796. Change the order of integration in the integral:
1) $\left.I_{1}=\int_{-2}^{2} d x \int_{x^{2}}^{4} f(x, y) d y ; 2\right) I_{2}=\int_{1}^{3} d y \int_{0}^{2 y} u d x$.

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-268.jpg?height=128&width=454&top_left_y=322&top_left_x=144)
Solution. 1) First, we determine the region of integration from the limits of integration. By setting $x$ equal to the limits of integration with the variable $x$, and $y$ equal to the limits of integration with the variable $y$, we obtain the equations of the lines bounding this region: $x=-2, x=2, y=x^{2}, y=4$.

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-268.jpg?height=319&width=261&top_left_y=703&top_left_x=102)

Fig. 152

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-268.jpg?height=257&width=351&top_left_y=760&top_left_x=398)

Fig. 153

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-268.jpg?height=360&width=411&top_left_y=675&top_left_x=773)

Fig. 154

By plotting these lines in Fig. 152, we obtain the parabolic segment $O A B$, symmetric about the axis $O \varphi$.

Integrating in the other order—first with respect to $x$, then with respect to $y$—the limits of the inner integral are found by solving the equation of the parabola for $x$: $x=-\sqrt{y}$ and $x=\sqrt{y}$. The limits of the outer integral $y=0$ and $y=4$ are found as the minimum and maximum values of $y$ in the entire region $OAB$. Therefore,

$$
\int_{-2}^{2} d x \int_{x^{2}}^{4} f(x, y) d y = \int_{0}^{4} d y \int_{-\sqrt{y}}^{\sqrt{y}} f(x, y) d x
$$

2) Here, the region of integration is bounded by the lines $y=1$, $y=3$, $x=0$, and $x=2y$. In Fig. 153, it represents the trapezoid $A B C D$.

When integrating in the other order, first with respect to $y$, it is necessary to divide the region $A B C D$ by the line $B H$, parallel to the $O y$ axis, into two parts, as the lower boundary line of this region consists of two parts $A B$ and $B C$, which have different equations: $y=1$ and $y=\frac{x}{2}$.

As a result, the integral $I_{2}$, when changing the order of integration, will be equal to the sum of two integrals:

$$
I_{2}=\int_{0}^{2} d x \int_{1}^{3} u d y + \int_{2}^{6} d x \int_{\frac{x}{2}}^{3} u d y
$$

3) Writing the equations of the lines bounding the region of integration: $x=-\sqrt{3}$, $x=1$, $y=-\sqrt{4-x^{2}}$, $y=0$, and plotting them in Fig. 154, we obtain the curvilinear trapezoid $A B C D$.

If integrating in the other order, first with respect to $x$, the region $A B C D$ must be divided by the lines $B B_{1}$ and $C C_{1}$, parallel to the $O x$ axis, into three regions, in each of which the left and right boundary lines are each defined by one equation.

In the region $C_{1} P C$, enclosed in the horizontal strip between the lines $y=-2$ and $y=-\sqrt{3}$, the left boundary line $C_{1} P$ has the equation $x=-\sqrt{4-y^{2}}$, and the right boundary line $P C$ has the equation $x=\sqrt{4-y^{2}}$.

In the region $B C_{1} C B_{1}$, which is enclosed in the horizontal strip between the lines $y=-\sqrt{3}$ and $y=-1$, the left boundary line is defined by the equation $x=-\sqrt{4-y^{2}}$, and the right boundary line by the equation $x=1$.

In the region $A B B_{1} D$, enclosed in the horizontal strip between the lines $y=-1$ and $y=0$, the equation of the left boundary line is $x=-\sqrt{3}$, and the equation of the right boundary line is $x=1$.

Therefore, with the new order of integration, the integral $I_{3}$ will be equal to the following sum of three integrals:

$$
I_{3}=\int_{-2}^{-\sqrt{3}} d y \int_{-\sqrt{4-y^{2}}}^{\sqrt{4-y^{2}}} v d x + \int_{-\sqrt{3}}^{-1} d y \int_{-\sqrt{4-y^{2}}}^{1} v d x + \int_{-1}^{0} d y \int_{-\sqrt{3}}^{1} v d x
$$