The following exercise is presented in my book as an example of linear approximation and calculating the rate of change using derivatives. The rate of change for the volume gives $ \approx-25.13 cm^3 $
Taking aside the purpose of the example, I'm wondering if there's any practical reason to use calculus instead of taking the volume of the ball when it has $ 5cm $, the volume when it has $ 4.92cm $ and then calculating the difference of those two values. With this approach, the solution is $ \approx-24.73 cm^3 $
Is this approach also correct?