The circumference of a sphere was measured to be 70.000 cm with a possible error of 0.50000 cm. Using linear approximation, what is the maximum error of the surface area

Question

Then estimate the relative error in the calculated surface area.

Answer 1

C = 2πr; S = 4πr²; dS = 8πrdr; dC = 2πdr; dr = dC/(2π), r = C/(2π)·. So, dS = 8π·C/(2π)·dC/(2π) = 2CdC/π =

2·70·0.5/π cm² = 22.282 cm²