Example 3. Find the mass of the plate $D$ with surface density $\mu=x / y^{5}$, bounded by the curves

$$
\frac{x^{2}}{16}+y^{2}=1, \quad \frac{x^{2}}{16}+y^{2}=3, \quad y=\frac{x}{4}, \quad x=0 \quad\left(y \geq \frac{x}{4}, x \geq 0\right)
$$
Solution.

1. The mass of the plate $D$ with surface density $\mu=x / y^{5}$ is determined by the formula

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

2. We calculate the obtained double integral:

a) define the region $D$ by inequalities in Cartesian coordinates

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

Since the region $D$ is bounded by ellipses and lines passing through the origin, it is easier to solve the problem in 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 sought mass is determined by the formula

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

To find the region $D^{\prime}$, we replace $x$ with $4 \varrho \cos \varphi$ and $y$ with $\varrho \sin \varphi$ in the inequalities defining the region $D$:

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

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

$$
D^{\prime}=\left\{\begin{array}{lc} 
& (\varrho, \varphi): \begin{array}{c}
1 \leq \varrho \leq \sqrt{3} \\
\frac{\pi}{4} \leq \varphi \leq \frac{\pi}{2}
\end{array}
\end{array}\right\}
$$

b) we transition from the double integral to the repeated integral:

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

c) integrating sequentially, we get

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

Answer. $m=4$ units of mass.

Conditions of the Problem. Find the mass of the plate $D$ with surface density $\mu$, where $D$ is bounded by the given lines.

1. $\mu=2 x+y^{2}, \quad x=4, y=0, y=\sqrt{x}$.
2. $\mu=x^{2}+y, \quad x=1, y=0, y=2 \sqrt{x}$.
3. $\mu=x^{2}+2 y, \quad x=0, y=4, y=x^{2}(x \geq 0)$.
4. $\mu=x+y^{2}, \quad x=0, y=1, y=x^{2} / 4(x \geq 0)$.
5. $\quad \mu=\frac{x-y}{x^{2}+y^{2}}, \quad x=0, \quad y=0, \quad x^{2}+y^{2}=4, \quad x^{2}+y^{2}=9$

$$
(x \geq 0, y \leq 0)
$$

6. $\quad \mu=\frac{2 y-x}{x^{2}+y^{2}}$

$x=0, \quad y=0, \quad x^{2}+y^{2}=3, \quad x^{2}+y^{2}=5$ $(x \leq 0, y \geq 0)$.
7. $\quad \mu=\frac{y-x}{x^{2}+y^{2}}, \quad x=0, \quad y=0, \quad x^{2}+y^{2}=4, \quad x^{2}+y^{2}=16$

$$
(x \leq 0, \quad y \geq 0)
$$

8. $\quad \mu(x, y)=y, \quad y=0, \quad y=x \sqrt{3}, \quad x^{2}+\frac{y^{2}}{4}=1, \quad x^{2}+\frac{y^{2}}{4}=9$

$$
(y \geq 0, \quad y \leq x \sqrt{3})
$$

9. $\quad \mu(x, y)=\frac{y}{x^{2}}, \quad y=0, \quad y=x, \quad \frac{x^{2}}{4}+y^{2}=1, \quad \frac{x^{2}}{4}+y^{2}=4$ $(y \geq 0, \quad y \leq x)$.
10. $\mu(x, y)=\frac{x}{y^{2}}, \quad x=0, \quad y=x, \quad \frac{x^{2}}{9}+\frac{y^{2}}{4}=1, \quad \frac{x^{2}}{9}+\frac{y^{2}}{4}=25$

$(x \geq 0, y \geq x)$.

Answers. 1. $m=448 / 15$. 2. $m=11 / 7$. 3. $m=448 / 15$. 4. $m=11 / 7$. 5. $m=2$. 6. $m=3(\sqrt{5}-\sqrt{3})$. 7. $m=4$. 8. $m=52 / 3$. 9. $m=(\sqrt{2}-1) / 2$. 10. $m=18(\sqrt{2}-1)$.

## 12.9. Triple Integral

in Cartesian coordinates

Problem Statement. Calculate the triple integral

$$
\iiint_{\Omega} f(x, y, z) d x d y d z
$$

where the region $\Omega$ is bounded by certain surfaces.

Plan of Solution.

1. Define the region $\Omega$ by a system of inequalities, for example,

$$
\left\{\begin{aligned}
a & \leq x \leq b \\
y_{1}(x) & \leq y \leq y_{2}(x) \\
z_{1}(x, y) & \leq z \leq z_{2}(x, y)
\end{aligned}\right.
$$

2. Transition from the triple integral to the repeated integral:

$$
\iiint_{\Omega} f(x, y, z) d x d y d z=\int_{a}^{b} d x \int_{y_{1}(x)}^{y_{2}(x)} d y \int_{z_{1}(x, y)}^{z_{2}(x, y)} f(x, y, z) d z
$$

3. Using the properties of the definite integral, integrate sequentially first with respect to $z$ (considering $x$ and $y$ as constants), then with respect to $y$ (considering $x$ as a constant), and finally with respect to $x$.

Write the answer.