620. Find the volume of a triaxial ellipsoid $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1$.
Solution. A plane section of the ellipsoid, parallel to the $x O z$ plane and at a distance $y=h$ from it, as shown in Fig. 98, represents an ellipse

$$
\frac{x^{2}}{a^{2}}+\frac{z^{2}}{c^{2}}=1-\frac{h^{2}}{b^{2}}, \quad \frac{x^{2}}{a_{1}^{2}}+\frac{z^{2}}{c_{1}^{2}}=1
$$

with semi-axes

$$
a_{1}=\frac{a}{b} \sqrt{b^{2}-l^{2}}, \quad c_{1}=\frac{c}{b} \sqrt{b^{2}-l^{2}}
$$

The area of this section, as the area of an ellipse, is found by

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-196.jpg?height=51&width=853&top_left_y=1662&top_left_x=90)

$$
S(h)=\pi a_{1} c_{1}=\frac{\pi a c}{z^{2}}\left(b^{2}-l^{2}\right) .
$$

Substituting into formula (*), we obtain the volume of the entire ellipsoid

$$
V=\frac{2 \pi a c}{b^{2}} \int_{0}^{b}\left(b^{2}-h^{2}\right) d h=\left.\frac{2 \pi a c}{b^{2}}\left(b^{2} h-\frac{h^{3}}{3}\right)\right|_{0} ^{b}=\frac{4}{3} \pi a b c
$$

For $a=b=c$, the formula for the volume of the ellipsoid transforms into the formula for the volume of a sphere $V=\frac{4}{3} \pi a^{3}$.