Question

Show that the relative error in computing the volume of a sphere, due to an error in measuring the radius, is approximately equal to three times the relative error in the radius ?

Answer 1

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

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

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

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

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

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

\[ \Rightarrow ∆ y = 3 \times y \times \frac{∆ x}{x}\]

\[ \Rightarrow \frac{∆ y}{y} = 3\frac{∆ x}{x}\]

Hence proved.