Comment by OP:
> If $f$ is nonnegative, the set $S$ of points $(x,y,z)$ in 3-space with $(x,y)$ in $[0,1]^2$ and $0\le z\le f(x,y)$ is called the ordinate set of $f$ over $[0,1]$.
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My interpretations is as follows. The picture represents a region $R$ bounded by $y=2x^2$, $y=x^2$ (with $0\le x\le 1$) and $y=1$.
!enter image description here
$$y=2x^2\text{ (blue)}, y=x^2\text{ (green)}$$
The volume bounded by $R$ and $0\le z\le x+y$ is given by
\begin{eqnarray*} I &=&\iint_{R}x+y\,dA=\int_{0}^{\sqrt{2}/2}\left( \int_{x^{2}}^{2x^{2}}x+y\,dy\right) dx+\int_{\sqrt{2}/2}^{1}\left( \int_{x^{2}}^{1}x+y\,dy\right) dx \\\ &=&\int_{0}^{\sqrt{2}/2}x^{3}+\frac{3}{2}x^{4}\,dx+\int_{\sqrt{2}% /2}^{1}x-x^{3}+\frac{1}{2}-\frac{1}{2}x^{4}\,dx \\\ &=&\frac{1}{16}+\frac{3}{80}\sqrt{2}+\frac{37}{80}-\frac{19}{80}\sqrt{2} \\\ &=&\frac{21}{40}-\frac{1}{5}\sqrt{2}. \end{eqnarray*}