Area bounded by absolute value equation

The region is symmetric in all four quadrants. In the first quadrant ($x > 0, y > 0$):

• If $x > y$: $(x+y) + (x-y) = 4 \implies 2x = 4 \implies x = 2$.
• If $y > x$: $(x+y) + (y-x) = 4 \implies 2y = 4 \implies y = 2$. Thus, in the first quadrant, the region is bounded by $0 \le x \le 2$ and $0 \le y \le 2$, forming a square of side 2 with area $2 \times 2 = 4$. By symmetry, the total area enclosed in all 4 quadrants $= 4 \times 4 = 16$.