A cylindrical tin has an internal radius of 6 cm and an internal height of 12 cm. A solid ball is dropped in and fits exactly, touching the curved side and both ends. How much space, in cubic cm, is left empty inside the tin? (Use π = 3.14.)

The ball touches both ends, so its radius is 6 cm and the tin's height is 12 cm. Tin = π$r^{2}h=3.14\times 36\times 12=1356.48$ cubic cm; ball = (4/3)π$r^3=(4/3)\times 3.14\times 216=904.32$ cubic cm; the gap = 1356.48 - 904.32 = 452.16 cubic cm, which is exactly one third of the tin — a tidy check, since a ball that just fits always takes two thirds of the tin. The trap 904.32 reports the ball rather than the space around it; 1356.48 is the empty tin, with the ball never put in; 678.24 drops the 4/3 and uses π$r^3$ for the ball.