3. Task: Plot the figure on the coordinate plane defined by the inequality, and calculate its area $x^{2}+y^{2}+4(x-|y|) \leq 0$.
Solution: $x^{2}+y^{2}+4(x-|y|) \leq 0$

$$
\begin{aligned}
& x^{2}+4 x+y^{2}-4|y| \leq 0 \\
& x^{2}+4 x+4+y^{2}-4|y|+4 \leq 8 \\
& (x+2)^{2}+(|y|-2)^{2} \leq(2 \sqrt{2})^{2}
\end{aligned}
$$

The inequality defines a set of points symmetric with respect to the $O x$ axis. For $y \geq 0$, this is the interior of a circle with center at point $B(-2 ; 2)$ and radius $2 \sqrt{2}$.

The circles intersect the $O x$ axis at points $A(-4 ; 0)$ and $C(0 ; 0)$.

Segments $A C$ and $B D$ are equal and perpendicular, thus $A B C D$ is a square with diagonals of length 4, side length $2 \sqrt{2}$, and area equal to 8.

The square $A B C D$ cuts off sectors with angles of $90^{\circ}$ from the circles, the areas of the remaining parts of the circles are $\frac{3}{4} \pi(2 \sqrt{2})^{2}=6 \pi$. As a result, the desired area is $12 \pi+8$.

![](https://cdn.mathpix.com/cropped/2024_05_06_d04b19ad82f79b6d81e2g-18.jpg?height=945&width=848&top_left_y=1321&top_left_x=250)

Answer: $12 \pi+8$.