The problem is to integrate $x^{-3}$ on the region $D=\lbrace (x,y)\in\mathbb{R}^2 : \sqrt{x}\leq y \leq 2\sqrt{x},\ x^2\leq y \leq 4x^2 \rbrace$. I want to know if my attempt is good: Since the most 'lefted' point of $D$ is $(1/16)^{1/3}$ (this comes from the common point of $y=4x^2$ and $y=\sqrt{x}$) and the most 'righted' is $4^{1/3}$ (this come from the common point of $y=2\sqrt{x}$ and $y=x^2$).
$$ \int \int _D \frac{1}{x^3}\ dxdy = \int_{(1/16)^{1/3}} ^{4^{1/3}} \int _{\sqrt{x}}^{2\sqrt{x}} \frac{1}{x^3} \ dydx$$
Am I right? thanks in advance.
EDIT: I don't care about the solution, I can do that integral by myself. I only want to assure if the limits that I put in both integrals are correct.