Question 152: There is a solid metal sphere with a radius of 20. A circular hole with a radius of 12 is drilled through the sphere along a diameter. The remaining part is then melted and cast into a new solid sphere. Try to find the radius of the new sphere.
Question 152, Solution: As shown in the figure, according to the conditions, the removed part is a cylinder and two spherical caps, where the base radius of the cylinder is 12, and the height is \(16 \times 2 = 32\); the height of the spherical cap is 4. Therefore, the volume of the remaining part is:
$$
\frac{4 \pi \cdot 20^{3}}{3} - \pi \cdot 12^{2} \cdot 32 - 2 \times \frac{\pi \cdot 4^{2}(3 \times 20 - 4)}{3} = \frac{4 \pi \cdot 16^{3}}{3}
$$

This indicates that the radius of the new sphere is 16.
Note: The volume formula for a spherical cap: If the radius of the sphere is \(R\), and the height of the spherical cap is \(h\), then the volume of the spherical cap is \(V = \frac{\pi h^{2}(3 R - h)}{3}\).
Proof: As shown in the figure, the differential element of the shaded part is \(\pi \left(R^{2} - x^{2}\right) d x\). Therefore, the volume of the spherical cap is:
$$
\int_{R-h}^{h} \pi \left(R^{2} - x^{2}\right) d x = \left. \frac{\pi \left(3 R^{2} x - x^{3}\right)}{3} \right|_{R-h}^{R} = \frac{\pi h^{2}(3 R - h)}{3}
$$