Help with change of variables in a double integral

Solution 1:

$S$ is a union of two disjoint regions $S_1$ and $S_2$ where $S_1$ is in the first quadrant and $S_2$ is in the second quadrant.

There is no need to express $x,y$ in terms of $u,v$ in this problem. Observe that $$\frac{\partial(x,y)}{\partial(u,v)}=\frac{1}{{\partial(u,v) \over \partial(x,y)}}=-\frac{1}{\frac{x^2}{y^2}+\frac{4x}{y}}=-\frac{y^2}{x^2+4xy}$$ If $(x,y)\in S_1,$ $$\begin{eqnarray*}\Big[\frac{1}{xy}+\frac{2}{2xy+x^2}\Big]\Bigg|\frac{\partial(x,y)}{\partial(u,v)}\Bigg| &=& \frac{y}{x^2(x+2y)} &=& {1 \over uv}\end{eqnarray*}$$ Now if $(x,y)\in S_2,$ $$\begin{eqnarray*}\Big[\frac{1}{xy}+\frac{2}{2xy+x^2}\Big]\Bigg|\frac{\partial(x,y)}{\partial(u,v)}\Bigg| &=& -\frac{y}{x^2(x+2y)} &=& -{1 \over uv}\end{eqnarray*}$$ Because $(x,y)\mapsto \Big(x+2y,{x^2 \over y}\Big)$ maps of each $S_1$ and $S_2$ bijectively to $(1,2)^2$ we can say $$\begin{eqnarray*}\iint_{S}\Big[\frac{1}{xy}+\frac{2}{2xy+x^2}\Big]\mathrm{d}A&=&\iint_{S_1}\Big[\frac{1}{xy}+\frac{2}{2xy+x^2}\Big]\mathrm{d}A+\iint_{S_2}\Big[\frac{1}{xy}+\frac{2}{2xy+x^2}\Big]\mathrm{d}A \\ &=& \iint_{(1,2)^2}\frac{\mathrm{d}u\mathrm{d}v}{uv}+\iint_{(1,2)^2}-\frac{\mathrm{d}u\mathrm{d}v}{uv} \\ &=& 0\end{eqnarray*}$$ In fact, $$\iint_{S_1}\Big[\frac{1}{xy}+\frac{2}{2xy+x^2}\Big]\mathrm{d}A=-\iint_{S_2}\Big[\frac{1}{xy}+\frac{2}{2xy+x^2}\Big]\mathrm{d}A=\ln^2(2)$$