My question is about the integration limits of a double integral

Question

I have to solve the following exercise :

Find the area of the $R$ region that is enclosed by the parable $y=x^2$ and the line $y=x+2$.

I did the following shape:

And I solved the double integral:

(the integration limits to $x$ axe: $y=x^2$ to $y=x+2$ and $-1\leq x\leq 2$ )

$A= \int_{R}^{}\int_{}{}dA=\int_{-1}^{2}\int_{x^2}^{x+2}dydx=\int_{-1}^{2}[y]_{x^2}^{x+2}dx=\int_{-1}^{2}(x+2-x^2)dx=[\frac{x^2}{2}]_{-1}^{2}+[2x]_{-1}^{2}-[\frac{x^3}{3}]_{-1}^{2}=\frac{9}{2} $

If we reverse the sequence of integration (from $dydx$ to $dxdy$), I am confused about what the integration limits are

Answer 1

You have to split the region into two parts. For $y$ between $0$ and $1$ we have $-\sqrt y <x <\sqrt y$ and for $1<y<4$ it is $y-2<x<\sqrt y$.

Answer 2

The possible values for $y$ are the numbers from the interval $[0,4]$. If you reverse the order, then you get the sum of two integrals:$$\int_0^1\int_{-\sqrt y}^{\sqrt y}\,\mathrm dx\,\mathrm dy+\int_1^4\int_{y-2}^{\sqrt y}\,\mathrm dx\,\mathrm dy.$$

Answer 3

If you reverse the order of integration you have two integrals to .$$ A=\int_{0}^{1}\int_{-\sqrt y}^{\sqrt y}dxdy+\int_{1}^{4}\int_{y-2}^{\sqrt y}dxdy$$

You can take over from here and compare your result with your first attempt.