Sketch of the ordinate set of $f$

Question

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)

Answer 1

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 $y=1$.

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

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*}