The surface, [itex]z = x^2 + y^2[/itex], can be written in terms of parameters,u, v, as
x= u, y= v, [itex]z= u^2+ v^2[/itex] so that any point on the surface can be written in terms of the position vector [itex]\vec{r(u, v)}= u\vec{ix}+ v\vec{j}+ (u^2+ u^2)\vec{k}[/itex].
The derivatives with respect to the parameters are
[itex]\vec{r_u}= \vec{i}+ 2u\vec{j}[/itex] and [itex]\vec{r_v}= \vec{j}+ 2v\vec{k}[/itex]
Those are two vectors tangent to the surface, and their cross product,
[tex]\vec{r_v}\times\vec{r_u}= 2u\vec{i}+ 2v\vec{j}+ \vec{k}[/tex]
times dudv, is the "vector differential of surface area". Its length, [itex]dS= \sqrt{4u^2+ 4v^2+ 1}dudv[/itex] is the "differential of surface area".
So [itex]\int |xyz|dS= \int\int |uv(u^2+ v^2)(\sqrt{4u^2+ 4v^2+ 1}dudv[/itex]
Seeing all those sum of squares I would be inclined to put it in "polar coordinates with [itex]u= rcos(\theta)[/itex], [itex]v= r sin(\theta)[/itex]. z goes from 0 up to [itex]z= u^2+ v^2= r^2= 2[/itex] so that becomes
[tex]\int_{r= 0}^\sqrt{2}\int_{\theta= 0}^{2\pi} |r^4cos(\theta)sin(\theta)|\sqrt{4r^2+ 1}r drd\theta[/tex]
To handle the absolute value, you will need to determine where [itex]cos(\theta)sin(\theta)[/itex] is positive and where negative.