Reverse the order of integration of $\int_{\frac{1}{3}}^\frac{2}{3} \int_{y^2}^{\sqrt y} f(x,y) \,dx\,dy$. Q: Reverse the order of integration of $\int_{\frac{1}{3}}^\frac{2}{3} \int_{y^2}^{\sqrt y} f(x,y) \, dx\,dy$.
I imagined that $x$ should varie between $\dfrac{1}{9}$ and $\sqrt{\dfrac{2}{3}}$ and $y$ varie between $x^2$ and $\sqrt x$, then we get:
$$\int_{\frac{1}{3}}^\frac{2}{3} \int_{y^2}^{\sqrt y} f(x,y) \,dx\,dy = \int_{\frac{1}{9}}^{\sqrt{\frac{2}{3}}} \int_{x^2}^{\sqrt x} f(x,y) \, dy \, dx$$
But that is not the answer cuz i tested it with some $f(x,y)$ and got different values on the integral.
Can somebody help me? Thanks!

The 'x' limit $y^2$ means $x=y^2$, so we have $y=\sqrt{x}$
The 'x' limit $\sqrt{y}$ means $x=\sqrt{y}$, so we have $y=x^2$
Notice you have exchanged lower and upper 'x' limits
For the 'y' limit $y=\frac{1}{3}$ we must get the $x=y^2 = \frac{1}{9}$ function because $\left(\frac{1}{3} \right)^2 < \sqrt{\frac{1}{3}}$.
Same reasoning for $y=\frac{2}{3}$ we use the other, $x=\sqrt{\frac{2}{3}}$
Thus, we arrive to
$$\int_{\frac{1}{3}}^\frac{2}{3} \int_{y^2}^{\sqrt y} f(x,y) dxdy= 
\int_{\frac{1}{9}}^{\sqrt{\frac{2}{3}}} \int_{\sqrt x}^{x^2} f(x,y) dydx$$
Being that said, draw both functions (square and root), because perhaps you have to decompose the region into three pieces.
