Region Bounded by Inequalities

Rewrite $-4 \le x^2 + y^2 + 4x - 4y \le 0$ by completing squares: $-4 \le (x+2)^2 - 4 + (y-2)^2 - 4 \le 0$ $4 \le (x+2)^2 + (y-2)^2 \le 8$ This region represents an annulus bounded by concentric circles centered at $(-2, 2)$ with radii $r = 2$ and $R = 2\sqrt{2}$.

The line $y = x + 4$ passes directly through the center $(-2, 2)$. The inequality $y \ge x + 4$ restricts the area to half of the annulus.

$\text{Area} = \frac{1}{2} \pi (R^2 - r^2) = \frac{1}{2} \pi (8 - 4) = 2\pi$