Change the order

We are given a double integral $\iint_R dA$ of a function $f(x,y)$ over a region $R$ , we can write write it as two different iterated integrals. We can integrate with respect to $x$ first or we can integrate with respect to $y$ first. The integral on the left shows integration by $y$ first, otherwise the one on the right shows integration by $x$ first. In this problem, we are not concerned about $f(x,y)$ , we don't need to specify what the function is, rather we need to focus on the limits of integration.

Notice in the left integral, the limits of the outer $dx$ integral mean that $0 \le x \le 1$ ,and the limits of the inner $dy$ integral means that each value of $y$ satisfies $0 \le y \le x(2-x)$ . To change these limits for an integral of $dxdy$ , means $y$ is the variable of the outer integral, its limits must be constant and correspond to the total range of $y$ over the region $R$ . The total range of $y$ is $0 \le y \le 1$ . To determine the limits of $x$ for the inner integration, we can rewrite the equation $y=x(2-x)$ and solve for $x$ .

Solving for $x$ in $2x-x^2=y$ $\implies -x^{2}+2x -y = 0$ $\implies, x=1-\sqrt{1-y}$ or $x=1+\sqrt{1-y}$ ; therefore, $p=x=1-\sqrt{1-y}$ because we are taking the lower bound.

Also, adding a graph will show the region of integration ( the highlighted one), the range and the domain of this region will create our integration boundaries depending on the order of integration whether it is $dxdy$ or $dydx$ .