Area enclosed by what?

Preform the coordinate transformation $u=y-x,v=y+x$ . Note the Jacobian has magnitude $\bigg|\text{det}\begin{pmatrix}1&1\\-1&1\end{pmatrix}\bigg|=2$ .

In the new coordinates, the region is mapped to the region bound by $u^2=4,v^2=4$ . This is a square of side length four centered at the origin, so has area $16$ .

Therefore, in the $xy$ -plane the original region has area $\frac{16}{2}=\boxed{8}$ .

$(y-x)^2=4,~~~\implies~~y=x+2 ......(1) \\~~and~~ ~~~y=x-2 ......(2) \\ (y+x)^2=4,~~~~~~\implies~~y=-x+2 ......(3) \\ ~~ and~~ ~~~y=-x-2 ......(4) \\ (1)~ and~(3) ~intersect~at~x=0~and~y=2. \\ (2)~ and~(4) ~intersect~at~x=0~and~y=-2. \\ (1)~ and~(4) ~intersect~at~x=-2~and~y=0. \\ (2)~ and~(3) ~intersect~at~x=2~and~y=0. \\ So~Common ~area ~is~a~square~ABCD(0,2),(-2,0) ,(0,-2)~and~ (2,0) .\\ The~ diagonal ~AC=4,\\ So~area~ABCD~=(diagonal)^2/2=\Large 8.\\$

$(y-x)^{2}=4 =>$ $y=x+2$ or $y=x-2$ $(y+x)^{2}=4 => y=-x+2$ or $y=-x-2$ So we are asked for the area of $4$ times the right triangle formed by $y=x+2$ and the coordinate axes. $4\cdot \frac{2\cdot2}{2} = 8$