Example. Compute the surface integral

$$
\iint_{\Sigma}\left(x^{2}+y^{2}\right) d \sigma
$$

where $\Sigma$ is the part of the surface $x^{2}+y^{2}=1$, cut off by the planes $z=0, \quad z=2$.
SOLUTION.

1. We 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

$$
\Sigma=\left\{\begin{array}{ll}
\varrho=1 \\
(\varrho, \varphi, z): & 0 \leq \varphi \leq 2 \pi \\
& 0 \leq z \leq 2
\end{array}\right\}
$$

2. Since $d \sigma=1 \cdot d \varphi d z$ and $x^{2}+y^{2}=1$, we have

$$
\iint_{\Sigma}\left(x^{2}+y^{2}\right) d \sigma=\int_{0}^{2 \pi} d \varphi \int_{0}^{2} d z
$$

3. We compute the double integral:

$$
\iint_{\Sigma}\left(x^{2}+y^{2}\right) d \sigma=\int_{0}^{2 \pi} d \varphi \int_{0}^{2} d z=4 \pi
$$

Answer. $\iint_{\Sigma}\left(x^{2}+y^{2}\right) d \sigma=4 \pi$.

Conditions of the PROBLEM. Compute the surface integrals.

1. $\iint_{\Sigma} x d \sigma$
$\Sigma=\left\{(x, y, z): \begin{array}{l}x^{2}+y^{2}=1, \\ z=0, z=2\end{array}\right\}$.
2. $\iint_{\Sigma}(x+y) d \sigma$,
$\Sigma=\left\{(x, y, z): \begin{array}{l}x^{2}+y^{2}=4 \\ z=0, z=4\end{array}\right\}$.
3. $\iint_{\Sigma}(x+y+z) d \sigma$,
$\Sigma=\left\{(x, y, z): \begin{array}{l}x^{2}+y^{2}=1, \\ z=0, z=1\end{array}\right\}$.
4. $\iint_{\Sigma}\left(y^{2}-1\right) d \sigma$,
$\Sigma=\left\{(x, y, z): \begin{array}{l}x^{2}+y^{2}=2, \\ z=0, z=1\end{array}\right\}$.
5. $\iint_{\Sigma}\left(2-x^{2}-y^{2}\right) d \sigma$,
$\Sigma=\left\{(x, y, z): \begin{array}{l}x^{2}+y^{2}=1, \\ z=0, z=2\end{array}\right\}$.
6. $\iint_{\Sigma}\left(x^{2}+z\right) d \sigma$,
$\Sigma=\left\{(x, y, z): \begin{array}{c}x^{2}+y^{2}=1, \\ z=0, z=3\end{array}\right\}$.
7. $\iint_{\Sigma} \sqrt{x^{2}+y^{2}} d \sigma$,
$\Sigma=\left\{(x, y, z): \begin{array}{l}x^{2}+y^{2}=4, \\ z=0, z=1\end{array}\right\}$.
8. $\iint_{\Sigma}\left(x y+z^{2}\right) d \sigma$
$\Sigma=\left\{(x, y, z): \begin{array}{c}x^{2}+y^{2}=1 \\ z=0, z=1\end{array}\right\}$.
9. $\iint_{\Sigma} \sqrt[3]{9-x^{2}-y^{2}} d \sigma, \quad \Sigma=\{(x, y, z):$
$\left.\begin{array}{l}x^{2}+y^{2}=1 \\ z=0, z=1\end{array}\right\}$
10. $\iint_{\Sigma} \sqrt{z+x^{2}+y^{2}} d \sigma, \quad \Sigma=\left\{(x, y, z): \begin{array}{l}x^{2}+y^{2}=2, \\ z=0, z=2\end{array}\right\}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_5878ce417e2ca061000bg-340.jpg?height=39&width=1057&top_left_y=592&top_left_x=120)
8. $2 \pi / 3$. $\quad 9.4 \pi$. $\quad 10.32 \pi(\sqrt{2}-1) / 3$.

## 13.3. Integral

## over a spherical surface

PROBLEM STATEMENT. Compute the surface integral

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

where $\Sigma-$ is the upper hemisphere

$$
x^{2}+y^{2}+z^{2}=r^{2}, \quad z \geq 0
$$

## PLAN OF SOLUTION.

1. We introduce curvilinear coordinates on the given surface (sphere)

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

In these coordinates, the surface is defined by the conditions

$$
\Sigma=\left\{\begin{array}{cc} 
& \varrho=r \\
(\varrho, \theta, \varphi): & 0 \leq \theta \leq \pi / 2 \\
& 0 \leq \varphi \leq 2 \pi
\end{array}\right\}
$$

2. Since $d \sigma=r^{2} \sin \theta d \theta d \varphi$, we have $\iint_{\Sigma} f(x, y, z) d \sigma=r^{2} \int_{0}^{2 \pi} d \varphi \int_{0}^{\pi / 2} f(r \cos \varphi \sin \theta, r \sin \varphi \sin \theta, r \cos \theta) \sin \theta d \theta$.
3. We compute the double integral and write down the answer.