Calculate the area of this function.

Solution 1:

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, $$\color{red}{\boxed{A(D)=A(D^{*})}}=\frac{8\sqrt{2}}{3}.$$