Calculate: $$ \begin{aligned} &\iiint\limits_Gf(x;y;z)dxdydz,\ \text{where}\\ &G\ \text{is bounded by the planes}\ y=-x,\ z=x,\ z=y,\ z=1,\ \text{and}\\ &f(x;y;z)=x^2-z^2 \end{aligned} $$
I've been struggling to plot $G$, and, unfortunately, it didn't help much since I couldn't get much out of it. So, I thought that there might be some sort of analytical solution to this problem. I understand that the most difficult part is to go to the iterated integral, but I don't really know how to do that. So, I would appreciate any help!
In the $xy$-plane, the triple integral is over the two shaded regions, for which the $z$-integral is bounded by $z=x$ and $z=1$ in the purple region and by $z=y$ and $z=1$ in the blue region.
Thus, the integral can be set up as
$$I = \int_0^1dx\int_{-x}^{x}dy\int_x^1(x^2-z^2)dz+\int_0^1dy\int_{-y}^{y}dx\int_y^1(x^2-z^2)dz$$
which can then be integrated analytically.