Double integral; changing order of integration I'm attempting some exercises on double integration in Schaum's Outline of Calculus. The integral is
$$I = \int_{y=1}^{2}\int_{x=0}^{y^{3/2}} \frac{x}{y^2}\ dx\ dy.$$ 
I can do it in this direction and it turns out to be $I = \frac{3}{4}$ but for some reason I'm struggling to swap the order of integration and I think I may be confused.
I presume it'll be in the form of
$$I = \int_{x = \text{const}}^\text{const}\int_{y = \text{const}}^{y(x)} \frac{x}{y^2}\ dy \ dx$$
 A: Originally, you have $y$ as independent variable such that $y \in (1,2)$ and $x$ as dependent variable such that $x \in (0,y^{3/2})$, or thus $0 < x < y^{3/2}$.
If you now want $x$ to be the indepent variable, note that the maximal value of $y^{3/2}$ is $2^{3/2}$. So $x \in (0,2^{3/2})$. Then, since $x < y^{3/2}$ or thus $x^{2/3} < y$. Thus $y \in (x^{2/3},2)$. But, at the same time, $y \in (1,2)$ so $y \in (\max\{1,x^{2/3}\},2)$.
A: After drawing the region I think the Integral will be broken in 2 parts. The limit of 1st part should be $y=1$ to $y=2$ and $x=0$ to $x=y^{1.5}$ . The limit of second part should be $y=x^{\frac{2}{3}}$ to $y=2$ and $x=1$ to $x=2^{1.5}$.
A: $$\begin{align}\int_{1}^{2}\int_{0}^{y^{3\over2}}{x\over y^2}dxdy \\\end{align}$$
$$\begin{align}=\int_{0}^{1}\int_{y=1}^{y=2}{x\over y^2}dydx+\int_{1}^{2\sqrt{2}}\int_{y=x^{2\over3}}^{y=2}{x\over y^2}dydx\end{align}$$
$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$$\space$
I hope it can help you.
