Example 2. Change the order of integration:

$$
I=\int_{0}^{1} d x \int_{0}^{x^{2}} f(x, y) d y+\int_{1}^{\sqrt{2}} d x \int_{0}^{\sqrt{2-x^{2}}} f(x, y) d y
$$
Solution. We will construct the region of integration (Fig. 3.5), express it as a system of inequalities, and transition to another iterated integral:

![](https://cdn.mathpix.com/cropped/2024_05_22_446938f1e4808e7d6e07g-051.jpg?height=304&width=454&top_left_y=1337&top_left_x=73)

$$
\begin{aligned}
D:\left\{\begin{array}{l}
0 \leqslant x \leqslant 1 \\
0 \leqslant y \leqslant x^{2}
\end{array}\right. & \cup\left\{\begin{array}{l}
1 \leqslant x \leqslant \sqrt{2} \\
0 \leqslant y \leqslant \sqrt{2-x^{2}}
\end{array} \Leftrightarrow\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
0 \leqslant y \leqslant 1 \\
\sqrt{y} \leqslant x \leqslant \sqrt{2-y^{2}}
\end{array}\right.
\end{aligned}
$$

Fig. 3.5

$$
\text { Answer. } I=\int_{0}^{1} d y \int_{\sqrt{y}}^{\sqrt{2-y^{2}}} f(x, y) d x
$$