In the XY-plane, the area, in sq. units, of the region defined by the inequalities $y \geq x+4$ and $-4 \leq x^{2}+y^{2}+4(x-y) \leq 0$ is

$y \geq x+4$

$-4 \leq x^{2}+y^{2}+4(x-y) \leq 0$

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The second condition contains two inequalities:

$x^{2}+y^{2}+4(x-y) \geq -4$

$x^{2}+y^{2}+4(x-y) \leq 0$

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Starting with: $x^{2}+y^{2}+4(x-y) \geq -4$

Expand: $x^{2}+y^{2}+4x-4y \geq -4$

Rearrange: $x^{2}+y^{2}+4x-4y+4 \geq 0$

Complete the square:

For $x$ terms: $x^{2}+4x = (x+2)^{2}-4$

For $y$ terms: $y^{2}-4y = (y-2)^{2}-4$

Substituting: $(x+2)^{2}-4+(y-2)^{2}-4+4 \geq 0$

Simplifying: $(x+2)^{2}+(y-2)^{2} \geq 4$

This is the region outside or on a circle with center $(-2, 2)$ and radius $2$.

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Starting with: $x^{2}+y^{2}+4(x-y) \leq 0$

Expand: $x^{2}+y^{2}+4x-4y \leq 0$

Add $8$ to both sides: $x^{2}+y^{2}+4x-4y+8 \leq 8$

Complete the square: $(x+2)^{2}+(y-2)^{2} \leq 8$

This is the region inside or on a circle with center $(-2, 2)$ and radius $\sqrt{8} = 2\sqrt{2}$.

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We need the region satisfying all three conditions:

Above the line $y = x+4$

Outside the smaller circle: $(x+2)^{2}+(y-2)^{2} \geq 4$

Inside the larger circle: $(x+2)^{2}+(y-2)^{2} \leq 8$

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Check if the line $y = x+4$ passes through center $(-2, 2)$:

When $x = -2$: $y = -2+4 = 2$ ✓

The line passes through the center of both circles, dividing each into two equal halves.

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Area of larger circle $= \pi \times (\sqrt{8})^{2} = 8\pi$

Area of smaller circle $= \pi \times 2^{2} = 4\pi$

Area between circles $= 8\pi - 4\pi = 4\pi$

Since we only want the half above the line:

Required area $= \dfrac{1}{2} \times 4\pi = 2\pi$

Therefore, the area is $2\pi$ square units.