i'm having a small problem with changing variables.
u = x and v = xy, i am having trouble getting y in terms of v.
if my region of integration is needed here it is: from x = 1 to x= 2 and xy = 1 to xy = 2
In the uv-plane you will have the lines u=x=1 and u=x=2 and v=xy=1 v=xy=2
You will get the square with corners (1,1) (1,2) (2,1) and (2,2)
y will be in terms of u and v
$\displaystyle y=\frac{v}{x}=\frac{v}{u}$
The jacobian matrix would be
$\displaystyle \frac{\partial{(x,y)}}{\partial{(u,v)}}=
\begin{vmatrix}
1 && 0 \\
-\frac{v}{u^2} && \frac{1}{u} \\
\end{vmatrix}
$
The transformation is $\displaystyle \bold{G}(u,v) = (u,v/u)$. If $\displaystyle f$ is the function you are integrating then $\displaystyle \int_R f = \int_{\bold{G}^{-1} (R) } f\circ \bold{G}$. Where $\displaystyle R = \{ (x,y)\in \mathbb{R}^2 | 1/x \leq y\leq 2/x \mbox{ and }1\leq x\leq 2\}$. And $\displaystyle \bold{G}^{-1}(R) = \{ (u,v) \in \mathbb{R}^2 | \bold{G}(u,v) \in R\}$. Thus, for a point $\displaystyle (u,v)$ to we be in $\displaystyle \bold{G}^{-1}(R)$ we require that $\displaystyle (u,v/u) \in R$, thus, $\displaystyle 1\leq u\leq 2 \mbox{ and }1/u \leq v/u\leq 2/u \implies 1\leq u\leq 2 \mbox{ and }1\leq v\leq 2$. Thus, the this new set you are integrating over is simply the rectangle $\displaystyle [1,2]\times [1,2]$.