In a manufacturing process, ball bearings must be made with radius of 0.5 mm, with a maximum error in the radius of ±0.019 mm. Estimate the maximum error in the volume of the ball bearing.
Solution: The formula for the volume of the sphere is 4/3r^3Pi . If an error Dr is made in measuring the radius of the sphere, the maximum error in the volume is DV= 4(r+delta(r))^2pi-4^2pi
Rather than calculating DV, approximate DV with dV, where dV= 4r^2Pidr .
Replacing r with 0.5 and dr=Dr with ± 0.019 gives dV=± _______
The maximum error in the volume is about ________ mm3
The text in red are the answers i got, and the ones that are blank i dont know how to do. Can anyone tell me if the answers i obtained are correct....?