Compute $\iint _S (y^2+x^2)\, dA$ by change of variables. 
Let $S$ be the region in the first quadrant bounded by the curves $xy=1$, $xy=3$, $x^2-y^2=1$, and $x^2-y^2=4$. Compute $$\iint _S (y^2+x^2)\, dA.$$

I have tried to use the change of variables $u=xy$ and $v=x+y$. Then $$\begin{align}x^2+y^2=v^2-2u \\ x^2-y^2=v^2-2u \end{align}$$
Since $v^2=x^2-y^2+2u$, $$0 \le v^2 \le 4-1=3$$, so $$0 \le v \le \sqrt{3}$$
The integral becomes
$$\begin{align} \iint _S (y^2+x^2)\, dA= \int_0^\sqrt{3} \int_1^3 (v^2-2u) \left|J \right| \,du\,dv \end{align}$$
but I can't solve $x,y$ in terms of $u,v$.
I have tried another one, $u=x^2+y^2$ and $v=x^2-y^2$.
I get $x=\sqrt{\frac{u+v}{2}}$ and $y=\sqrt{\frac{u-v}{2}}$
Then $1\le xy=\frac{\sqrt{u^2-v^2} }{2} \le 3$, and $$4\le u^2-v^2 \le 36$$
The integral becomes $$\begin{align} \iint _S (y^2+x^2)\, dA= \int_1^4 \int_{\sqrt{4+v^2}}^{\sqrt{36+v^2}} u \left|J \right| \,du\,dv \end{align}$$
Hope someone could give me some suggestion how to do it.
 A: EDIT
The substitution $u=xy$, $v=x^2-y^2$ does the job. The Jacobian with respect to $x$ and $y$ is
$$
J=\left(\begin{matrix}
y & x\\
2x & -2y
\end{matrix}\right)
$$
so $|\det(J)|=2(x^2+y^2)$, which means $dudv=2(x^2+y^2)dxdy$ and your integral becomes
$$\begin{align} \iint _S (y^2+x^2) dxdy= \int _1^3 \int_1^4 2 \;dudv \end{align}$$

As Ian said, this is no more than a special case of the "inverse method", used also in order to calculate one-dimensional integrals. E.g., if you have
$$
\int \tan(x)^4 dx
$$
and you apply the substitution $t=\tan(x)$, then it is easy to obtain $dt=(1+\tan(x)^2)dx=(1+t^2)dx$, and then $dx=\frac{dt}{1+t^2}$ (of course, in this case it is equally easy to start from $x=\arctan(t)$ and so on, but sometimes you can avoid some annoying calculation).
A: Using the first method you have a standard situation for the use of the "inverse method": the Jacobian of $(u,v) \mapsto (x,y)$ is the inverse of the Jacobian of $(x,y) \mapsto (u,v)$ and so the determinants of these are reciprocals of each other. You should find the Jacobian is easy to find going the other way and then also easy to write in terms of only $u$ and $v$.
That said your bounds are not right in this method; since $x^2-y^2=v^2-2u$ is held between two constants, the bounds on $v$ as the inner variable will depend on $u$. This issue goes away if you select $v=x^2-y^2$ instead, and the Jacobian is still easily evaluated using the inverse method. This is what is detailed in user6530's answer.
Your second method at least in principle works without the use of the inverse method though I am not so confident that the integrals are easy to do in the end that way.
