11.8 Find the area of the figure defined on the coordinate plane by the inequality $x^{2}+y^{2} \leq 2(|x|-|y|)$.
Answer: $2 \pi-4$.

![](https://cdn.mathpix.com/cropped/2024_05_06_7a0ab87e7d282e4dcdc0g-2.jpg?height=731&width=766&top_left_y=140&top_left_x=288)

Solution. It is obvious that the figure

is symmetric with respect to the coordinate axes and the origin (since the inequality does not change when the signs of $x, y$ are changed). Therefore, it is sufficient to consider the part of the figure in the first quadrant and multiply the area of this part by 4. For non-negative $x, y$, the inequality can be written as $(x-1)^{2}+(y+1)^{2} \leq 2$. Thus, in the first quadrant, we have a part of a circle with radius $\sqrt{2}$ and center at $(1; -1)$ (see the figure). This part of the circle is a segment with a central angle of $90^{\circ}$ (this is the angle between the two radii drawn to the boundary points $(0; 0)$ and $(2; 0)$ of this segment). Therefore, the area of the segment is $\pi(\sqrt{2})^{2} / 4 - 1$, and the area of the figure is

$2 \pi-4$.