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Let $R$ be the region in $\mathbb{R}^2$ below the line $y = x + 2$ and above the parabola $y = x^2$. Check the integral of these $2$ functions in terms of $dx\cdot dy$ and then $dy\cdot dx$ ' I am having an issue figuring out what the integrals will range from. I have: $$G =\{(y,x) : -1 < x < 2 \text{ and } x^2 < y < (x + 2)\}\to dy.dx$$ $$H =\{(x,y) : 0 < y < 4 \text{ and } y -2 < x < \sqrt{y}\} \to dx.dy$$

However when I create the integrals in terms of $dx\cdot dy$ and $dy\cdot dx$ they differ? Any help please, did i get the range of the $G$ and $H$ wrong? Its a parabola cut with a line

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For $dydx$ the integral over $1$ is given by $$\int_{-1}^2\int_{x^2}^{x+2}dydx=\int_{-1}^2-x^2+x+2dx=[\frac13x^3+\frac12x^2+2x]_{-1}^2=\frac92$$ with the region that you correctly evaluated. But for the second integral we need to split the region into two parts - for $0\le y\le1$ we have that $-\sqrt{y}\le x\le\sqrt{y}$ and when $1\le y\le4$ we have $y-2\le x\le\sqrt{y}$. So the integral over the function $1$ is $$\int_0^1\int_{-\sqrt{y}}^\sqrt{y}dxdy+\int_1^4\int_{y-2}^\sqrt{y}dxdy=\int_0^12\sqrt{y}\,dy+\int_1^4\sqrt{y}-y+2\,dy$$ $$=[\frac43y^{\frac32}]_0^1+[\frac23y^{\frac32}-\frac12y^2+2y]_1^4=\frac92$$ So the two regions are now equal.

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  • $\begingroup$ Thank you this helps loads. My text book does not describe splitting them up and i did not think of it. Much appreciated! $\endgroup$ – Shaun Weinberg Mar 31 at 10:36
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For the region $H$, the lower limit of $x$ shouldn't be $y-2$.

It should be the maximum of $-\sqrt{y}$ and $y-2$. In fact, when $0 \le y \le 1$, the lower limit is $-\sqrt{y}$.

That is

$$H =\{(x,y): 0 \le y \le 4 , \max(-\sqrt{y}, y-2) < x < \sqrt{y} \} $$

That is from the picture below, the left limit of the region consists of the green color and blue color part.

enter image description here

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  • $\begingroup$ I originally wrote an answer for the duplicate, so I copied it here. (+1) btw $\endgroup$ – Peter Foreman Mar 30 at 19:13
  • $\begingroup$ I did the same thing. ;) $\endgroup$ – Siong Thye Goh Mar 30 at 19:14

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