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In a sphere of radius $\sqrt{3}$, a cylindrical hole is drilled; the axis of the cylinder passes through the center of the sphere, and the diameter of the cylinder's base is equal to the radius of the sphere. Find the volume of the remaining part of the sphere.
Let $R=\sqrt{3}$ be the radius of the sphere, $r$ be the radius of the hole, $H$ be the height of the cylindrical part of the hole, $h$ be the height of each of the two spherical segments that, together with the cylindrical part, form the drilled part of the sphere, $V$ be the volume of the sphere, $V_1$ be the volume of the cylindrical part of the hole, $v$ be the volume of each of the specified segments, and $V_2$ be the volume of the remaining part of the sphere after drilling. Then,

$$
\begin{gathered}
r=\frac{R}{2}, H=\sqrt{(2 R)^{2}-R^{2}}=R \sqrt{3}, \\
h=\frac{2 R-H}{2}=\frac{2 R-R \sqrt{3}}{2}=\frac{R(2-\sqrt{3})}{2}, \\
V=\frac{1}{3} \pi R^3, V_1=\pi r^2 H=\pi\left(\frac{R}{2}\right)^2 \cdot R \sqrt{3}=\frac{\pi R^3 \sqrt{3}}{4}, \\
v=\pi h^2\left(R-\frac{h}{3}\right)=\pi\left(\frac{R(2-\sqrt{3})}{2}\right)^2 \cdot\left(R-\frac{R(2-\sqrt{3})}{6}\right)= \\
=\pi R^3(2-\sqrt{3})^2 \cdot \frac{1+\sqrt{3}}{6}=\frac{\pi R^3(7-4\sqrt{3}) \cdot(1+\sqrt{3})}{6}= \\
=\frac{\pi R^3(16-9\sqrt{3})}{12}, \\
V_2=V-V_1-2v=\frac{1}{3} \pi R^3-\frac{\pi R^3 \sqrt{3}}{4}-\frac{\pi R^3(16-9\sqrt{3})}{6}= \\
=\pi R^3\left(\frac{1}{3}-\frac{\sqrt{3}}{4}-\frac{16-9\sqrt{3}}{6}\right)=\frac{\pi R^3 \cdot(\sqrt{3}}{12}=\pi R^3 \cdot \frac{\sqrt{3}}{2}= \\
=\pi(\sqrt{3})^3 \cdot \frac{\sqrt{3}}{2}=\pi \cdot 3\sqrt{3} \cdot \frac{\sqrt{3}}{2}=\frac{9\pi}{2}.
\end{gathered}
$$

Answer

## $\frac{9\pi}{2}$