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
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this communityCalculate 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
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.
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:
$$\frac13\int_{1/2}^1\int_{1/2}^1 u\sin(uv)\,du\,dv=\frac{-2 \sin (1/4)+3 \sin (1/2)-\sin (1)}3.$$
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}