Example 8. Calculate the volume of the body bounded above by the surface $z=x y^{2}$, and below by the rectangle $0 \leqslant x \leqslant 1,0 \leqslant y \leqslant 2$.
Solution. The required volume can be calculated using the formula

$$
V=\int_{D} \int_{2} z d x d y, \text { where } D-\text { is the given rectangle, and } z=x y^{2} .
$$

The double integral can be reduced to the product of definite integrals, whose limits of integration are numbers:

$$
V=\int_{0}^{1} x d x \cdot \int_{0}^{2} y^{2} d y=\left.\left.\frac{x^{2}}{2}\right|_{0} ^{2} \cdot \frac{y^{3}}{3}\right|_{0} ^{2}=\frac{4}{3}
$$

Answer. $V=\frac{4}{3}$.

## Exercises

Change the order of integration.

1. $\int_{0}^{1} d x \int_{2 x}^{3 x} f(x, y) d y$.
2. $\int_{0}^{1} d y \int_{-\sqrt{1-y^{2}}}^{1-y} f(x, y) d x$.
3. $\int_{0}^{1} d y \int_{y^{2} / 2}^{\sqrt{3-y^{2}}} f(x, y) d x$.
4. $\int_{a / 2}^{a} d x \int_{0}^{\sqrt{2 a x-x^{2}}} f(x, y) d y$.
5. $\int_{0}^{\pi} d x \int_{0}^{\sin x} f(x, y) d y$.
6. $\int_{0}^{48} d y \int_{y / 12}^{\sqrt{y / 3}} f(x, y) d x$.

Calculate the double integrals.
7. $\iint_{D}\left(x^{2}+2 y\right) d x d y, D: 0 \leqslant x \leqslant 1,0 \leqslant y \leqslant 2$.
8. $\iint_{D}(x+2 y) d x d y, D: y^{2}-4 \leqslant x \leqslant 5,-3 \leqslant y \leqslant 3$.
9. $\iint_{D} \frac{d x d y}{(x+y)^{2}}, D: 3 \leqslant x \leqslant 4, \quad 1 \leqslant y \leqslant 2$.
10. $\iint_{D} \frac{x d x d y}{x^{2}+y^{2}}, \quad D$ is bounded by the parabola $y=x^{2}$ and the line $y=x$.
11. $\iint_{D} x y d x d y, D$ is bounded by the $O x$ axis and the circle $(x-2)^{2}+$ $+y^{2}=1(y \geqslant 0)$.
12. $\int_{D} \int_{x-y^{2}}^{x y} d x d y, D-$ is a triangle with vertices $O(0,0)$, $A(10,1), B(1,1)$.

Calculate the areas of figures bounded by the given lines.
13. $x=y, x=2 y, x+y=8, x+3 y=8$.
14. $x+3 y=0, y^{2}=4+x$.
15. $y^{2}=10 x+25, y^{2}=-6 x+9$.
16. $y=x^{2}, 4 y=x^{2}, y=4$.

Calculate the volumes of bodies bounded by the given surfaces.
17. $y=\sqrt{x}, y=2 \sqrt{x}, x+z=6, z=0$.
18. $x+y+z=9,3 y+y=3,3 x+2 y=18, y=0, z=0$.
19. $z=0, z=\frac{x^{2}}{y^{2}}, x=2, y=x, x y=1$.

20. $z=0, z=1-\frac{x}{2}-\frac{y}{2}, y=x, y=x^{2}$.
Answers
21. $\int_{0}^{2} d y \int_{y / 3}^{y / 2} f(x, y) d x+\int_{2}^{3} d y \int_{y / 3}^{1} f(x, y) d x$.
22. $\int_{-1}^{0} d x \int_{0}^{\sqrt{1-x^{2}}} f(x, y) d y+\int_{0}^{1} d x \int_{0}^{1-x} f(x, y) d y$.
23. $\int_{0}^{1 / 2} d x \int_{0}^{\sqrt{2 x}} f(x, y) d y+\int_{1 / 2}^{\sqrt{2}} d x \int_{0}^{1} f(x, y) d y+\int_{\sqrt{2}}^{\sqrt{3}} d x \int_{0}^{\sqrt{3-x^{2}}} f(x, y) d y$.
24. $\int_{0}^{a \sqrt{3} / 2} d y \int_{a / 2}^{a} f(x, y) d x+\int_{a \sqrt{3} / 2}^{a} d y \int_{a-\sqrt{a^{2}-y^{2}}}^{a} f(x, y) d x$. 5. $\int_{0}^{1} \int_{\arcsin y}^{\pi-\arcsin y} f(x, y) d x .6 . \int_{0}^{4} d x \int_{3 x^{2}}^{2 \pi} f(x, y) d y .7 . \frac{14}{3} .8 .50,4.9 . \ln \frac{25}{24}$.
25. $\ln 2$
26. 
27. $\frac{125}{6}$.
28. $\frac{16 \sqrt{15}}{3}$.
29. $\frac{32}{3}$.
30. $\frac{48 \sqrt{6}}{5}$.
31. $40,5.19 .2,25.20 . \frac{11}{120}$.

## § 2. Change of Variables in a Double Integral

The task is to simplify the integration procedure by simplifying the description of the integration domain or the integrand. In any case, we deal with a change of variables.

## 2.1. Curvilinear Coordinates

$1^{\circ}$. Suppose in the plane $O u v$ there is a region $G$ (Fig. 3.10), and in it are defined continuous and differentiable functions

$$
\left\{\begin{array}{l}
x=x(u, v) \\
y=y(u, v)
\end{array}\right.
$$

which transform points of the region $G$ into the region $D$ of the plane $O x y$ (Fig. 3.11).

![](https://cdn.mathpix.com/cropped/2024_05_22_446938f1e4808e7d6e07g-054.jpg?height=363&width=473&top_left_y=1301&top_left_x=129)

Fig. 3.10

![](https://cdn.mathpix.com/cropped/2024_05_22_446938f1e4808e7d6e07g-054.jpg?height=369&width=434&top_left_y=1298&top_left_x=697)

Fig. 3.11
The rectangle $A B C D$ of the region $G$ with vertices $A(u, v), B(u+$ $+\Delta u, v), C(u+\Delta u, v+\Delta v), D(u, v+\Delta v)$ and with area equal to $d \sigma=d u d v$, is transformed by functions (1) into the curvilinear parallelogram $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$. $2^{\circ}$. Let $z=f(x, y)$ be a function continuous in the region $D$ of the plane $O x y, G$ - the region of the plane $O u v$, which is transformed into $D$ by the system of functions (1). Then

$$
\iint_{D} f(x, y) d x d y=\iint_{G} f[x(u, v), y(u, v)]|I(u, v)| d u d v
$$

where $I(u, v)=\left|\begin{array}{ll}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{array}\right|$

This is the formula for changing variables in a double integral, and $I(u, v)$ is called the Jacobian of the transformation, or the Jacobian determinant.

## 2.2. Polar Coordinates

