Example. Calculate the volume of the body bounded by the surfaces

$$
\frac{x^{2}}{16}+\frac{y^{2}}{9}+\frac{z^{2}}{196}=1, \quad z=0, \quad z=7
$$
Solution. If $S=S(z)$ is the area of the cross-section of a body by a plane perpendicular to the $O Z$ axis and intersecting it at a point with ordinate $z$, then the volume of the part of the body enclosed between the planes $z=z_{1}$ and $z=z_{2}$ is determined by the formula

$$
V=\int_{z_{1}}^{z_{2}} S(z) d z
$$

1. The cross-section of the given body by the plane $z=$ const is defined by the inequality

$$
\frac{x^{2}}{16}+\frac{y^{2}}{9} \leq 1-\frac{z^{2}}{196}
$$

i.e., for $|z|<14$ it is an ellipse

$$
\frac{196 x^{2}}{16\left(196-z^{2}\right)}+\frac{196 y^{2}}{9\left(196-z^{2}\right)} \leq 1
$$

with semi-axes

$$
a=\frac{4 \sqrt{196-z^{2}}}{14}, \quad b=\frac{3 \sqrt{196-z^{2}}}{14}
$$

The area of this ellipse is

$$
S=\pi a b=\frac{3 \pi}{49}\left(196-z^{2}\right)
$$

Thus, for $0 \leq z \leq 7$

$$
S(z)=\frac{3 \pi}{49}\left(196-z^{2}\right)
$$

2. Substituting $S(z)$ into formula (2) and computing the definite integral:

$$
V=\frac{3 \pi}{49} \int_{0}^{7}\left(196-z^{2}\right) d z=77 \pi(\text { units of length })^{3}
$$

Answer. $77 \pi$ (units of length) ${ }^{3}$.

Conditions of the Problems. Calculate the volumes of bodies bounded by the specified surfaces.

1. $z=4 x^{2}+9 y^{2}, \quad z=6$.
2. $z=9 x^{2}+4 y^{2}, \quad z=6$.
3. $z=2 x^{2}+8 y^{2}, \quad z=4$.
4. $\quad x^{2}+\frac{y^{2}}{9}+\frac{z^{2}}{4}=1, \quad z=0, \quad z=2$.
5. $\quad \frac{x^{2}}{16}+y^{2}+\frac{z^{2}}{9}=1, \quad z=0, \quad z=3$.
6. $\quad \frac{x^{2}}{9}+y^{2}-3 z^{2}=1, \quad z=0, \quad z=1$.
7. $x^{2}+\frac{y^{2}}{9}-2 z^{2}=1, \quad z=0, \quad z=1$.
8. $\frac{x^{2}}{3}+\frac{y^{2}}{27}-z^{2}=1, \quad z=0, \quad z=2$.
9. $\frac{x^{2}}{9}+\frac{y^{2}}{4}-z^{2}=-1, \quad z=2$.
10. $\frac{x^{2}}{4}+\frac{y^{2}}{25}-\frac{z^{2}}{9}=-1, \quad z=6$.

Answers. 1. $3 \pi$. 2. $3 \pi$. 3. $2 \pi$. 4. $4 \pi$. 5. $8 \pi$. 6. $6 \pi$. 7. $5 \pi$. 8. $42 \pi$. $9.8 \pi$. $10.50 \pi$.

### 8.12. Calculation of Volumes of Solids of Revolution

Problem Statement. Calculate the volume of the solid formed by the rotation of the region bounded by the graphs of the functions $y=f_{1}(x)$ and $y=f_{2}(x)$, possibly by the lines $x=a$ and $x=b$, around the $O X$ axis.

Solution Plan. The volume of the solid formed by the rotation of the region bounded by the curves $y=u(x)$ and $y=v(x)$ and the lines $x=a, x=b$, where $u(x) \leq v(x)$, i.e., the region defined by the system of inequalities

$$
\left\{\begin{aligned}
a & \leq x \leq b \\
u(x) & \leq y \leq v(x)
\end{aligned}\right.
$$

is calculated using the formula

$$
V=\pi \int_{a}^{b}\left(v(x)^{2}-u(x)^{2}\right) d x
$$

1. Determine the region $D$. If the inequalities defining the region $D$ are unknown, i.e., if $a$ and $b$ and/or which of the functions $f_{1}(x)$ and $f_{2}(x)$ is greater on the interval $[a, b]$ are unknown, then perform the following operations.

a) find $a$ and $b$ as the abscissas of the points of intersection of the graphs of the functions $f_{1}(x)$ and $f_{2}(x)$, i.e., solve the equation

$$
f_{1}(x)=f_{2}(x)
$$

b) investigate the sign of the difference $f_{1}(x)-f_{2}(x)$ on $[a, b]$. It is sufficient to compute the value of $f_{1}(x)-f_{2}(x)$ at some point in $(a, b)$. If it is positive, then $f_{1}(x) \geq f_{2}(x)$ and, consequently, $u(x)=f_{2}(x)$ and $v(x)=f_{1}(x)$. If it is negative, then $f_{1}(x) \leq f_{2}(x)$ and, consequently, $u(x)=f_{1}(x)$ and $v(x)=f_{2}(x)$.

2. Compute the volume using formula (1).

Write the answer, not forgetting the units of measurement.

Remark. The problem is solved similarly if the solid is formed by the rotation of the region around the $O Y$ axis or the $O Z$ axis.