Example. Calculate the surface integral

$$
\iint_{\Sigma}(-x+3 y+4 z) d \sigma
$$

where $\Sigma-$ is the part of the plane

$$
x+2 y+3 z=1
$$

located in the first octant (i.e., $x \geq 0, y \geq 0, z \geq 0$ ).
SOLUTION.

1. The unit normal vectors $\vec{n}_{0}=\{\cos \alpha, \cos \beta, \cos \gamma\}$ to the surface defined by the equation $F(x, y, z)=0$ are given by the formula

$$
\vec{n}_{0}= \pm \frac{\operatorname{grad} F}{|\operatorname{grad} F|}
$$

In this case, $F(x, y, z)=x+2 y+3 z-1$. Therefore,

$$
\vec{n}_{0}= \pm \frac{\{1,2,3\}}{\sqrt{14}}, \quad|\cos \gamma|=\frac{3}{\sqrt{14}}
$$

2. The surface $\Sigma$ is defined by the conditions

$$
\Sigma=\left\{(x, y, z): \begin{array}{l}
x+2 y+3 z=1 \\
x \geq 0, y \geq 0, z \geq 0
\end{array}\right\}
$$

We find the projection $D$ of $\Sigma$ onto the $X O Y$ plane by eliminating $z$ from the conditions defining $\Sigma$:

$$
\begin{gathered}
\Sigma=\left\{(x, y, z): \begin{array}{l}
z=(1-x-2 y) / 3 \\
x \geq 0, y \geq 0, z \geq 0
\end{array}\right\} \Longrightarrow \\
\Longrightarrow D=\left\{(x, y): \begin{array}{l}
x \geq 0, y \geq 0 \\
0 \leq(1-x-2 y) / 3
\end{array}\right\}
\end{gathered}
$$

From this,

$$
D=\left\{(x, y): \begin{array}{l}
0 \leq y \leq 12 \\
0 \leq x \leq 1-2 y
\end{array}\right\}
$$

3. From the equation $x+2 y+3 z-1=0$, we find $z(x, y)=(1-x-2 y) / 3$.
4. We transition from the surface integral to a double integral using formula (1) and compute the double integral by reducing it to a repeated integral:

$$
\begin{aligned}
& \iint_{\Sigma}(-x+4 y+3 z) d \sigma=\left.\iint_{D}(-x+4 y+3 z)\right|_{z=(1-x-2 y) / 3} \frac{\sqrt{14}}{3} d x d y= \\
= & \frac{\sqrt{14}}{3} \iint_{D}(1-2 x+2 y) d x d y=\frac{\sqrt{14}}{3} \int_{0}^{1 / 2} d y \int_{0}^{1-2 y}(1-2 x+2 y) d x=\frac{\sqrt{14}}{18} . \\
& \text { Answer. } \int_{\Sigma} \int_{0}(-x+4 y+3 z) d \sigma=\frac{\sqrt{14}}{18} .
\end{aligned}
$$

Conditions of the Problems. Compute the surface integrals.

1. $\iint_{\Sigma} x y z d \sigma$
$\Sigma=\{(x, y, z):$
$\left.\begin{array}{l}x+y+z=1 \\ x \geq 0, y \geq 0, z \geq 0\end{array}\right\}$
2. $\iint_{\Sigma}(y+z) d \sigma, \quad \Sigma=\left\{(x, y, z): \begin{array}{l}x+y+z=1, \\ x \geq 0, y \geq 0, z \geq 0\end{array}\right\}$.
3. $\iint_{\Sigma}\left(3 x+\frac{2}{3} z\right) d \sigma, \quad \Sigma=\{(x, y, z)$ :
$\left.\begin{array}{l}x+y / 2+z / 3=1 \\ x \geq 0, y \geq 0, z \geq 0\end{array}\right\}$.
4. $\iint_{\Sigma}(x+18 y+24 z) d \sigma, \quad \Sigma=\{(x, y, z)$ : $\left.\begin{array}{l}x+2 y+3 z=1 \\ x \geq 0, y \geq 0, z \geq 0\end{array}\right\}$
5. $\iint_{\Sigma} z d \sigma$

$$
\Sigma=\left\{(x, y, z): \begin{array}{l}
x^{2}+y^{2}+z^{2}=1 \\
z \geq 0
\end{array}\right\}
$$

6. $\iint_{\Sigma}(x+y+z) d \sigma$,
$\Sigma=\{(x, y, z):$
$\left.\begin{array}{l}x+y+z=1 \\ x \geq 0, y \geq 0, z \geq 0\end{array}\right\}$.
7. $\iint_{\Sigma} z d \sigma$

$$
\Sigma=\left\{\begin{array}{ll}
(x, y, z): \begin{array}{l}
x^{2}+y^{2}-2 z=0 \\
0 \leq z \leq 1
\end{array}
\end{array}\right\}
$$

8. $\iint_{\Sigma}\left(x^{2}+y^{2}\right) d \sigma$
$\Sigma=\{(x, y, z):$
$\left.\begin{array}{l}x^{2}+y^{2}-z^{2}=0 \\ 0 \leq z \leq 1\end{array}\right\}$
9. $\iint_{\Sigma}(x y+y z+z x) d \sigma$
$\Sigma=\{(x, y, z):$
$\left.\begin{array}{l}x^{2}+y^{2}-z^{2}=0 \\ 0 \leq z \leq 1\end{array}\right\}$
10. $\iint_{\Sigma} \sqrt{x^{2}+y^{2}} d \sigma$
$\Sigma=\{(x, y, z):$ $\left.\begin{array}{l}x^{2}+y^{2}-4 z^{2}=0 \\ 0 \leq z \leq 1 / 2\end{array}\right\}$

Answers. 1. $\sqrt{3} / 120$. 2. $\sqrt{3} / 3$. 3. $35 / 6$. 4. $\sqrt{14} / 2$. 5. $2 \pi$. 6. $\sqrt{3} / 2$.
7. $12(1+6 \sqrt{3}) \pi / 15$.
8. $\sqrt{2} \pi / 2$.
9. $0$. 10. $\sqrt{5} \pi / 3$.

## 13.2. Integral

## over a cylindrical surface

PROBLEM STATEMENT. Compute the surface integral

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

where $\Sigma$ is the part of the surface $x^{2}+y^{2}=r^{2}$ cut off by the planes $z=0$ and $z=h$.

## PLAN OF SOLUTION.

1. Introduce curvilinear coordinates on the given surface (cylinder)

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

In these coordinates, the surface is defined by the conditions

$$
\Sigma=\left\{\begin{array}{ll}
\varrho=r \\
(\varrho, \varphi, z): & 0 \leq \varphi \leq 2 \pi \\
& 0 \leq z \leq h
\end{array}\right\}
$$

2. Since

$$
d \sigma=r d \varphi d z
$$

then

$$
\iint_{\Sigma} f(x, y, z) d \sigma=r \int_{0}^{2 \pi} d \varphi \int_{0}^{h} f(r \cos \varphi, r \sin \varphi, z) d z
$$

3. Compute the repeated integral and write the answer.