Calculate the following integral $\iint_{T}\frac{x^2\sin(xy)}{y}\,dx\,dy$ Calculate the following integral $$\iint_{T}\frac{x^2\sin(xy)}{y}\,dx\,dy\,,$$ where $$T=\{(x,y)\in\mathbb{R}^2:x^2<y<2x^2,y^2<x<2y^2\}$$
I found the $1/2\leq x\leq 1,1/2\leq y\leq 1$ but i got stuck on how set the limits of the integral
 A: Your region is bounded by $y=x^2$, $y=2x^2$, $y=\sqrt x$, and $y=\sqrt{x/2}$. See the figure below. So you need to calculate the intersections.


*

*Intersection $y=x^2$ with $y=\sqrt x$:
$$x^2=\sqrt x$$
This yields $x=0$ or $x=1$. At $x=1$ you have $y=1$

*Intersection $y=x^2$ with $y=\sqrt{x/2}$
$$x^2=\sqrt\frac x2$$
This yields $x=0$ and $x=2^{-\frac13}\approx 0.79$. At this $x$ you have $y=2^{-\frac 23}\approx 0.63$

*Intersection of $y=2x^2$ with $y=\sqrt{x/2}$ yields $x=0$ and $x=0.5$, with $y=0.5$

*intersection of $y=2x^2$ with $y=\sqrt x$ yields $x=4^{-\frac13}=2^{-\frac23}$ and $y=2\cdot 4^{-\frac23}=2^{-\frac 13}$
This means that you can split the $x$ integral in three regions, from $1/2$ to $2^{-\frac23}$ to $2^{-\frac 13}$ to $1$. The corresponding limits for $y$ will be given by the curves that bound your region.
A: Let $x= (u^2v)^{1/3}$ and $y= (uv^2)^{1/3}$. Then, $u=\frac{x^2}y,\>v=\frac{y^2}x$ and the Jacobian is $\frac13$. As a result, the integral becomes
\begin{align}
\iint_{T}\frac{x^2\sin(xy)}{y}\,dx\,dy
&=\frac13\int_{\frac12}^1 \int_{\frac12}^1 u\sin(uv) dudv \\
&= -\frac23\sin\frac14+\sin\frac12-\frac13\sin1
\end{align}
A: Consider the folowing transformation of coordinates:
$$
u=\frac{x^2}y,\quad v=\frac{y^2}x.
$$
A simple calculation will show that the Jacobian of the transformation is $\frac13$, so that the integral will be simply:

$$\frac13\int_{1/2}^1\int_{1/2}^1 u\sin(uv)\,du\,dv=\frac{-2 \sin (1/4)+3 \sin (1/2)-\sin (1)}3.$$

