Question

if the error in the measurement of the radius is 2% what is the error in the volume of the sphere

Answer 1

Given error 2% For a sphere, V = 4\3 × π r^3 Taking log both sides, Log V = 3 log r + log (4\3 π) Diff. Both sides w.r.to V d\dV log V = 3 d\dV r + d\dV log (4\3 π) 1\V = 3\r dr\dV + 0 dV\V = 3×2% = 6%. 

Answer 2

Step-by-step explanation:

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