Example 3. Find the mass of the body $\Omega$ with density $\mu=20 z$, bounded by the surfaces

$$
z=\sqrt{1-x^{2}-y^{2}}, \quad z=\sqrt{\frac{x^{2}+y^{2}}{4}}
$$
Solution.

1. The mass of the body $\Omega$ with density $\mu=20 z$ is determined by the formula

$$
m=\iiint_{\Omega} 20 z d x d y d z
$$

Since $\Omega$ is the 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 mass is determined by the formula

$$
m=\iiint_{\Omega^{\prime}} 20 \varrho^{3} \sin \theta \cos \theta d \varphi d \theta d \varrho
$$

Substituting $x$ 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

$$
\varrho=1, \quad \operatorname{tg} \theta=2
$$

The region $\Omega^{\prime}$ is bounded by these surfaces.

2. We define the region $\Omega^{\prime}$ by the system of inequalities

$$
\Omega^{\prime}=\left\{\begin{array}{ll}
0 \leq \varrho \leq 1 \\
(\varrho, \theta, \varphi): & 0 \leq \theta \leq \operatorname{arctg} 2 \\
& 0 \leq \varphi \leq 2 \pi
\end{array}\right\}
$$

3. We compute $m$ by reducing the triple integral to a repeated integral:

$$
m=20 \int_{0}^{2 \pi} d \varphi \int_{0}^{\operatorname{arctg} 2} \sin \theta \cos \theta d \theta \int_{0}^{1} \varrho^{3} d \varrho=\left.\left.40 \pi \frac{\sin ^{2} \theta}{2}\right|_{0} ^{\operatorname{arctg} 2} \frac{\varrho^{4}}{4}\right|_{0} ^{1}=4 \pi
$$

Here we used the formula

$$
\left.\sin ^{2} \theta\right|_{\theta=\operatorname{arctg} 2}=\left.\frac{\operatorname{tg}^{2} \theta}{1+\operatorname{tg}^{2} \theta}\right|_{\theta=\operatorname{arctg} 2}=\frac{4}{5}
$$

Answer. $m=4 \pi$ units of mass.

Conditions of the problems. Find the mass of the body with density $\mu$, bounded by the given surfaces.

1. $\mu=2 y^{2} e^{x y}, \quad x=0, \quad y=1, \quad y=x, \quad z=0, \quad z=1$.
2. $\mu=y^{2} e^{-x y}, \quad x=0, \quad y=-2, \quad y=4 x, \quad z=0, \quad z=1$.
3. $\mu=y^{2} e^{x y / 2}, \quad x=0, \quad y=2, \quad y=2 x, \quad z=0, \quad z=-1$.
4. $\mu=\frac{x^{2}}{x^{2}+y^{2}}, \quad x^{2}+y^{2}-4 y=0, \quad z=4-x^{2}, \quad z=0$.
5. $\mu=\frac{y^{2}}{x^{2}+y^{2}}, \quad x^{2}+y^{2}-4 x=0, \quad z=10-y^{2}, \quad z=0$.
6. $\mu=\frac{x^{2}}{x^{2}+y^{2}}, \quad z=\sqrt{36-x^{2}-y^{2}}, \quad z=\frac{x^{2}+y^{2}}{9}$.
7. $\mu=\frac{y^{2}}{x^{2}+y^{2}}, \quad z=\sqrt{1-x^{2}-y^{2}}, \quad z=\frac{2\left(x^{2}+y^{2}\right)}{3}$.
8. $\mu=32 z, \quad z=\sqrt{1-x^{2}-y^{2}}, \quad z=\sqrt{x^{2}+y^{2}}$.
9. $\mu=5 z$,

$z=\sqrt{16-x^{2}-y^{2}}, \quad z=\sqrt{\frac{x^{2}+y^{2}}{9}}$.
10. $\mu=10 z$

$z=\sqrt{4-x^{2}-y^{2}}, \quad z=\frac{\sqrt{x^{2}+y^{2}}}{2}$.

Answers. 1. $m=e-2$. 2. $m=2 e^{-1}$. 3. $m=4 e-8$. 4. $m=5 \pi / 2$. 5. $m=17 \pi / 2$. 6. $m=171 \pi / 4$. 7. $m=19 \pi / 96$. 8. $m=4 \pi$. 9. $m=288 \pi$. 10. $m=32 \pi$.

## Chapter 13

## SURFACE INTEGRALS

In the topic of SURFACE INTEGRALS, you will learn about the concept of an integral over a surface of a function of three variables and how to reduce it to a double (and then to a repeated) integral by projecting the given surface onto one of the coordinate planes. Additionally, you will learn how to compute integrals over parts of cylindrical and spherical surfaces.

Using the package SOLVER.CA, you can find the coordinates of the unit normal vector to the surface (compute partial derivatives and the length of the vector), compute the resulting repeated integrals, perform all numerical calculations, and verify the correctness of your results.

## 13.1. Surface Integral of the First Kind

PROBLEM STATEMENT. Compute the surface integral

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

where $\Sigma$ is a part of the surface described by the equation $F(x, y, z)=0$ and some inequalities.

PLAN OF SOLUTION. The surface integral is reduced to a double integral by projecting $\Sigma$ onto the coordinate plane $X O Y$ using the formula

$$
\iint_{\Sigma} f(x, y, z) d \sigma=\iint_{D} f(x, y, z(x, y)) \frac{d x d y}{|\cos \gamma|}
$$

where $D$ is the projection of $\Sigma$ onto the plane $X O Y, \gamma$ is the angle between the normal to the surface $\Sigma$ and the $O Z$ axis; $z(x, y)$ is determined from the surface equation $F(x, y, z)=0$.

NOTE. If the equation $F(x, y, z)=0$ does not uniquely define the function $z=z(x, y)$, then project $\Sigma$ onto another coordinate plane or use curvilinear coordinates (you can also divide the surface into parts and use the additivity of the integral).

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

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

2. The projection $D$ of the surface $\Sigma$ onto the plane $X O Y$ is found by eliminating $z$ from the conditions defining $\Sigma$.
3. We find $z=z(x, y)$ by solving the equation $F(x, y, z)=0$.
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.

Write the answer.