Finding the limits of integration Given the function $f_{X,Y,Z}(x,y,z)=\begin{cases}\frac{1}{2^3} & \quad\text{if }0\le x,y,z\le 2\\0 &\quad\text{otherwise} \end{cases}$
and $A:=\{x+y+z:1\le x+y+z <2\}$, 
I am trying to find $\iiint_A f_{X,Y,Z}(x,y,z)\; \mathrm{d}z\; \mathrm{d}y\; \mathrm{d}x$ .
But I have trouble in finding the appropriate limits. How does one compute the limits of integration?
 A: I might consider this integral as $\iiint_{A_2} -\iiint_{A_1}$, where $A_r=\{(x,y,z):0\leq x+y+z\leq r\}$.
In $A_r$, $z$ can range from $0$ to $r$.
For a given $z$ in this range, y can range from $0$ to $r-z$.
For given $z$ and $y$ in these ranges, $x$ can range from $0$ to $r-y-z$.
It is extremely helpful to draw a picture to figure this out.
A: The integration domain is $A:=\{(x,y,z): 1\leq x+y+z\leq 2\}$
The support for the function is: $S:=\{(x,y,z): 0\leq x\leq 2,0\leq y\leq 2,0\leq z\leq 2\}$
You seek the intersection $S\cap A$, expressed to be used as bounds for the integration $\iiint_A f(x,y,z)\mathsf d z\,\mathsf d y\,\mathsf d x$
$S$ is the cube $(0;2)^3$. $A$ is bounded by two planes each of which conveniently contains three apex of this cube. So $S\cap A$ neatly truncates corners off this cube thusly:

$$S\cap A = S\smallsetminus \lower{1.5ex}{\left(\raise{1.5ex}{{\{(x,y,z):0\leq x\leq 1, 0\leq y\leq 1-x, 0\leq z\leq 1-x-y\}}\cup{\{(x,y,z):1\leq x\leq 2, 2-x\leq y\leq 2, 2-x-y\leq z\leq 2\}}}\right)}$$



 $$\iiint_A f(x,y,z)\mathsf d z\,\mathsf d y\,\mathsf d x={{\int_0^2\int_0^2\int_0^2\tfrac 1{2^3}\mathsf d z\,\mathsf d y\,\mathsf d x}\\-{\int_0^1\int_0^{1-x}\int_0^{1-x-y}\tfrac 1{2^3}\mathsf d z\,\mathsf d y\,\mathsf d x}\\-{\int_1^2\int_{2-x}^2\int_{2-x-y}^2\tfrac 1{2^3}\mathsf d z\,\mathsf d y\,\mathsf d x}}$$

