# 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)

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"compute the volume" or "compute the area"? – Américo Tavares Apr 7 '13 at 20:29
@AméricoTavares compute the volume (according to my book) We are talking about volume since we are consider two integrals. In the single variable case, we got the area. In the two variable case, it's a volume. – Jean-Francois Rossignol Apr 7 '13 at 20:37
I was wrong. Please forget my comment. – Américo Tavares Apr 7 '13 at 20:43
@Jean-FrancoisRossignol What's the ordinate set of a function? – Git Gud Apr 7 '13 at 21:47
@GitGud If f is nonnegative, the set S of points (x,y,z) in 3-space with (x,y) in $[0,1]^2$ and $0 \leq z \leq f(x,y)$ is called the ordinate set of f over [0,1] – Jean-Francois Rossignol Apr 7 '13 at 21:50

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*}

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Thank you Monsieur – Jean-Francois Rossignol Apr 7 '13 at 22:00
You are welcome! I would like to give credit to @Fly by Night by his idea in his answer deleted in the meantime. If there's any error, the fault is mine. – Américo Tavares Apr 7 '13 at 22:06
@Jean-FrancoisRossignol Could you tell me how does one translate "ordinate set" into French? – Américo Tavares Apr 8 '13 at 16:48