Let's consider the curves

$$
\begin{aligned}
& y=-x^{2}+4 \ldots \\
& \text { and } 16 x^{2}+25 y^{2}=400 \ldots
\end{aligned}
$$

when rotated about the $Y$-axis. What volume is cut out from the solid of revolution (1) by the solid of revolution (2)?
1) The parabola, whose main axis is the $Y$-axis; by rotating the parabola around this, we obtain a rotational paraboloid.

2) The ellipse, one of whose main axes is the $Y$-axis; by rotating it around this, we obtain a rotational ellipsoid.

The parabola touches the ellipse at the point $x=0, y=+4$; in addition, it intersects at two points that are symmetric with respect to the $Y$-axis. For these points,

$$
16 x^{2}+25\left(-x^{2}+4\right)^{2}=400, \quad \text { or } \quad x^{2}\left(25 x^{2}-184\right)=0
$$

Here, $x^{2}=0$ indicates that the two curves touch at the point $x=0, y=+4$.

For the intersection points $P_{1}, P_{2}$, we have $25 x^{2}-184=0$,

$$
x^{2}=\frac{184}{25}, \quad x= \pm \frac{2}{25} \sqrt{46}
$$

Thus,

$$
y=-\frac{184}{25}+4=-\frac{84}{25}=-3.36
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_72d87c077498e73b9d2cg-1.jpg?height=385&width=414&top_left_y=797&top_left_x=842)

The volume in question consists of two parts: one part is the slice of the paraboloid cut by a plane perpendicular to the $Y$-axis through points $P_{1}$ and $P_{2}$; the other part is the smaller slice of the ellipsoid cut by the same plane. The volume of the first part, if $x^{2}$ is expressed from the equation of the parabola,

$$
\begin{gathered}
K_{1}=\int_{-3.36}^{+4} x^{2} \pi \cdot d y=\pi \int_{-3.36}^{4}(4-y) d y= \\
=\pi\left[4 y-\frac{y^{2}}{2}\right]_{-3.36}^{4}= \\
=\pi\left(16-8+4 \cdot 3.36+\frac{3.36^{2}}{2}\right)= \\
=\pi(8+13.144+5.6448)=27.0848 \pi
\end{gathered}
$$

The volume of the second part (if $x^{2}$ is expressed from the equation of the ellipse),

$$
\begin{gathered}
K_{2}=\int_{-4}^{-3.36} x^{2} \pi d y=\pi \int_{3.36}^{4}\left(25-\frac{25}{16} y^{2}\right) d y=\pi\left[25 y-\frac{25}{48} y^{3}\right]_{3.36}^{4}= \\
=\pi\left[100-\frac{25}{48} 64-25 \cdot 3.36+\frac{25}{48} 3.36^{3}\right]= \\
=\pi\left(100-\frac{100}{3}-84+\frac{84}{48} \cdot 3.36^{2}\right)=-\pi(-17.3333+3.5 \cdot 5.6448)= \\
=2.4235 \pi \\
K=K_{1}+K_{2}=27.0848 \pi+2.4235 \pi \sim 29.51 \pi \quad \text { volume units. }
\end{gathered}
$$

B. Major Pál (Ref. rg. VIII. o. Csurgó).