Question: I need to calculate:
$$
\iint \_D \frac{2y^2+x^2}{xy}~\mathrm dx~\mathrm dy
$$
over the set $D$ which is:
$$
y\leq x^2 \leq 2y , \quad 1\leq x^2 +y^2 \leq 2 , \quad x\geq 0 
$$
can someone help me understand what possible change of variables can I do here?

Thanks a lot in advance .

Answer: 
You have $$\begin{aligned}\vert f(x)-f(y) \vert &= \left\vert \sqrt{x^2+1}-\sqrt{y^2+1} \right\vert \\
&= \left\vert (\sqrt{x^2+1}-\sqrt{y^2+1}) \frac{\sqrt{x^2+1}+\sqrt{y^2+1}}{\sqrt{x^2+1}+\sqrt{y^2+1}} \right\vert \\
&= \left\vert \frac{x^2-y^2}{\sqrt{x^2+1}+\sqrt{y^2+1}} \right\vert \\
&\le \frac{\vert x-y \vert (\vert x \vert + \vert y \vert )}{ \sqrt{x^2+1}+\sqrt{y^2+1}} \\
&\le \vert x-y \vert
\end{aligned}$$ hence choosing $\delta = \epsilon$ will work.
Note that $f'(x)= x/\sqrt {x^2+1}$ for all $x.$ We thus have $|f'(x)|<1$ everywhere. Suppose $x,y\in \mathbb R.$ Then the mean value theorem gives

$$f(y)-f(x) = f'(c)(y-x) \implies |f(y)-f(x)| \le |y-x|.$$

This easily gives the uniform continuity of $f$ on $\mathbb R.$