## 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 = \({4\over 3}\pi r^3\) \(\implies\) \(dV\over dr\) = \(4\pi r^2\) \(\implies\) \(({dV\over dr})_{r = 9}\) = …