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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

$$I = \int_{x = \text{const}}^\text{const}\int_{y = \text{const}}^{y(x)} \frac{x}{y^2}\ dy \ dx$$

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    See if you can draw out the region that you are integrating over. With that done, it's not too hard to restate the bounds with the variables switched. – 2017-01-04

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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 $y \in (\max\{1,x^{2/3}\},2)$.

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    Should the result necessarily be the same in either "direction"? – 2017-01-04
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    Since $\frac{x}{y^2}$ is bounded on the area of integration, and you are integrating over a bounded area: YES. If your function becomes infinite on the area of integration, or you are integration over an infinite area, things may get tricky. You should read up on [Fubini's theorem](https://en.wikipedia.org/wiki/Fubini%27s_theorem) if you're dealing with such situations. – 2017-01-04
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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 $x=2^{1.5}$.

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    The first part is just the integral that I did in the "direction" $dxdy$? – 2017-01-04
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    @Jahambo99 For the 1st part x=y^(1.5) is equal to 1 – 2017-01-04
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$$\begin{align}\int_{1}^{2}\int_{0}^{y^{3\over2}}{x\over y^2}dxdy \\\end{align}$$

$$\begin{align}=\int_{0}^{1}\int_{y=1}^{y=2}{x\over y^2}dydx+\int_{1}^{2\sqrt{2}}\int_{y=x^{2\over3}}^{y=2}{x\over y^2}dydx\end{align}$$

$\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$enter image description here

I hope it can help you.