Acylindrical hole of radius 1 is bored through a solid ball of radius 2. find the volume and outer surface area.

Outer surface area = outer surface area of the cylindrical ball - surfase area of the two circle top and bottom of the cylindrical hole = 4πr^2 - 2(πR^2) = 4π(2)^2 - 2(π(1)^2) = 16π - 2π = 43.98  square units

Volume = 4/3(πr^3) - πR^2h; where h is the height of the cylinder given by 2(r - sqrt(r^2 - R^2))
V = 4/3π(2)^3 - π(1)^2(2(2 - sqrt(2^2 - 1^2))) = 4/3π(8) - π(2(2 - sqrt(4 - 1))) = 32/3π - π(2(2 - sqrt(3))) = 32/3π - π(4 - 2sqrt(3)) = 31.83 cubic units.

Hmmm i had this before , give me a sec