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 is4/3r^3Pi. If an error Dris 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 DVwithdV, wheredV=4r^2Pidr.

Replacingrwith0.5anddr=Drwith ±0.019givesdV=± _______

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....?