Double Integration with change of variables I am having trouble with the following double integral:
$$\iint\limits_D(x^2+y^2) \;dA$$
where $D$ is given by the region enclosed by the curves 


*

*$xy=1$

*$xy=2$

*$x^2-y^2 =1$

*$x^2-y^2 =2$


I have tried changing variables to $u=x^2+y^2$, $v=x^2-y^2$ (and other changes of variables) but the Jacobian looks grim.

The change of variables $u = xy$, $v = x^2 - y^2$ works. Check that it's one-to-one in the first quadrant. (I assume you want only the first quadrant region bounded by those curves.)
EDIT:
First we prove that the the transformation is one-to-one on the first quadrant. Let $x, y > 0$, and define $u$, $v$ as above.  $x^2$ and $-y^2$ are the roots of the polynomial $f(T) = T^2 -vT - u^2$. Since one is positive and the other negative, $x^2$ and $-y^2$, hence $x$ and $y$, are uniquely determined by $u$ and $v$. 
Moreover the image of the transformation is the entire half-plane $u > 0$. For given $u$ and $v$ with $u > 0$, write the polynomial $f(T)$ as above. Since $f(0) < 0$, $f$ has one negative root $-y^2$ and one positive root $x^2$ (for some $x, y > 0$), and we have $v = x^2 - y^2$ and $- u^2 = -x^2 y^2$.  It follows that $u = xy$.
(Here is an alternative proof: the mapping $x + iy \mapsto v + 2iu$ is really the mapping $z \mapsto z^2$ from the first quadrant to the upper half-plane.) 
The Jacobian is calculated as 
$$\frac{\partial(u,v)}{\partial(x,y)} = \left| \matrix{y & x \\ 2x & -2y} \right| = -2(x^2 + y^2).$$ Thus $du \, dv = 2(x^2 + y^2) \, dx \, dy$.
Now the image of $D$ under the transformation is the set defined by the inequalities $1 \leq u \leq 2$ and $1 \leq v \leq 2$, so
$$\iint_D (x^2 + y^2) \, dx \, dy = \int_1^2 \int_1^2 \frac{1}{2} \, du \, dv = \frac{1}{2}.$$
