217. Pierced Sphere. The axis of the cylindrical hole drilled in the sphere is 10 cm long and lies on the diameter of this sphere. What is the volume of the remaining part of the sphere?
217. Let's imagine a sphere of radius $k$, bounded by a thin stretchable membrane that holds a liquid inside the sphere according to the laws of surface tension. Now, let's puncture the sphere along its diameter and insert a cylindrical tube of length $2 k$ that can stretch radially in such a way that it does not lose a single drop of liquid. After the diameter of the tube (but not its length) increases, surface tension will give the outer part of the membrane the shape of a segment of a sphere of a larger radius $R$; the volume of the resulting "ring" will remain unchanged. The length of the inner circumference of such a "ring," $2 \pi \sqrt{R^{2}-k^{2}}$, will increase compared to what it was before stretching, while the thickness of the ring, $R-\sqrt{R^{2}-k^{2}}$, will decrease. Therefore, the volume of the part of the sphere of radius $R$ that remains after a cylindrical hole of length $2 k$ is drilled through it is equal to the volume of the sphere of radius $k$, which is $\frac{4 \pi k^{3}}{3}$. Thus, the volume of the remaining part does not depend on the radius of the original sphere. In our case, $V=\frac{4 \pi\left(5^{3}\right)}{3} \approx 523.6 \mathrm{~cm}^{3}$.