I'm interested in computing the volume content of the region $A \in \mathbb{R}^3$ which is enclosed by the following three surfaces : $$0<x+y+z<1$$ $$0<y+z<1$$ $$0<z<1$$
Order of integration $x \rightarrow y \rightarrow z :$
$$\iiint_A\,dx\,dy\,dz=\int_{0}^{1}\int_{-z}^{1-z}\int_{-y-z}^{1-y-z}\,dx\,dy\,dz=1$$
Order of integration $x \rightarrow z \rightarrow y :$
$$\iiint_A\,dx\,dy\,dz=\int_{-1}^{1}\int_{-y}^{1-y}\int_{-y-z}^{1-y-z}\,dx\,dz\,dy=2$$
Order of integration $z \rightarrow x \rightarrow y :$
$$\iiint_A\,dx\,dy\,dz=\int_{-1}^{1}\int_{-1}^{1}\int_{-x-y}^{1-x-y}\,dz\,dx\,dy=4$$
What kind of dark magic is this$?$ Am I doing anything wrong in taking the limits of integration$?$
There are errors in the limits.
For example, in the 2nd integral, $y$ goes from $-1$ to $1$ and then $z$ does go from $-y$ to $1-y$ (to satisfy $0\lt y+z\lt 1$) but it also must keep between $0$ and $1$ (because of $0\lt z \lt 1$), i.e. really it will go between $\max(-y, 0)$ and $\min(1-y, 1)$.
The 3rd integral has even more problems. The first one seems correct.