Question

If there is an error of 0.1% in the measurement of the radius of a sphere, find approximately the percentage error in the calculation of the volume of the sphere ?

Answer 1

Let x be the radius and y be the volume of the sphere.

\[y = \frac{4}{3}\pi x^3 \]

\[\text { Let } ∆ x \text { be the error in the radius and } ∆ \text { y be the error in the volume }. \]

\[\text { Then,} \frac{∆ x}{x} \times 100 = 0 . 1\]

\[ \Rightarrow \frac{dx}{x} = \frac{1}{1000}\]

\[\text { Now,} y = \frac{4}{3}\pi x^3 \]

\[ \Rightarrow \frac{dy}{dx} = 4 \pi x^2 \]

\[ \Rightarrow dy = 4 \pi x^2 dx\]

\[ \Rightarrow \frac{dy}{y} = \frac{4 \pi x^2 dx}{\frac{4}{3}\pi x^3} = \frac{3}{x}dx\]

\[ \Rightarrow \frac{dy}{y} = \frac{3}{1000}\]

\[ \Rightarrow \frac{∆ y}{y} \times 100 = 0 . 3\]

Hence, the percentage error in the calculation of the volume of the sphere is 0.3.