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 \(\delta\)r be the error in measuring the radius. Then, r = 9 cm and \(\delta\)r = 0.03 cm.

Let V be the volume of the sphere. Then,

V = \({4\over 3}\pi r^3\)  \(\implies\) \(dV\over dr\) = \(4\pi r^2\)

\(\implies\) \(({dV\over dr})_{r = 9}\) = \(4\pi \times 9^2\) = \(324 \pi\)

Let \(\delta\)V be the error in V due to error in V due to error \(\delta\)r in r. Then,

\(\delta\)V = \(dV\over dr\) \(\delta\)r  \(\implies\)  \(\delta\)V =  \(324 \pi\times 0.03\) = \(9.72 \pi cm^3\)


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