evaluate a double integral $\int_D (x^2+y^2)$ in $D$ Use change of variables to compute $\int_D (x^2+y^2)$, where $D$ is the region in the first quadrant between the hyperbolas $xy=1, xy=2, x^2-y^2=1, x^2-y^2=4$. 
Can anyone give me a hint about how to choose the new variables? 
I tried $u=x^2, v=y^2$, but it did not seem to work. 
 A: This is really meant to be a comment rather than an answer. When doing these sorts of questions, I think it's good to visualise what you're doing, rather than trying to compute blindly. This is not just extra amusement - it's actually important that we verify that our new coordinates parametrise the integration domain in a one-to-one matter.
So I've plotted lines of constant $u = xy$ for various values of $u \in [1,2]$ (purple), and lines of constant $v = x^2 - y^2$ for various values of $v \in [1,4]$ (blue). I don't know what those straight grey lines are - that's just what Mathematica produced!
Anyway, your integration domain is where the blue and purple lines intersect. You can see from the picture that your $u$ and $v$ coordinates parametrise your domain in a one-to-one manner.

For the purposes of the doing the computation, you want to change variables by writing
$$ dx \ dy = \left| \det \frac{\partial(x,y)}{\partial(u,v)}\right| du \ dv$$
But in this particular example, it's actually easier to compute
$$ du \ dv = \left| \det \frac{\partial(u,v)}{\partial(x,y)}\right| dx \ dy  = (x^2 + y^2) \ dx \ dy $$
(You should check the algebra! You should also think about why the determinant turned out to be negative in this case...)
Since the original integral includes a factor of $x^2 + y^2$ in the integrand, this is very convenient! I'll let you finish off from here.
