The radius of a sphere was measured and found to be 18 cm with a possible error in measurement of at most 0.01 cm. What is the maximum error in using this value of the radius to compute the volume of the sphere? SOLUTION If the radius of the sphere is r then its volume is If the error in the measured value of r is denoted by dr = Ar, then the corresponding error in the calculated value of V is AV, which can be approximated by the differential dV = 4ar dr. When r= 18 and dr = 0.01, this becomes dV = 4x( |). (round to the nearest integer). The maximum error in the calculated volume is about |cm3.