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If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, then find the approximating error in calculating its volume.

Solution :

Let r be the radius of a sphere and δr be the error in measuring the radius. Then, r = 9 cm and δr = 0.03 cm.

Let V be the volume of the sphere. Then,

V = 43πr3  dVdr = 4πr2

(dVdr)r=9 = 4π×92 = 324π

Let δV be the error in V due to error in V due to error δr in r. Then,

δV = dVdr δ  δV =  324π×0.03 = 9.72πcm3


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