Example. Evaluate the double integral

$$
\iint_{D} \frac{x}{y^{5}} d x d y
$$

where the region $D$ is defined by the inequalities

$$
1 \leq \frac{x^{2}}{16}+y^{2} \leq 3, \quad y \geq \frac{x}{4}, \quad x \geq 0
$$
SOLUTION.

1. The region $D$ is defined by inequalities in the Cartesian coordinate system:

$$
D=\left\{(x, y): \begin{array}{c}
1 \leq \frac{x^{2}}{16}+y^{2} \leq 3 \\
\\
y \geq \frac{x}{4}, \quad x \geq 0
\end{array}\right\}
$$

2. Since the region $D$ is bounded by ellipses and lines passing through the origin, it is easier to solve the problem using generalized polar coordinates

$$
\left\{\begin{array}{l}
x=4 \varrho \cos \varphi \\
y=\varrho \sin \varphi
\end{array}\right.
$$

In this case, $(\varrho, \varphi) \in D^{\prime}$, and the desired integral is given by the formula

$$
\iint_{D} \frac{x}{y^{5}} d x d y=\iint_{D^{\prime}} \frac{4 \varrho \cos \varphi}{\varrho^{5} \sin ^{5} \varphi} 4 \varrho d \varrho d \varphi
$$

3. To find the region $D^{\prime}$, we replace $x$ with $a \varrho \cos \varphi$ and $y$ with $b \varrho \sin \varphi$ in the inequalities defining the region $D$:

$$
\left\{\begin{array}{l}
1 \leq \frac{16 \varrho^{2} \cos ^{2} \varphi}{16}+\varrho^{2} \sin ^{2} \varphi \leq 3 \\
\varrho \sin \varphi \geq \frac{4 \varrho \cos \varphi}{4}, \varrho \cos \varphi \geq 0
\end{array}\right.
$$

Solving these inequalities with respect to $\varrho$ and $\varphi$, we get

![](https://cdn.mathpix.com/cropped/2024_05_22_5878ce417e2ca061000bg-300.jpg?height=126&width=561&top_left_y=1443&top_left_x=344)

4. Transitioning from the double integral to the repeated integral and integrating sequentially, we obtain

$$
\iint_{D} \frac{x}{y^{5}} d x d y=\iint_{D^{\prime}} \frac{4 \varrho \cos \varphi}{\varrho^{5} \sin ^{5} \varphi} 4 \varrho d \varrho d \varphi=\int_{\pi / 4}^{\pi / 2} d \varphi \int_{1}^{\sqrt{3}} 16 \varrho^{-3} \cdot \frac{\cos \varphi}{\sin ^{5} \varphi} d \varrho=
$$

$$
=16 \int_{\pi / 4}^{\pi / 2} \frac{d \sin \varphi}{\sin ^{5} \varphi} \int_{1}^{\sqrt{3}} \varrho^{-3} d \varrho=\left.\left.16\left(-\frac{1}{4 \sin ^{4} \varphi}\right)\right|_{\pi / 4} ^{\pi / 2}\left(-\frac{1}{2 \varrho^{2}}\right)\right|_{1} ^{\sqrt{3}}=4
$$

Answer. $\iint_{D} \frac{x}{y^{5}} d x d y=4$.

PROBLEM CONDITIONS. Compute the double integrals.

1. $\iint_{D} \frac{y}{x} d x d y$

$$
D=\left\{1 \leq \frac{x^{2}}{9}+\frac{y^{2}}{4} \leq 2, y \geq 0, y \leq \frac{2 x}{3}\right\}
$$

2. $\int_{D}^{D} \int^{x} \frac{x}{y} d x d y$ $D=\left\{1 \leq \frac{x^{2}}{4}+\frac{y^{2}}{9} \leq 4, x \geq 0, y \geq \frac{3 x}{2}\right\}$.
3. $\iint_{D} \frac{x}{y} d x d y$ $D=\left\{1 \leq \frac{x^{2}}{16}+\frac{y^{2}}{4} \leq 4, x \geq 0, y \geq \frac{x}{2}\right\}$.
4. $\int_{D}^{D} \int^{3} y d x d y$ $D=\left\{1 \leq \frac{x^{2}}{4}+\frac{y^{2}}{9} \leq 1, x \geq 0, y \geq 0\right\}$.
5. $\iint_{D} \frac{x}{y} d x d y$
$D=\left\{1 \leq \frac{x^{2}}{4}+\frac{y^{2}}{16} \leq 5, x \geq 0, y \geq 2 x\right\}$.
6. $\iint_{D} \frac{x}{y} d x d y$
$D=\left\{1 \leq \frac{x^{2}}{9}+\frac{y^{2}}{4} \leq 5, x \geq 0, y \geq \frac{2 x}{3}\right\}$.
7. $\iint_{D} \frac{x}{y^{3}} d x d y$
$D=\left\{1 \leq \frac{x^{2}}{4}+y^{2} \leq 25, x \geq 0, y \geq \frac{x}{2}\right\}$.
8. $\int_{D}^{D} \int^{y} \frac{y}{x^{3}} d x d y$
$D=\left\{1 \leq x^{2}+\frac{y^{2}}{16} \leq 9, y \geq 0, y \leq 4 x\right\}$.
9. $\iint_{D} \frac{8 y}{x^{3}} d x d y$
$D=\left\{1 \leq \frac{x^{2}}{4}+y^{2} \leq 4, y \geq 0, y \leq \frac{x}{2}\right\}$.
10. $\iint_{D} \frac{9 x}{y^{3}} d x d y$
$D=\left\{1 \leq \frac{x^{2}}{4}+\frac{y^{2}}{9} \leq 36, x \geq 0, y \geq \frac{3 x}{2}\right\}$.

Answers. 1. $I=\ln 2$.
2. $I=3 \ln 2$.
3. $I=12 \ln 2$.
4. $I=6$.

5. $I=4 \ln 2$.
6. $I=9 \ln 2$.
7. $I=2 \ln 5$.
8. $I=8 \ln 3$.
9. $I=\ln 2$.
10. $I=2 \ln 6$.

## 12.5. Calculation of Volumes Using Double Integrals

PROBLEM STATEMENT. Find the volume of the body bounded by the surfaces

$g_{i}(x, y)=0(i=1,2, \ldots), \quad z=f_{1}(x, y), z=f_{2}(x, y)\left(f_{2}(x, y) \geq f_{1}(x, y)\right)$.

## PLAN OF SOLUTION.

1. The volume of the cylindrical prism bounded by the given surfaces is determined by the formula

$$
V=\iint_{D}\left[f_{2}(x, y)-f_{1}(x, y)\right] d x d y
$$

where $D$ is the projection of the body onto the $X O Y$ plane.

2. To find $D$, we define the body using inequalities and eliminate $z$.

For example, suppose the coordinates of the points of the body satisfy the inequalities $0 \leq z \leq f(x, y), g_{1}(x, y) \geq 0$ and $g_{2}(x, y) \leq 0$. Then the body is defined by the system of inequalities

$$
\left\{\begin{array}{l}
g_{1}(x, y) \geq 0 \\
g_{2}(x, y) \leq 0 \\
0 \leq z \leq f(x, y)
\end{array}\right.
$$

Eliminating $z$, we get

$$
D=\left\{\begin{array}{ll} 
& g_{1}(x, y) \geq 0 \\
(x, y): & g_{2}(x, y) \leq 0 \\
0 \leq f(x, y)
\end{array}\right\}
$$

3. We compute the double integral using the formula (1) with $f_{2}=f(x, y)$ and $f_{1}=0$.

We write the answer, not forgetting the dimension.