# 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

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

• You might try changing to parabolic coordinates Jun 18, 2020 at 14:20
• So after i change to parabolic coordinates it's the same calculation for rectangel? Jun 18, 2020 at 14:23
• After you change to parabolic coordinates you have to 1) Find f(u(x,y), v(x,y)) by subtistuting the new values in your old f(x,y) 2) Find the new rectangle area (radius and $\theta$ angle) 3) Find the Jacubi. Then calculate the new integral. Jun 18, 2020 at 14:33
• @Dimitris I've tried if you could write an answer with the main ideas and some of the way it will be really helpfull thanks :) Jun 18, 2020 at 14:35

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.

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:

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

• is this parabolic coordiants? Jun 18, 2020 at 15:54
• Not in the usual meaning of the term.
– user
Jun 18, 2020 at 16:07
• Any intuition when to use this coordiantes? Jun 18, 2020 at 16:08
• In this case it was intuitively obvious, as the boundaries have the form $x^2/y=$const.
– user
Jun 18, 2020 at 17:03

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}

• is this parabolic coordiants? Jun 18, 2020 at 15:54