Calculate $\iiint_Ax^2y^2z dx dydz$ where $A=\{(x,y,z)\in \mathbb R^3 | 0\leq z\leq x^2+y^2,|x|+|y|=1\}$ Calculate 
$$\iiint_Ax^2y^2z  dx dydz$$
where $A=\{(x,y,z)\in \mathbb R^3 | 0\leq z\leq x^2+y^2,|x|+|y|=1\}$
so I tried:
$$\iint_Dx^2y^2 (\int_0^{x^2+y^2}zdz) dx dy$$ where $D=\{|x|+|y|<=1\}$
$$\frac{1}{2}\iint_Dx^2y^2 (x^2+y^2)^2 dx dy$$
and to solve now this integral, I can't figure a change of variables to simplify things, because I'd arrive at a very long polynomial. I've thinking of doing that integral considering $D++$ which is $D$ but with $x,y\geq 0$, and then multiplying the result by 4, but it would be very long without a change of variables I believe. Any hint? 

Notice that the integrand function $f(x,y)$ in your double integral only depends on $x^2$ and $y^2$, so that $f(x,y)=f(\pm x,\pm y)$. Due to this symmetry both of the function and of the domain of integration $D$, the four integrals over the four quadrants are equal. So your idea of integrating only over the part of $D$ within the first quadrant is great — and then you don't have to do that four times, because you can simply multiply the result by $4$.
UPDATE (after the original question was edited). Another option could be to substitute $u=x+y$, $v=x-y$. It will simplify the original domain $D$ into a rectangle with sides parallel to the axes; but simplifying the integrand again will be a little bit of work. This probably will be faster.
