### 11.222 The height of the cylinder is equal to the radius of its base and has a length

 a. Through the axis of the cylinder, another cylindrical surface is drawn, dividing the circumference of the base into two arcs, the lengths of which are in the ratio 2:1. This cylindrical surface divides the given cylinder into two parts. Find the lateral surface area and volume of the larger part of the cylinder.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-274.jpg?height=295&width=380&top_left_y=569&top_left_x=103)

Let the base of the given cylinder, a circle with center \(O_{1}\), be divided by another circle passing through \(O_{1}\) into two arcs

\(\cup A C B, \cup A D B\), the lengths of which are in the ratio \(2: 1 \Rightarrow \cup A C B=240^{\circ}, \cup A D B=120^{\circ} \Rightarrow \angle A O_{1} B=120^{\circ}\)

(a central angle subtending an arc of \(120^{\circ}\)).

If now \(O_{2}\) is the midpoint of \(\cup A D B\), then \(\angle A O_{1} O_{2}=60^{\circ}\),

\(O_{1} A=O_{1} O_{2}=a \Rightarrow A O_{2}=a\), i.e., \(O_{2}\) is the center of the second circle \(\Rightarrow\)

\(\Rightarrow \cup A O_{1} B=\cup A O_{2} B \Rightarrow|\cup A C B|+\left|\cup A O_{1} B\right|=2 \pi a \Rightarrow\) the lateral surface area of the larger part \(T\) of the given cylinder, divided by a new cylindrical surface (the section of which by the base plane of the given cylinder is the circle \(O_{2}\)), is \(2 \pi a^{2}\).

Next, \(S_{O_{1} A B}=\frac{a^{2} \sin 120^{\circ}}{2}=\frac{\sqrt{3} a^{2}}{4}, S_{O_{1} A D B}=\frac{\pi a^{2}}{3} \Rightarrow\)

\(\Rightarrow S_{A D B}=\frac{\pi a^{2}}{3}-\frac{\sqrt{3} a^{2}}{4} \Rightarrow\) the area of the common part of the circles \(O_{1}\) and \(O_{2}\) is \(2 S_{A D B}\). Therefore, the area of the part of the circle \(O_{1}\) minus \(2 S_{A D B}\) is \(\frac{a^{2}}{6}(2 \pi+3 \sqrt{3})\), from which \(V_{T}=\frac{a^{3}}{6}(2 \pi+3 \sqrt{3})\).

Answer: \(2 \pi a^{2} ; \frac{a^{3}}{6}(2 \pi+3 \sqrt{3})\).