Double integral and change of variable $$
\iint_D \left(x^2-y^2\right)\ dxdy
$$
over $D$ which is bounded by region enclosed by the four curves $y = x, y = x + 1, xy = 1$ and $xy = 2$ in the first quadrant.
What will be a suitable change of variable for this question?
 A: Notice that
$$D=\{(x,y)\in \mathbb{R}^{2}: y=x, y=x+1, xy=1, xy=2\}$$
With the change of variables $$u=x-y\quad \text{and}\quad v=xy$$
We can re-write the region $D$ as
$$D^{*}=\{(u,v)\in \mathbb{R}^{2}: -1\leqslant u \leqslant 0, \quad 1\leqslant v\leqslant 2\}=[-1,0]\times[1,2]$$
Hence by the change of variables theorem for double integration, we have
$$\iint_{D}f(x,y)\, {\rm d}A=\iint_{D^{*}}f(x(u,v),y(u,v))\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\, {\rm d}A^{*}$$
where $f(x,y)=x^{2}-y^{2}$.
Well,

*

*We get $\displaystyle \left|\frac{\partial(u,v)}{\partial(x,y)}\right|=x+y>0$ over first quadrant. Hence since $\displaystyle \left|\frac{\partial(x,y)}{\partial(u,v)}\frac{\partial(u,v)}{\partial(x,y)}\right|=1$ we get $\displaystyle \left|\frac{\partial(x,y)}{\partial(u,v)}\right|=\frac{1}{x+y}$.


*Also we have $f(x,y)=x^{2}-y^{2}=(x-y)(x+y)$.
Therefore,
\begin{align*}
\iint_{D}(x^{2}-y^{2})\, {\rm d}A&=\iint_{D^{*}}(x+y)(x-y)\frac{1}{x+y}\, {\rm d}A^{*},\\ &=\int_{1}^{2}\int_{-1}^{0}u\, {\rm d}u\, {\rm d}v,\\&=\int_{1}^{2}-\frac{1}{2}\, {\rm d}v,\\&=-\frac{1}{2}.
\end{align*}
Just a small remark:

*

*First notice that we we are not integrating with density $1$ so there is no reason to expect a positive quantity for the double integral.


*I think that the selected transformations satisfy the injectivity of the change of variable, in addition to being of class  $C^1$, so everything seems to work fine out there.


*Of course a double integral can be negative, in this case notice that $u\in [-1,0]$.


*I can see that we do not agree with David G. Stork, please indicate me if my answer is incorrect and I will proceed to correct it and if it is hopelessly broken I will proceed to eliminate it.
A: No need to change variables:

$$\int\limits_{x=(\sqrt{5}-1)/2}^1 (x+1) - 1/x\ dx + \int\limits_{x=1}^\sqrt{2} 2/x - x\ dx = \frac{3}{4}-\frac{\sqrt{5}}{4}+\log \left(\sqrt{5}-1\right) = 0.402918$$
The first integral expresses the green area and the second the red area.
(This does not address the full $x^2 - y^2$ integrand question, which was not clear at the outset.)
