Finding integral over inconvenient set Put $F = \{ (x,y) \in \mathbb{R}^2 : |x^2-y^2| \leq 1, 2|xy| \leq 1 \}$. How do we find the following integral?
$$\int_F (x^2 + y^2) \,d(x,y)$$
I'm sure we need to use Jacobi's transformation theorem to get rid of that $F$ in the integral. But I can't see how.
 A: HINT: Consider the change of variables $u=x^2-y^2$, $v=xy$.
Elaborating:
By symmetry, we get $4$ times the integral over the region $S=\{(x,y): x,y\ge 0,\ -1\le x^2-y^2\le 1,\ xy\le 1/2\}$. Consider  $R=\{(u,v): -1\le u\le 1,\ 0\le v\le 1/2\}$ and $g\colon R\to S$, where $g^{-1}(x,y) = (x^2-y^2,xy) = (u,v)$. Note that $\det(Dg^{-1})(x,y) = 2(x^2+y^2)$.
Let $f(x,y)=x^2+y^2$. Then the Change of Variables Theorem gives us
\begin{align*}
\int_S f(x,y)\,dx\,dy &= \int_R f\circ g(u,v)|\det Dg(u,v)|\,du\,dv \\
&= \int_R \left(f\cdot \frac1{|\det (Dg^{-1})|}\right)\circ g(u,v)\,du\,dv \\
&= \int_R \frac 12\,du\,dv = \frac12\int_0^{1/2}\int_{-1}^1 du\,dv = \frac12.
\end{align*}
Thus, the final answer is $4\cdot\dfrac12 = 2$.
A: Let we set $x=\rho\cos\theta$ and $y=\rho\sin\theta$. The first condition translates into $|\cos 2\theta|\leq\frac{1}{\rho^2}$, the second one into $|\sin(2\theta)|\leq\frac{1}{\rho^2}$. If we fix $\rho>0$, the Lebesgue measure of the set:
$$ E_{\rho}=\left\{\theta\in[0,2\pi): \max(|\cos(2\theta)|,|\sin(2\theta)|)\leq\frac{1}{\rho^2}\right\} $$
is $2\pi$ if $\rho\leq 1$, $0$ if $\rho\geq 2^{1/4}$ and
$$ -2\pi+8\arcsin\frac{1}{\rho^2} $$
if $\rho\in[1,2^{1/4}]$. Our integral equals:
$$\begin{eqnarray*} I=\int_{0}^{2^{1/4}} \rho^3 \mu(E_\rho)\,d\rho&=&2\pi\int_{0}^{1}\rho^3\,d\rho-2\pi\int_{1}^{2^{1/4}}\rho^3\,d\rho+8\int_{1}^{2^{1/4}}\rho^3\arcsin\frac{1}{\rho^2}\,d\rho\\&=&\frac{\pi}{2}-\frac{\pi}{2}+2=\color{red}{2}.\end{eqnarray*}$$
