Example 3.29. Express the double integral of the function $f(x, y)$ over the region $D$, bounded by the parabola $y=3-x^{2}$ and the line $y=-1$, as iterated integrals with different orders of integration.
Solution. Solving the system of equations $\begin{aligned} & y=3-x^{2} \\ & y=-1\end{aligned}$, we find the points of intersection of the parabola and the line. These points are $A(-2, -1)$ and
$B(2, -1)$. The vertex of the parabola is at point $C(0, 3)$; the $O y$ axis is the axis of symmetry of the parabola (Fig. 49).

![](https://cdn.mathpix.com/cropped/2024_05_22_e915d089cc29fd681be5g-178.jpg?height=456&width=534&top_left_y=316&top_left_x=461)

Fig. 49

First, consider the repeated integral over the region $D$, integrating in the inner integral with respect to $y$, and in the outer integral with respect to $x$. For this, we will consider the region $D$ as enclosed between the lines $x=-2$ and $x=2$ (the region $D$ projects onto the $O x$ axis as the segment $[-2, 2]$). The region is bounded below by the segment $A B$ with the equation $y=-1 \ (-2 \leq x \leq 2)$, and above by the arc of the parabola $y=3-x^{2} \ (-2 \leq x \leq 2)$. Therefore, the double integral will be $\int_{-2}^{2} d x \int_{-1}^{3-x^{2}} f(x, y) d y$.

Now, change the order of integration, i.e., integrate in the inner integral with respect to $x$, and in the outer integral with respect to $y$. For this, we will consider the region $D$ as enclosed between the lines $y=-1$ and $y=3$ (the region $D$ projects onto the $O y$ axis as the segment $[-1, 3]$). The region is bounded on the left by the arc of the parabola $x=-\sqrt{3-y} \ (-1 \leq y \leq 3)$, and on the right by the arc of the parabola $x=\sqrt{3-y} \ (-1 \leq y \leq 3)$. These equations are obtained by solving the equation $y=3-x^{2}$ for $x$. Therefore, the double integral will now be $\int_{-1}^{3} d y \int_{-\sqrt{3-y}}^{\sqrt{3-y}} f(x, y) d x$.

Thus,

$$
\iint_{(D)} f(x, y) d x d y=\int_{-2}^{2} d x \int_{-1}^{3-x^{2}} f(x, y) d y=\int_{-1}^{3} d y \int_{-\sqrt{3-y}}^{\sqrt{3-y}} f(x, y) d x
$$

The area $S$ of a plane region $G$ is expressed, depending on the coordinate system considered, by the following integrals:

$$
S=\iint_{G} d x d y
$$

in Cartesian rectangular coordinates, and in polar coordinates $x=r \cos \phi, y=r \sin \phi$ we have

$$
S=\iint_{G} r d r d \phi
$$