Example. Calculate the triple integral

$$
\iint_{\Omega} \int^{2} \operatorname{sh}(x y) d x d y d z
$$

where $\Omega$ is bounded by the planes

$$
x=2, \quad y=\frac{x}{2}, \quad y=0, \quad z=0, \quad z=1
$$
Solution.

1. Let's define the region $\Omega$ by inequalities. Clearly, $0 \leq z \leq 1$. For $y$, the possible inequalities are $0 \leq y \leq x / 2$ or $x / 2 \leq y \leq 0$. If $0 \leq y \leq x / 2$, then $x \geq 0$ and for $x$ we have $0 \leq x \leq 2$. If $x / 2 \leq y \leq 0$, then $x \leq 0$ and the region does not approach the plane $x=2$. Therefore, we must accept that $0 \leq y \leq x / 2$ and define $\Omega$ by the system of inequalities

$$
\left\{\begin{array}{c}
0 \leq x \leq 2 \\
0 \leq y \leq x / 2 \\
0 \leq z \leq 1
\end{array}\right.
$$

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

$$
\iiint_{\Omega} x^{2} \operatorname{sh}(x y) d x d y d z=\int_{0}^{2} d x \int_{0}^{x / 2} d y \int_{0}^{1} x^{2} \operatorname{sh}(x y) d z
$$

3. Using the properties of the definite integral, we 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 $x$:

$$
\begin{aligned}
& \iiint_{\Omega} x^{2} \operatorname{sh}(x y) d x d y d z=\int_{0}^{2} d x \int_{0}^{x / 2} d y \int_{0}^{1} x^{2} \operatorname{sh}(x y) d z= \\
& =\int_{0}^{2} d x \int_{0}^{x / 2} x^{2} \operatorname{sh}(x y) d y \int_{0}^{1} d z=\int_{0}^{2} d x \int_{0}^{x / 2} x^{2} \operatorname{sh}(x y) d y= \\
& \quad=\left.\int_{0}^{2} x^{2} \frac{\operatorname{ch}(x y)}{x}\right|_{0} ^{x / 2} d x=\int_{0}^{2} x\left(\operatorname{ch} \frac{x^{2}}{2}-\operatorname{ch} 0\right) d x=
\end{aligned}
$$

$$
=\left.\operatorname{sh} \frac{x^{2}}{2}\right|_{0} ^{2}-2=\operatorname{sh} 2-2
$$

Answer. $\iiint_{\Omega} x^{2} \operatorname{sh}(x y) d x d y d z=\operatorname{sh} 2-2$.

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

1. $\iiint_{\Omega} 4 y^{2} z e^{x y} d x d y d z, \quad x=0, y=1, y=x, z=0, z=1$.
2. $\iiint_{\Omega} 3 y^{2} z^{2} e^{-x y} d x d y d z, \quad x=0, y=-2, y=4 x, z=0, z=1$.
3. $\iiint_{\Omega} 3 x^{2} z^{2} e^{x y / 2} d x d y d z, \quad x=0, y=2, y=2 x, z=0, z=-1$.
4. $\iiint_{\Omega} 2 y^{2} z \cos (\pi x y) d x d y d z, x=0, y=1, y=2 x, z=0, z=\pi^{2}$.
5. $\iiint_{\Omega} x^{2} z \sin (\pi x y) d x d y d z, x=1, y=2 x, y=0, z=0, z=4 \pi$.
6. $\iiint_{\Omega} y^{2} z \operatorname{ch}(x y) d x d y d z, \quad x=0, y=-1, y=x, z=0, z=2$.
7. $\iiint_{\Omega} y z^{3} \cos (x y) d x d y d z, x=1, y=\pi, z=2, x=0, y=0, z=0$.
8. $\iint_{\Omega} \int^{2} z \operatorname{sh}(3 x y) d x d y d z, \quad x=1, y=2 x, y=0, z=0, z=1$.
9. $\iiint_{\Omega} y^{2} z \operatorname{ch}(2 x y) d x d y d z, \quad x=0, y=1, y=x, z=0, z=4$.
10. $\iiint_{\Omega} 3 x z^{2} \sin (x y) d x d y d z, x=2, y=\pi, z=1, x=0, y=0, z=0$.

Answers. 1. $I=e-2.2 . I=2 e^{-1} .3 . I=4 e-8.4 . I=4.5 . I=16 \pi^{2}$. 6. $I=2 \operatorname{ch} 1-2$. $7 . I=8.8 . I=(\operatorname{sh} 6-6) / 72.9 . I=\operatorname{ch} 2-1 . \quad 10 . I=2$.

### 12.10. Triple Integral

## in Cylindrical Coordinates

Problem Statement. Compute the triple integral

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

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

$$
z=g_{1}\left(x^{2}+y^{2}\right), \quad z=g_{2}\left(x^{2}+y^{2}\right)
$$

Plan of Solution.

1. Since $\Omega$ is a solid 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 given by the formula

$$
\iint_{\Omega} \int f(x, y, z) d x d y d z=\iiint_{\Omega^{\prime}} f(\varrho \cos \varphi, \varrho \sin \varphi, z) \varrho d \varrho d \varphi d z
$$

2. Define the region $\Omega^{\prime}$ by 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 $g_{1}\left(\varrho^{2}\right) \leq z \leq g_{2}\left(\varrho^{2}\right)$ or $g_{2}\left(\varrho^{2}\right) \leq z \leq g_{1}\left(\varrho^{2}\right)$.

To choose the correct inequalities, we solve the equation $g_{1}\left(\varrho^{2}\right)=g_{2}\left(\varrho^{2}\right)$ with respect to $\varrho$. If it has two solutions $\varrho_{1}$ and $\varrho_{2}\left(0 \leq \varrho_{1}<\varrho_{2}\right)$, we investigate which of the functions $g_{1}\left(\varrho^{2}\right)$ or $g_{2}\left(\varrho^{2}\right)$ is greater than the other on the interval $\varrho_{1} \leq \varrho \leq \varrho_{2}$. Suppose for definiteness that $g_{1}\left(\varrho^{2}\right) \leq g_{2}\left(\varrho^{2}\right)$ for $\varrho_{1} \leq \varrho \leq \varrho_{2}$. Then the region $\Omega^{\prime}$ is defined by the system of inequalities

$$
\Omega^{\prime}=\left\{\begin{aligned}
\varrho_{1} & \leq \varrho \leq \varrho_{2} \\
(\varrho, \varphi, z): g_{1}\left(\varrho^{2}\right) & \leq z \leq g_{2}\left(\varrho^{2}\right) \\
0 & \leq \varphi \leq 2 \pi
\end{aligned}\right\}
$$

If the equation $g_{1}\left(\varrho^{2}\right)=g_{2}\left(\varrho^{2}\right)$ has a unique positive solution $\varrho=\varrho_{2}$, then the inequalities for $\varrho$ are $0 \leq \varrho \leq \varrho_{2}$.

3. Transition from the triple integral to the iterated integral:

$$
\begin{aligned}
& \iiint_{\Omega} f(x, y, z) d x d y d z=\iiint_{\Omega^{\prime}} f(\varrho \cos \varphi, \varrho \sin \varphi, z) \varrho d \varrho d \varphi d z= \\
&=\int_{0}^{2 \pi} d \varphi \int_{\rho_{1}}^{\varrho_{2}} \varrho d \varrho \int_{g_{1}(\varrho)}^{g_{2}(\varrho)} f(\varrho \cos \varphi, \varrho \sin \varphi, z) d z
\end{aligned}
$$

and integrate sequentially, using the properties of the definite integral.

Write the answer.