# Creating an integral from 2 functions

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

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.

• Thank you this helps loads. My text book does not describe splitting them up and i did not think of it. Much appreciated! – Shaun Weinberg Mar 31 at 10:36

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