6. The figure on the coordinate plane consists of points $(x, y)$, satisfying for any $t \in \mathbb{R}$ the two inequalities

$$
x^{2}+y^{2}<\pi^{2}+t^{2}, \quad \cos y<2+\cos 2 x+\cos x(4 \sin t-1)-\cos 2 t
$$

Find the area of this figure.
Answer: $\pi^{3}-2 \pi^{2}$.

Solution. The figure, the coordinates of the points of which satisfy the inequality $x^{2}+y^{2}\cos y+\cos x$

From the last inequality, it follows that $(x, y)$ satisfying this inequality for all $t$ are precisely $(x, y)$ satisfying the inequality

$$
\cos y+\cos x 0 } \\
{ \operatorname { cos } \frac { x - y } { 2 }  0 }
\end{array} }
\end{array} \Longleftrightarrow \left[\begin{array}{c}
\left\{\begin{array}{c}
-x-\pi<y<-x+\pi \\
x+\pi<y<x+3 \pi
\end{array}\right. \\
\left\{\begin{array}{c}
-x+\pi<y<-x+3 \pi \\
x-\pi<y<x+\pi
\end{array}\right.
\end{array}\right.\right.
$$

The figure defined by these inequalities represents two small squares, and considering the periodicity, it forms a "tessellation" of small squares (see Fig. 3).

![](https://cdn.mathpix.com/cropped/2024_05_06_024d7be6fbdc002fb8c6g-4.jpg?height=540&width=646&top_left_y=1626&top_left_x=705)

Fig. 3

The intersection of the circle with the "tessellation of small squares" consists of four circular segments, the total area of which is easier to find as the area of the complement to the square defined by the condition $|x|+|y| \leqslant \pi$, within the circle $\left.x^{2}+y^{2} \leqslant \pi^{2}\right)$. Therefore, the desired area is $\pi^{3}-2 \pi^{2}$.