find triple integral $\iiint_V (3x^{2}+2y+z )dxdydz$ where $\left\{ \begin{array}{c} |x-y| \le 1 \\ |y-z| \le 1 \\ |x+z| \le1 \end{array} \right.$
I try to change variables $x-y=u,y-z=v,x+z=w$ but I have problem with showing function $3x^2+2y+z$ in variables u,v,w.
Can you guys help me with this problem/exercise ? Thank u so much
If $x-y=u$, $y-z=v$, and $x+z=w$, then$$x=\frac{u+v+w}2,\ y=\frac{-u+v+w}2,\text{ and }z=\frac{-u-v+w}2,$$and therefore$$3x^2+2y+z=\frac{3 u^2+6 v u+6 w u-6 u+3 v^2+3 w^2+2 v+6 v w+6 }4.$$On the other hand, the Jacobian of the map$$(u,v,w)\mapsto\left(\frac{u+v+w}2,\frac{-u+v+w}2,\frac{-u-v+w}2\right)$$is constant and equal to $\frac12$. Therefore, your integral is equal to$$\int_{-1}^1\int_{-1}^1\int_{-1}^1\frac{3 u^2+6 v u+6 w u-6 u+3 v^2+3 w^2+2 v+6 v w+6 }8\,\mathrm dw\,\mathrm dv\,\mathrm du.$$