619. Find the volume of the part of the cylinder cut off by a plane passing through the diameter $2 R$ of its base at an angle $x$ to the base plane.
Solution. Depicting the given body's section, Fig. 97, we notice that any section of it by a plane parallel to the plane $A B C$ represents a right-angled triangle.

Let's find the area of the section at a distance $O P=x$ from point $O$. From the right-angled triangle $A M P$ we have $M P^{2}=R^{2}-(R-x)^{2}$. From the right-angled triangle $P N M$ we have $M N=:=M P \operatorname{tg} \alpha$.

The area of the section $S(x)$, as a right-angled triangle with legs $M P$ and $M N$:

$$
\begin{gathered}
S(x)=\frac{1}{2} M P \cdot M N=\frac{1}{2} M P^{2} \operatorname{tg} \alpha= \\
=\frac{1}{2}\left|R^{2}-(R-x)^{2}\right| \operatorname{tg} \alpha=\frac{1}{2}\left(2 R x-x^{2}\right) \operatorname{tg} \alpha
\end{gathered}
$$

When $x$ changes by an amount $d x$

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-196.jpg?height=337&width=340&top_left_y=376&top_left_x=859)

Fig. 97 1) the volume $v$ will change by an amount $\Delta v$, equivalent to the volume of a right cylinder (prism) with height $d x$ and base area $S(x)$:

$$
\Delta v \approx d v=S(x) d x==\frac{1}{2}\left(2 R x-x^{2}\right) \operatorname{tg} \alpha d x
$$

The entire sought volume corresponds to the change in $x$ from 0 to $2 R$, therefore

$$
V=\frac{\operatorname{tg} \alpha}{2} \int_{0}^{2 R}\left(2 R x-x^{2}\right) d x=\left.\frac{\operatorname{tg} a}{2}\left(R x^{2}-\frac{x^{3}}{3}\right)\right|_{0} ^{2 R}=\frac{2}{3} R^{3} \operatorname{tg} \alpha
$$