Suppose you had the double integral $\iint \limits_{A} \frac{y^{2}}{x^{4}}e^{xy} \ dx \ dy$, where $A$ is the region defined by $x>0, \ y>0$ satisfying $x^{2} \leq y \leq 2x^{2}, \ \frac{1}{x} \leq y \leq \frac{2}{x}$, and the coordinate transformation $u=xy, \ v= \frac{y}{x^{2}}$.
If we express $x$ and $y$ in terms of $u$ and $v$ we get $x=u^{\frac{1}{3}}v^{-\frac{1}{3}}, \ y=u^{\frac{2}{3}}v$.
Subbing this into the parameters for $A$, we get \begin{gather*} x^{2} \leq y \leq 2x^{2} \Rightarrow 1 \leq v \leq 2 \\ \\ \frac{1}{x} \leq y \leq \frac{2}{x} \Rightarrow 1\leq u \leq 2 \end{gather*}
Am I correct in thinking that $\iint \limits_{A} dA = \int_{1}^{2} \int_{1}^{2} du\ dv $?