If the radius of a sphere is measured as 7 m with an error of 0.02 m, then find the approximate error in calculating its volume.

Solution

Let r be the radius of the sphere and ∆r be the error in measuring the radius.
∴ r =7 m. ∆r = 0.02 m
Let V be volume of sphere
$therefore space space space space space space space straight V space equals space 4 over 3 πr cubed Now comma space space space space space space dV space equals space open parentheses dV over dr close parentheses increment straight r space$
$equals space 4 πr squared increment straight r space equals space 4 straight pi left parenthesis 7 right parenthesis squared space left parenthesis 0.02 right parenthesis space equals space 4 straight pi space cross times space 49 space cross times space 0.02 space equals space 3.92 space straight pi space straight m cubed$
$therefore$ approximate error in caluclating the volume = $3.92 space straight pi space straight m cubed.$

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Application Of Derivatives

If the radius of a sphere is measured as 7 m with an error of 0.02 m, then find the approximate error in calculating its volume.

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