Sketch of the ordinate set of $f$ Let $f$ be defined on $[0,1] \times [0,1]$ as follows:
$f(x,y)= \begin{cases} x+y \mbox{ if } x^2 \leq y \leq 2x^2 \\ 0 \mbox{ otherwise} \end{cases}$
I want to make a sketch of the ordinate set of $f$ over $[0,1] \times [0,1]$  and compute the volume of this ordinate set by double integration (Assuming the integral exists)

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 $[0,1]$.


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 $y=1$.

$$y=2x^2\text{ (blue)}, y=x^2\text{ (green)}$$
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*}
