### Question Description

The circumference of a sphere was measured to be 76 cm with a possible error of 0.5 cm.

What is the relative error? (Round your answer to three decimal places.)

(b) Use differentials to estimate the maximum error in the calculated volume. (Round your answer to the nearest integer.)

What is the relative error? (Round your answer to three decimal places.)

## Explanation & Answer

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S = 4*pi*r^2

delta S = 8*pi*r delta r = 8*pi*76*0.5 = 955 cm^2

relative error = dela S/S = 8 pi*r*deltar / (4*pi*r^2) = 2delta r/r = 1/76 = 0.013

b) V= 4/3 pi*r^3

dV = 4pir^2 dr = 4*pi*76^2 *0.5 = 36292

relative error = 4pi r^2 dr / (4/3*pir^3) = 3dr/r = 1.5/76 = 0.020

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