10.1. Find the area of the figure defined on the coordinate plane by the inequalities $|x|-1 \leq y \leq \sqrt{1-x^{2}}$
Answer: $\frac{\pi}{2}+1$. Solution: The figure has the appearance shown in the diagram: it is bounded from below by the graph $\quad y=|x|-1, \quad$ and from above

by the semicircle $y=\sqrt{1-x^{2}} \Leftrightarrow\left\{\begin{array}{l}x^{2}+y^{2}=1 \\ y \geq 0\end{array}\right.$.

The area of the figure consists of the area of a semicircle with a unit radius and half

![](https://cdn.mathpix.com/cropped/2024_05_06_0f550f224e2f9692eebdg-1.jpg?height=623&width=671&top_left_y=131&top_left_x=1041)
of a square with a diagonal equal to 2.