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.
 A: 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
$$\iint_D (x^2 + y^2) \, dx \, dy = \int_1^2 \int_1^2 \frac{1}{2} \, du \, dv = \frac{1}{2}.$$
A: Consider the variables
\begin{align*}
u = xy, && v = x^2-y^2.
\end{align*}
We obtain that $x=\frac{u}{y}$. Therefore:
$$y^2=x^2-v,$$
$$y^2 -\frac{u^2}{y^2}=-v,$$
$$\frac{y^4-u^2}{y^2} =-v,$$
$$y^4+vy^2-u^2=0,$$
$$y^2 = \frac{-v\pm \sqrt{v^2-4(-u^2)}}{2} = \frac{-v \pm \sqrt{4u^2+v^2}}{2}.$$
From which we obtain:
$$x^2=\frac{-v \pm \sqrt{4u^2+v^2}}{2}+v.$$
Thinking now on the integrand:
$$x^2+y^2 = \frac{-v\pm \sqrt{4u^2+v^2}}{2}+\frac{-v\pm \sqrt{4u^2+v^2}}{2}+v = \pm\sqrt{4u^2+v^2}.$$
Last equation gives us 2 options for the integrand
$i)$$\displaystyle \frac{v+\sqrt{4u^2+v^2}}{2}+\frac{v+\sqrt{4u^2+v^2}}{2}-v=\sqrt{4u^2+v^2}.$
$ii)$$\displaystyle \frac{v-\sqrt{4u^2+v^2}}{2}+\frac{v-\sqrt{4u^2+v^2}}{2}-v= -\sqrt{4u^2+v^2}.$
Of course, the volume of the solid must be positive, hence we take the first case. Then we have that:
$$\iint\limits_{D}(x^2+y^2)\;dA=\iint\limits_{S}\sqrt{4u^2+v^2}\;|\textbf{J}(T(u,v))|d\hat{A},$$
where $T(u,v)$ is the transform that takes us from $S$ to $D$. It is clear that $S$ is the rectangle:
\begin{align*}
S:\begin{cases}
u=1\\
u=2\\
v=1\\
v=2
\end{cases}.
\end{align*}
Now let's calculate the Jacobian:
$$\textbf{J}(T)= \left| {\begin{array}{cc}
\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\
\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}
\end{array}}\right|. $$
Para calcular $\displaystyle \frac{\partial x}{\partial u}$ considere las variables $u=xy$ y $v=x^2-y^2$, de donde se obtiene que:
\begin{align*}
0=\frac{\partial x}{\partial u}y+ \frac{\partial y}{\partial u}x, && 1 = 2x\frac{\partial x}{\partial u}-2y\frac{\partial y}{\partial u}.\\
\end{align*}
Multiplying the first equation times $2y$ and the second one times $x$ and adding the equations, we obtain:
\begin{align*}
2y=2y^2\frac{\partial x}{\partial u} +2xy \frac{\partial y}{\partial u}, && 0=2x^2\frac{\partial x}{\partial u}-2xy\frac{\partial x}{\partial u},
\end{align*}
$$2y = \frac{\partial x}{\partial u}(2y^2+2x^2),$$
$$\frac{\partial x}{\partial u} = \frac{2y}{2y^2+2x^2}.$$
In a similar fashion we obtain the rest of the partial derivatives:
$$\textbf{J}(T)= \left| {\begin{array}{cc}
\frac{y}{y^2+x^2} & \frac{x}{2x^2+2y^2}\\
 \frac{x}{x^2+y^2} & \frac{-y}{2x^2+2y^2} 
\end{array}}\right| = -\frac{y^2}{2(y^2+x^2)^2}-\frac{x^2}{2(y^2+x^2)^2} = -\frac{y^2+x^2}{2(y^2+x^2)^2}  = -\frac{1}{2\sqrt{4u^2+v^2}}. $$
Therefore, the original integral is now:
$$\iint\limits_{S}\sqrt{4u^2+v^2}\left(\frac{1}{2\sqrt{4u^2+v^2}}\right)\;d\hat{A} = \iint\limits_{S}\frac{1}{2}\;d\hat{A}.$$
Because of Fubini's Theorem we can write:
$$\int_1^2 \left(\int_1^2 \frac{1}{2}\;du\right)dv=\int_1^2 \left[\frac{1}{2}u\right]^2_1dv =\int_1^2 \frac{1}{2}\;dv =\boxed{\frac{1}{2}}.$$
