Calculate the area of this function. I want to calculate the region bounded by $x^2-2xy+y^2+x+y=0$ and $x+y+2=0$.
I have drawn a sketch of this region in Geogebra:

I noticed I can use $\iint1dD$ to solve for the integral. I can apply the mapping: $(x^2-2xy+y^2)=(x-y)^2=u$, and $x+y=v$, so $x^2-2xy+y^2+x+y=u+v=0$ and $x+y+2=v+2=0$.

This is the new equation I have. Then I can evaluate $\int_{0}^{2}\int_{-x}^{0}1dA$ yes?
Or maybe I need to add the Jacobian as well...
 A: The answer is: $\frac{8\sqrt{2}}{3}$.
Explanation:
We want to calculate the area of the region $D$ bounded by $$x^{2}-2xy+y^{2}+x+y=0\quad \text{and}\quad  x+y+2=0.$$
The area of the region $D$ is given by $$\color{red}{\boxed{A(D)=\iint_{D}{\rm d}A}}.$$
Setting the change of variable,
$$u\longmapsto u(x,y)=(x-y)^{2} \quad \text{and}\quad v\longmapsto v(x,y)=x+y$$
so the area of the new region $D^{*}$ given by the change of variable is $$\color{red}{\boxed{A(D^{*})=\iint_{D^{*}}\left|\frac{\partial (x,y)}{\partial (u,v)} \right|{\rm d}A^{*}}}.$$
We have that, $$x^{2}-2xy+y^{2}+x+y=0 \iff u+v=0$$ and
$$ x+y+2=0 \iff v+2=0.$$
Then,
$$\frac{\partial (x,y)}{\partial (u,v)}=\det \begin{bmatrix} x_{u} & x_{v}\\ y_{u} & y_{v}\end{bmatrix}=\det \begin{bmatrix}-\frac{1}{2\sqrt{u}} & 1\\ \frac{1}{2\sqrt{u}} & 1 \end{bmatrix}=-\frac{2}{2\sqrt{u}}=-\frac{\sqrt{u}}{u}.$$
Hence,
$$A(D^{*})=\iint_{D^{*}}\left|-\frac{\sqrt{u}}{u} \right|{\rm d}A^{*}=\int_{0}^{2}\int_{-2}^{-u}\frac{\sqrt{u}}{u}{\rm d}v{\rm d}u=\int_{0}^{2}\frac{2-u}{\sqrt{u}}{\rm d}u=\frac{8\sqrt{2}}{3}\approx 3.7712.$$
Therefore,
$$\color{red}{\boxed{A(D)=A(D^{*})}}=\frac{8\sqrt{2}}{3}.$$
