminate variable $z1$ and compute the following double integral: Consider the following integral \begin{equation} \int_{\Omega} f(y,My-z_2)\, g(z_2,y) ~ dz_2\, dy \end{equation} where $f(y,My-z_2) = 1$ and $g(z_2,y) = z_2$. The integration region $\Omega$ is defined as $\{(z_2, y) ~|~ 0 \leq z_2 \leq 1 - My; 0 \leq y \leq 1 \}$. What about this? $\endgroup$ – Statistics Optimization Real E Nov 11 '14 at 11:58
  • $\begingroup$ The interpretation makes sense. Also possible: scalar surface integral. More context will be useful. $\endgroup$ – Martín-Blas Pérez Pinilla Nov 11 '14 at 16:11

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