Example 1. Express the double integral $I=\int_{D} \int f(x, y) d x d y$ as iterated integrals, if $D-$ is the region depicted in Fig. 3.4, and the boundary is given by the equations

$$
y=\cos \frac{x}{2}, \quad x=-\frac{\pi}{2}, \quad y=\frac{x}{2}-\frac{\pi}{4}, \quad x=\frac{\pi}{2}
$$
Solution. Let's write $D$ as a system of inequalities and form repeated integrals.

The analysis of Fig. 3.4 shows that the region $D$ is bounded from below by the graph of the function $y=\frac{x}{2}-\frac{\pi}{4}$, and from above by the graph of the function $\cos \frac{x}{2}$, and these functions are defined on the entire interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Therefore, the region $D$ can be written as a system of inequalities

$$
D:\left\{\begin{aligned}
-\frac{\pi}{2} & \leqslant x \leqslant \frac{\pi}{2} \\
\frac{x}{2}-\frac{\pi}{4} & \leqslant y \leqslant \cos \frac{x}{2}
\end{aligned}\right.
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_446938f1e4808e7d6e07g-050.jpg?height=500&width=506&top_left_y=958&top_left_x=1974)

Fig. 3.4
In addition, it can be concluded that the region $D$ is bounded on the left by the graphs of two functions of the form $x=x(y): x=-\frac{\pi}{2}$ and $x=-2 \arccos y$ (this function is obtained from the equation $y=\cos \frac{x}{2}$), and on the right by the graphs of three functions: $x=2 y+\frac{\pi}{2}$ (the function is obtained from the equation $y=\frac{x}{2}-\frac{\pi}{4}$), $x=\frac{\pi}{2}$, and $x=2 \arccos y$. The graphs of all these functions form the boundary of the region $D$. Thus, the region $D$ can be written as a union of systems

4 K.N. Lungu, E.V. Makarov

$$
\begin{gathered}
D_{1}:\left\{\begin{array}{rl}
-\frac{\pi}{2} & \leqslant y \leqslant 0, \\
-\frac{\pi}{2} & \leqslant x \leqslant 2 y+\frac{\pi}{2}
\end{array} \cup \quad D_{2}:\left\{\begin{array}{c}
0 \leqslant y \leqslant \frac{\sqrt{2}}{2} \\
-\frac{\pi}{2} \leqslant x \leqslant \frac{\pi}{2}
\end{array}\right.\right. \\
\cup \quad D_{3}:\left\{\begin{array}{c}
\frac{\sqrt{2}}{2} \leqslant y \leqslant 1 \\
-2 \arccos y \leqslant x \leqslant 2 \arccos y
\end{array}\right.
\end{gathered}
$$

Therefore,

$$
\begin{aligned}
I= & \int_{-\pi / 2}^{\pi / 2} d x \int_{x / 2}^{\cos (x / 2)} f(x, y) d y=\int_{-\pi / 4}^{0} d y \int_{-\pi / 2}^{2 y+\pi / 2} f(x, y) d x+ \\
& +\int_{0}^{\sqrt{2} / 2} d y \int_{-\pi / 2}^{\pi / 2} f(x, y) d x+\int_{\sqrt{2} / 2}^{1} d y \int_{-2 \arccos y}^{2 \arccos y} f(x, y) d x
\end{aligned}
$$

The repeated integrals on either side of the second equality differ in the order of integration.