I am trying to solve the following exercise:
Consider the region $\Omega$, in the first quadrant, bounded by the curves: \begin{align} x^2 + y^2 = 4; \;\;\; x^2 + y^2 = 9; \;\;\; x^2 - y^2 = 1; \;\;\; x^2 - y^2 = 4 \end{align} By a change of variables calculate de mass of $\Omega$ if the density at each point $(x,y)\in \Omega$ is equal to the product of the distances of the point to the coordinate axis.
My attempt is the following. First, I define de density function $\rho\colon \Omega \rightarrow \mathbb{R}$ by \begin{align*} \rho(x,y) = x\cdot y. \end{align*} Then, I applied the substitution $x = r\cdot \cos(\theta)$ and $y = r \cdot \sin(\theta)$. For the first two equations I get $r = 2$ and $r=3$. My problem lies in the equations $x^2 - y^2 = 1$ and $x^2 - y^2 = 4$, I don't know how to continue from here. Of course, what I want is to find the limits of the integral
\begin{align*} m(\Omega) = \iiint_{\Omega} x\cdot y. \end{align*}
If $\require{cancel}u=x^2+y^2$ and $v=x^2-y^2$, then $4\leqslant u\leqslant9$ and $1\leqslant v\leqslant4$. Besides,$$xy=\sqrt{x^2y^2}=\sqrt{\frac{u+v}2\times\frac{u-v}2}=\frac{\sqrt{u^2-v^2}}2.$$Finally, $(x,y)=\left(\sqrt{\frac{u+v}2},\sqrt{\frac{u-v}2}\right)$. But if$$g(u,v)=\left(\sqrt{\frac{u+v}2},\sqrt{\frac{u-v}2}\right),$$then$$\bigl|\det Jg(u,v)\bigr|=\frac1{4\sqrt{u^2-v^2}}.$$So, since $\Omega=g([4,9]\times[1,4])$, you have\begin{align}\iint_\Omega xy\,\mathrm dx\,\mathrm dy&=\iint_{g([4,9]\times[1,4])}xy\,\mathrm dx\,\mathrm dy\\&=\iint_{[4,9]\times[1,4]}\frac{\sqrt{u^2-v^2}}2\left|\det Jg(u,v)\right|\,\mathrm du\,\mathrm dv\\&=\int_4^9\int_1^4\frac{\cancel{\sqrt{u^2-v^2}}}2\frac1{4\cancel{\sqrt{u^2-v^2}}}\,\mathrm dv\,\mathrm du\\&=\frac{15}8.\end{align}