Example. Calculate the triple integral

$$
\iiint_{\Omega} \frac{x^{2}}{x^{2}+y^{2}} d x d y d z
$$

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

$$
z=\frac{9}{2} \sqrt{x^{2}+y^{2}}, \quad z=\frac{11}{2}-x^{2}-y^{2}
$$
Solution.

1. Since $\Omega$ is a body of revolution around the $O Z$ axis, it is convenient to switch to cylindrical coordinates

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

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

$$
\iiint_{\Omega} \frac{x^{2}}{x^{2}+y^{2}} d x d y d z=\iiint_{\Omega^{\prime}} \cos ^{2} \varphi \varrho d \varrho d \varphi d z
$$

2. Let's define the region $\Omega^{\prime}$ using inequalities. For this, we first replace $x$ with $\varrho \cos \varphi$ and $y$ with $\varrho \sin \varphi$ in the equations of the surfaces. Then $\Omega^{\prime}$ is defined by the inequalities $9 \varrho / 2 \leq z \leq 11 / 2-\varrho^{2}$ or $11 / 2-\varrho^{2} \leq z \leq 9 \varrho / 2$. To choose the correct inequalities, we solve the equation

$$
\frac{9}{2} \varrho=\frac{11}{2}-\varrho^{2}
$$

This equation has a unique positive solution $\varrho=1$. Therefore, $0 \leq \varrho \leq 1$. For $0 \leq \varrho \leq 1$

$$
\frac{9}{2} \varrho \leq \frac{11}{2}-\varrho^{2}
$$

Thus, the region $\Omega^{\prime}$ is defined by the system of inequalities:

$$
\left\{\begin{array}{l}
0 \leq \varrho \leq 1 \\
9 / 2 \varrho \leq z \leq 11 / 2-\varrho^{2} \\
0 \leq \varphi \leq 2 \pi
\end{array}\right.
$$

3. We transition from the triple integral to a repeated integral:

$$
\begin{aligned}
& \iiint_{\Omega} \frac{x^{2}}{x^{2}+y^{2}} d x d y d z=\iiint_{\Omega^{\prime}} \frac{\varrho^{2} \cos ^{2} \varphi}{\varrho^{2} \cos ^{2} \varphi+\varrho^{2} \sin ^{2} \varphi} \varrho d \varrho d \varphi d z= \\
&=\int_{0}^{2 \pi} \cos ^{2} \varphi d \varphi \int_{0}^{1} \varrho d \varrho \int_{9 \varrho / 2}^{11 / 2-\varrho^{2}} d z
\end{aligned}
$$

Integrating sequentially, we get

$$
\int_{0}^{2 \pi} \cos ^{2} \varphi d \varphi \int_{0}^{1} \varrho d \varrho \int_{9 \varrho / 2}^{11 / 2-\varrho^{2}} d z=\pi \int_{0}^{1}\left(\frac{11}{2}-\varrho^{2}-\frac{9}{2} \varrho\right) \varrho d \varrho=\pi
$$

Answer. $\iiint_{\Omega} \frac{x^{2}}{x^{2}+y^{2}} d x d y d z=\pi$.

Conditions of the Problems. Compute the triple integral over the region $\Omega$ bounded by the given surfaces.

1. $\iiint_{\Omega} \frac{x}{x^{2}+y^{2}} d x d y d z, \quad x^{2}+y^{2}+4 x=0, z=8-y^{2}, z=0$.
2. $\iiint_{\Omega} \frac{y}{x^{2}+y^{2}} d x d y d z, \quad x^{2}+y^{2}-4 y=0, z=6-x^{2}, z=0$.
3. $\iiint \frac{y}{\sqrt{x^{2}+y^{2}}} d x d y d z, \quad x^{2}+y^{2}-4 y=0, z=4-x^{2}, z=0$.
4. $\iint_{\Omega}^{\Omega} \int \frac{x}{\sqrt{x^{2}+y^{2}}} d x d y d z, \quad x^{2}+y^{2}-4 x=0, z=10-y^{2}, z=0$.
5. $\iiint_{\Omega} \frac{x+y}{x^{2}+y^{2}} d x d y d z, \quad x^{2}+y^{2}-4 x=0, z=12-y^{2}, z=0$.
6. $\iiint_{\Omega} \frac{x^{2}}{x^{2}+y^{2}} d x d y d z, \quad z=\sqrt{36-x^{2}-y^{2}}, \quad z=\frac{x^{2}+y^{2}}{9}$.
7. $\iiint_{\Omega} \frac{y^{2}}{x^{2}+y^{2}} d x d y d z, \quad z=\sqrt{1-x^{2}-y^{2}}, \quad z=\frac{2\left(x^{2}+y^{2}\right)}{3}$.
8. $\iiint_{\Omega} \frac{y}{\sqrt{x^{2}+y^{2}}} d x d y d z, \quad z=\sqrt{64-x^{2}-y^{2}}, \quad z=\frac{x^{2}+y^{2}}{12}$.
9. $\iint_{\Omega} \frac{x}{\sqrt{x^{2}+y^{2}}} d x d y d z, \quad z=\sqrt{144-x^{2}-y^{2}}, \quad z=\frac{x^{2}+y^{2}}{18}$.
10. $\iint_{\Omega} \frac{2 x+3 y}{\sqrt{x^{2}+y^{2}}} d x d y d z, \quad z=\sqrt{\frac{4}{9}-x^{2}-y^{2}}, \quad z=x^{2}+y^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_5878ce417e2ca061000bg-322.jpg?height=92&width=1061&top_left_y=1147&top_left_x=118)
$\begin{array}{lllll}\text { 6. } 42 \frac{3}{4} \pi . & \text { 7. } \frac{19}{96} \pi . & \text { 8. } 0 . & \text { 9. } 0 . & 10.0 .\end{array}$

### 12.11. Triple Integral

in spherical coordinates

PROBLEM STATEMENT. Compute the triple integral

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

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

$$
x^{2}+y^{2}+z^{2}=R^{2}, \quad z= \pm \sqrt{\frac{x^{2}+y^{2}}{a^{2}}}
$$

PLAN OF SOLUTION.

1. Since $\Omega$ is bounded by a sphere and a circular cone, it is convenient to switch to spherical coordinates

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

The possible ranges for the spherical coordinates are

$$
0 \leq \varrho<\infty, \quad 0 \leq \theta \leq \pi, \quad 0 \leq \varphi \leq 2 \pi
$$

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

$$
\begin{aligned}
& \iiint_{\Omega} f(x, y, z) d x d y d z= \\
& \quad=\iiint_{\Omega^{\prime}} f(\varrho \cos \varphi \sin \theta, \varrho \sin \varphi \sin \theta, \varrho \cos \theta) \varrho^{2} \sin \theta d \varrho d \theta d \varphi
\end{aligned}
$$

2. Replace $x$ with $\varrho \cos \varphi \sin \theta$, $y$ with $\varrho \sin \varphi \sin \theta$, and $z$ with $\varrho \cos \theta$ in the equations of the surfaces. We get

$$
\varrho=R, \quad \operatorname{tg} \theta= \pm|a|
$$

3. Define the region $\Omega^{\prime}$ using a system of inequalities:

$$
\left\{\begin{array}{l}
0 \leq \varrho \leq R \\
\theta_{1} \leq \theta \leq \theta_{2} \\
0 \leq \varphi \leq 2 \pi
\end{array}\right.
$$

where the boundaries of $\theta$ are found by solving the equation $\operatorname{