Double Integral With a Change of Variables 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 .
 A: Use this change of variables
$$\begin{align}
u&=x^2+y^2\\
v&=\frac{x^2}{y}
\end{align}$$
So we may compute the Jacobian first
$$\frac{\partial(u,v)}{\partial(x,y)}=
\begin{vmatrix}
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\
\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}
\end{vmatrix}
=
\begin{vmatrix}
2x & 2y \\
2\frac{x}{y} & -\frac{x^2}{y^2}
\end{vmatrix}
=-2\frac{x^3}{y^2}-4xy = -2(\frac{x^3+2xy^3}{y^2})=-2\frac{x}{y^2}(x^2+2y^2)
$$
We also know that
$$\frac{\partial(x,y)}{\partial(u,v)}=\frac{1}{\frac{\partial(u,v)}{\partial(x,y)}}=\frac{y^2}{-2x(x^2+2y^2)}$$
Consequently, your integral becomes
$$\begin{align}
I&=\int_{1}^{2}\int_{1}^{2}\frac{x^2+2y^2}{xy} \cdot \frac{y^2}{-2x(x^2+2y^2)}dudv \\
&=-\frac{1}{2}\int_{1}^{2}\int_{1}^{2} \frac{y}{x^2} dudv \\
&=-\frac{1}{2}\int_{1}^{2}\int_{1}^{2} \frac{1}{v} dudv \\
&=-\frac{1}{2}\left(\int_{1}^{2}\frac{1}{v}dv\right)\left(\int_{1}^{2}du\right) \\
&=-\frac{1}{2}(\ln2)(1) \\
&=\boxed{-\ln\sqrt{2}}
\end{align}$$
The domain of integration is showed in the following figure.

A: Perhaps you could apply Green's Theorem.
You are trying to integrate the differential two-form
$$\alpha := \frac{2y^2+x^2}{xy}~\mathrm{d}x\wedge\mathrm{d}y$$
over the region $D$. Green's Theorem tells us that we can rewrite this as a line integral around the boundary of $D$, written $\partial D$. All we have to do is fine a one-form, say $\omega$, whose exterior derivative is $\alpha$, i.e. $\mathrm d \omega = \alpha$. Well, it's not difficult the see that
$$\omega = -\frac{y^2}{x}~\mathrm{d}x+\frac{x^2}{2y}~\mathrm{d}y$$
will work.Hence, by Green's Theorem:
$$\iint_D \frac{2y^2+x^2}{xy}~\mathrm{d}x\wedge\mathrm{d}y = \oint_{\partial D} -\frac{y^2}{x}~\mathrm{d}x+\frac{x^2}{2y}~\mathrm{d}y$$
All you need to do is parametrise the boundary $\partial D$. 
