Calculus 1 help!!!!!!! please i can use some!?

Question

The circumference of a sphere was measured to be 88 cm with a possible error of .8 cm. Use differentials to estimate the maximum error in the calculated surface area.
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Estimate the relative error in the relative error in the calculated surface area
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Answer 1

Since the circumference 2pir = 88 +/- 0.8 it follows r = (88 +/- 0.8)/2pi and |dr|max = 0.8/2pi cm. 
The surface S = 4pir^2 therefore dS = 8pir dr. Therefore |dS|max = 8pir |dr|max = 4r |dr|max = 44.84 cm^2.
The maximum relative surface error is |dS|max/S = (8pir/4pir^2) |dr|max = (2/r) |dr|max = 0.8/(88*pi)= 0.002895 = 0.2895 %.