4.1. Find the area of the figure defined on the coordinate plane by the system

$$
\left\{\begin{array}{l}
2 \pi\left(x^{2}+y^{2}\right) \leqslant 15 \\
x^{4}-y^{4} \leqslant x y-x^{3} y^{3}
\end{array}\right.
$$
Answer: 3.75.

Solution. The first set is the interior of a circle with radius $\sqrt{\frac{15}{2 \pi}}$. The second inequality is equivalent to

$$
x^{3}\left(x+y^{3}\right) \leqslant y\left(x+y^{3}\right) \quad \Leftrightarrow \quad\left(x^{3}-y\right)\left(x+y^{3}\right) \leqslant 0
$$

By plotting the curves $y=x^{3}$ and $x=-y^{3}$ on the plane, we obtain that the system is satisfied by the shaded region. Considering that the regions marked

![](https://cdn.mathpix.com/cropped/2024_05_06_11c83b0ebf7dcf5d9828g-2.jpg?height=337&width=400&top_left_y=143&top_left_x=1505)
with the number 1 are identical, as well as the regions marked with the number 2, we get that the desired area is equal to the area of a semicircle with radius $\sqrt{\frac{15}{2 \pi}}$, which is $\frac{1}{2} \pi\left(\sqrt{\frac{15}{2 \pi}}\right)^{2}=\frac{15}{4}$.