Double integral transformation - how to determine bounds of integration So here is the exact problem which i got stuck on.  

Evaluate
  $$ I= \iint_A \sqrt{xy}\ dx\ dy  \quad\text{where $A$ is bounded by $y^2=2x$, $y^2=x$, $xy=1$, $xy=2$}$$

I am supposed to use the transformation $u=xy, v=\frac{y^2}{x}$.
Now I know that I will end up with $\iint \sqrt{u} \frac{1}{3v} du\ dv$. 
So the bounds are supposed to be $\int_1^2 \int_1^2$ but I have no idea how were the bounds determined. Originally I assumed it was determined from the intersections of the $A$ but those are not $<1,2>$.
Also another thing I don't understand is why is the $G=\det|\cdot|$ used as $J =G^{-1}$ in the integral, when if we transform into polar coordinates we simply use the $J=G$ in  $\iint_B f(x,y)J\ dx\ dy$.  

Hint. Note that
$$A=\{(x,y): x\leq y^2\leq 2x,1\leq xy\leq 2\}=\{(x,y): 1\leq \frac{y^2}{x}\leq 2,1\leq xy\leq 2\}.$$
Moreover
$$dudv=\left|\frac{\partial(u,v)}{\partial(x,y)}\right|dxdy$$
where
$$\left|\frac{\partial(u,v)}{\partial(x,y)}\right|=\left|u_xv_y-u_yv_x\right|=|y(2y/x)-x(-y^2/x^2)|=\frac{3y^2}{x}=3v.$$
Hence, as you already remarked
$$ I= \iint_A \sqrt{xy}\ dx\ dy=\int_1^2 \int_1^2 \frac{\sqrt{u}}{3v} du\ dv=\left(\int_1^2\sqrt{u}du\right)\cdot \left(\int_1^2 \frac{1}{3v}\ dv\right).$$
