Example. Compute the triple integral

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

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

$$
z=\sqrt{36-x^{2}-y^{2}}, \quad z=\sqrt{\frac{x^{2}+y^{2}}{3}}
$$
Solution.

1. Since $\Omega$ is a region bounded by the upper hemisphere and the upper half-cone, it is convenient to switch to spherical coordinates

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

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

$$
\iint_{\Omega} \int \frac{x^{2}}{x^{2}+y^{2}} d x d y d z=\iint_{\Omega^{\prime}} \cos ^{2} \varphi \varrho^{2} \sin \theta d \varrho d \theta d \varphi
$$

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

$$
\varrho=6, \quad \operatorname{tg} \theta=\sqrt{3}
$$

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

$$
\Omega^{\prime}=\left\{\begin{array}{cc} 
& 0 \leq \varrho \leq 6 \\
(\varrho, \theta, \varphi): & 0 \leq \theta \leq \pi / 3 \\
& 0 \leq \varphi \leq 2 \pi
\end{array}\right\}
$$

4. Transition from the triple integral to the repeated integral and integrate sequentially, we get

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

$$
=\int_{0}^{2 \pi} \cos ^{2} \varphi d \varphi \int_{0}^{\pi / 3} \sin \theta d \theta \int_{0}^{6} \varrho^{2} d \varrho=36 \pi
$$

Answer. $\iiint_{\Omega} \frac{x^{2}}{x^{2}+y^{2}} d x d y d z=36 \pi$.

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

1. $\iiint_{\Omega} \frac{x^{2}}{x^{2}+y^{2}} d x d y d z$
$z=\sqrt{25-x^{2}-y^{2}}, \quad z=\sqrt{\frac{x^{2}+y^{2}}{99}}$.
2. $\iiint_{\Omega} \frac{y^{2}}{x^{2}+y^{2}} d x d y d z$

$z=\sqrt{36-x^{2}-y^{2}}, \quad z=\sqrt{\frac{x^{2}+y^{2}}{63}}$.
3. $\iiint_{\Omega} \frac{x}{x^{2}+y^{2}} d x d y d z, \quad z=\sqrt{16-x^{2}-y^{2}}, \quad z=\sqrt{\frac{x^{2}+y^{2}}{15}}$.
4. $\iiint_{\Omega} \frac{y}{x^{2}+y^{2}} d x d y d z, \quad z=\sqrt{9-x^{2}-y^{2}}, \quad z=\sqrt{\frac{x^{2}+y^{2}}{80}}$.
5. $\iiint_{\Omega} \frac{y}{\sqrt{x^{2}+y^{2}}} d x d y d z, \quad z=\sqrt{9-x^{2}-y^{2}}, \quad z=\sqrt{\frac{x^{2}+y^{2}}{35}}$.
6. $\iiint_{\Omega} \frac{x^{2}}{x^{2}+y^{2}} d x d y d z, \quad z=\sqrt{4-x^{2}-y^{2}}, \quad z=\sqrt{\frac{x^{2}+y^{2}}{15}}$.
7. $\iiint_{\Omega} \frac{y^{2}}{x^{2}+y^{2}} d x d y d z, \quad z=\sqrt{9-x^{2}-y^{2}}, \quad z=\sqrt{\frac{x^{2}+y^{2}}{8}}$.
8. $\iint_{\Omega} \frac{x}{\sqrt{x^{2}+y^{2}}} d x d y d z, \quad z=\sqrt{36-x^{2}-y^{2}}, \quad z=\sqrt{\frac{x^{2}+y^{2}}{3}}$.
9. $\iiint_{\Omega} \frac{2 x+3 y}{\sqrt{x^{2}+y^{2}}} d x d y d z, \quad z=\sqrt{25-x^{2}-y^{2}}, \quad z=\sqrt{\frac{x^{2}+y^{2}}{99}}$.
10. $\iiint_{\Omega} \frac{y}{\sqrt{x^{2}+y^{2}}} d x d y d z, \quad z=\sqrt{16-x^{2}-y^{2}}, \quad z=\sqrt{\frac{x^{2}+y^{2}}{15}}$.

Answers. 1. $75 \pi / 2.2 .63 \pi .3 .0 .4 .0 .6 .0 .6 .2 \pi .7 .6 \pi$. 8. 0. 9. 0. 10.0 .

### 12.12. Volume Calculation Using a Triple Integral

Problem Statement. Find the volume of the body $\Omega$ bounded by the given surfaces.

Solution Plan. The desired volume is equal to

$$
V=\iiint_{\Omega} 1 \cdot d x d y d z
$$

1. Define the region $\Omega$ using inequalities.
2. Compute the triple integral, reducing it to a repeated integral, and write the answer, not forgetting the units of measurement.