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!

Three piece decomposition $\int\limits_\frac{1}{9}^\frac{4}{9}\int\limits_\frac{1}{3}^\sqrt{x}+\int\limits_\frac{4}{9}^\sqrt{\frac{1}{3}}\int\limits_\frac{1}{3}^\frac{2}{3}+\int\limits_\sqrt{\frac{1}{3}}^\sqrt{\frac{2}{3}}\int\limits_{x^2}^\frac{2}{3}f(x,y)dydx$
Draw a picture.  You will see it immediately.
