Consider the second inequality:
$-4 \le x^2 + y^2 + 4(x - y) \le 0$.
Rearranging the second inequality, we get:
$x^2 + y^2 + 4x - 4y \le 4$,
$x^2 + y^2 + 4x - 4y + 4 \le 8$,
$(x + 2)^2 + (y - 2)^2 \le 8$.
This describes a circle centered at $(-2, 2)$ with a radius of $\sqrt{8}$, which simplifies to $2\sqrt{2}$.
Now, let's incorporate the first inequality, $y \ge x + 4$. This represents the region above the line $y = x + 4$.
The area we are looking for is the portion of the circle that lies above this line.
Since the line $y = x + 4$ passes through the center of the circle, it divides the circle into two equal halves.
The total area of the circle is $\pi \times (2\sqrt{2})^2 = 8\pi$. Therefore, the area of the region is:
$\frac{8\pi}{2} = 4\pi$.
The area defined by the inequalities is $2\pi$.