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$$\iint_Cy dxdy, \quad C=\{(x,y)\colon0\leqslant x\leqslant2+y-y^2\}.$$

It is simple to see $x=2+y-y^2$ is a parabola with the symmetry axes is $x$ and the vertex $(9/4,1/2)$. It is easy to find the limits of the $x$ : $0\le x\leq 9/4$. Why the limits of the $y$ are: $$\frac{1}{2}\left(1-\sqrt{9-4 x}\right)\leqslant y\leqslant \frac{1}{2}\left(\sqrt{9-4 x} +1\right)\quad ?$$

What is the technique to derive the limitations of $y$ (or $x$)? Is there a correlation with the limits of $x$ seen that we have $\sqrt{9-4x}$?

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    $\begingroup$ Have you tried to draw this parabola? you can immediately see where the boundaries come from. You have already given the formal answer yourself. Simpler boundaries will be obtained if you stand on the $Oy$. $\endgroup$
    – zkutch
    Commented May 5 at 13:14
  • $\begingroup$ @zkutch Hi. I have drawn the parabola and I have found the boundaries of $y$ for $x=0$ that are: $y=-1$ and $y=2$. I have understood your comment. But I not very able to found the generic boundaries of the $y$. $\endgroup$
    – Sebastiano
    Commented May 5 at 13:17
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    $\begingroup$ Simply solve $y$ from $x=2+y-y^2=\frac 9 4 -\left( y - \frac 1 2 \right)^2$ and you'll have two branches for upper and lower functions. $\endgroup$
    – zkutch
    Commented May 5 at 13:23
  • $\begingroup$ @zkutch Ok. Done! $-y^2+y+(2-x)=0 \iff y^2-y+(x-2)=0 \implies y=1/2\pm \sqrt{9-4x}/2$. Hence if we have $-y^2+y+(2-x)\geq 0 \iff y^2-y+(x-2)\leq 0$ I will have the solution. Thank you very much. $\endgroup$
    – Sebastiano
    Commented May 5 at 13:30

1 Answer 1

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$$\int\limits_{-1}^{2}dy\int\limits_{0}^{\frac 9 4 -\left( y - \frac 1 2 \right)^2}dx=\int\limits_{0}^{\frac 9 4}dx\int\limits_{\frac 1 2 - \sqrt{\frac 9 4 - x}}^{\frac 1 2 + \sqrt{\frac 9 4 - x}}dy$$

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