A cylindrical drill with radius 3 is used to bore a hole through the center of a sphere of radius 5 . Find the volume of the ring shaped solid that remains.
Answer
Final Answer: The volume of the ring-shaped solid that remains is \(\boxed{240.85543677521747}\) cubic units.
Step 1 ：Given a sphere with radius 5 and a cylindrical drill with radius 3.
Step 2 ：The volume of the sphere is given by the formula \(V = \frac{4}{3} \pi r^3\). Substituting the given radius of the sphere, we get \(V_{sphere} = \frac{4}{3} \pi (5)^3 = 523.5987755982989\) cubic units.
Step 3 ：The volume of the cylindrical hole is given by the formula \(V = \pi r^2 h\). In this case, the height of the cylinder is the diameter of the sphere, which is \(2 * 5 = 10\). Substituting the given radius of the drill and the height, we get \(V_{hole} = \pi (3)^2 * 10 = 282.7433388230814\) cubic units.
Step 4 ：The volume of the ring-shaped solid that remains is the volume of the sphere minus the volume of the cylindrical hole. So, \(V_{ring} = V_{sphere} - V_{hole} = 523.5987755982989 - 282.7433388230814 = 240.85543677521747\) cubic units.
Step 5 ：Final Answer: The volume of the ring-shaped solid that remains is \(\boxed{240.85543677521747}\) cubic units.
