The distance between the foci of the ellipse is $F_{1} F_{2}=2 c$, the major axis is $2 a$; the chord $M_{1} M_{2}(=2 x)$ is parallel to the major axis. Let's examine and plot the volume of the solid formed by the rotation of the trapezoid $F_{1} F_{2} M_{1} M_{2}$ around the major axis, as $x$ varies from 0 to $a$!
The volume of the solid generated by the rotation of a trapezoid is twice the volume of the solid generated by the rotation of $O C M F_{2}$. If the projection of $M$ on the major axis $A A^{\prime}$ is $H$, the solid of revolution of $O C M F_{2}$ can be composed of the solid of revolution of the rectangle $O C M H$ and the solid of revolution of the triangle $M H F_{2}$.

The volume of the cylinder is $v_{1}=\pi \cdot \overline{M H^{2}} \cdot \overline{O H}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_752c6635ee6b0cb329a2g-1.jpg?height=277&width=379&top_left_y=344&top_left_x=862)

Here, $O H = C M = x (0 \leq x \leq a)$ and $\overline{M H}$ is the ordinate of point $M$, which can be expressed from the equation of the ellipse:

$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1
$$

$$
\overline{M H^{2}}=y^{2}=\frac{b^{2}}{a^{2}}\left(a^{2}-x^{2}\right)=\frac{a^{2}-c^{2}}{a^{2}}\left(a^{2}-x^{2}\right)
$$

The volume of the cone is: $v_{2}=\frac{1}{3} \pi \overline{M H^{2}} \cdot H F$

$$
H F_{2}=O F_{2}-O H=c-x \geq 0, \quad v_{2}=\frac{1}{3} \pi \frac{a^{2}-c^{2}}{a^{2}}\left(a^{2}-x^{2}\right)(c-x)
$$

The volume of the solid generated by the rotation of the trapezoid is

$$
\begin{aligned}
& v=2\left(v_{1}+v_{2}\right)=2 \pi \overline{M H^{2}}\left(O H+\frac{H F_{2}}{3}\right)=2 \pi \frac{a^{2}-c^{2}}{a^{2}}\left(a^{2}-x^{2}\right)\left(x+\frac{c-x}{3}\right)= \\
& =2 \pi \frac{a^{2}-c^{2}}{3 a^{2}}\left(a^{2}-x^{2}\right)(2 x+c)
\end{aligned}
$$

For $\quad x=0, \quad v_{0}=\frac{2 \pi}{3 a^{2}}\left(a^{2}-c^{2}\right) a^{2} c=\frac{2 \pi c}{3}\left(a^{2}-c^{2}\right)$.

If $\quad x=a, \quad v=0$. Now, the derivative of $v$ is

$$
v^{\prime}=\frac{2 \pi}{3 a^{2}}\left(a^{2}-c^{2}\right)\left[-2 x(2 x+c)+2\left(a^{2}-x^{2}\right)\right]=-\frac{4 \pi}{3 a^{2}}\left(a^{2}-c^{2}\right)\left(3 x^{2}+c x-a^{2}\right)
$$

$v^{\prime}=0$, if $3 x^{2}+c x-a^{2}=0$. This equation has real roots of opposite signs; only the positive one is valid, i.e.,

$$
x_{1}=\frac{-c+\sqrt{c^{2}+12 a^{2}}}{6}
$$

At this point, the function $3 x^{2}+c x-a^{2}$ transitions from negative to positive values, so $v^{\prime}$ transitions from positive to negative, and thus $v$ has a maximum at this point. At $x=a$, $v^{\prime}=0$. Therefore, one of the roots of the equation lies between 0 and $a$. The other root is negative.