Q: Finding a volume of a region defined by |x-y+z|+|y-z+x|+|z-x+y|=1 Find the volume of the region definded by |x-y+z|+|y-z+x|+|z-x+y|=1.
I'm having trouble approaching this problem. Could someone maybe give me a hint or a solution, it would be so helpful. 
Thanks in advance and sorry for my bad english.

A: Let's find the volume of the related region
$$|x|+|y|+|z| \leq 1$$
By the symmetry of the region we can reduce it to an integral in the first octant only
$$\iiint_EdV = 8 \iiint_{E\:\cap\:\text{First Octant}} dV$$
Then setting up and doing the integral is not that hard
$$ = 8\int_0^1 \int_0^{1-x} \int_0^{1-x-y} dz\:dy\:dx = 8\int_0^1 \frac{1}{2} - x + \frac{1}{2}x^2\:dx = \frac{4}{3}$$
Now how does this help us with this problem? We can use the substitution
$$\begin{cases}u = x-y+z \\ v = x+y-z \\ w = -x+y+z \\ \end{cases} \implies J^{-1} = \begin{vmatrix} 1 & -1 & 1 \\ 1 & 1 & -1 \\ -1 & 1 & 1 \\ \end{vmatrix} = 4$$
Thus with this change of variables we get that
$$\iiint_{|x-y+z|+|y-z+x|+|z-x+y|\leq 1} dV = \iiint_{|u|+|v|+|w|\leq 1}\frac{1}{4} \:dV' = \frac{1}{3}$$

A: This may help you visualize the region:

