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.
$$\begin{gathered} y=\frac{4}{3}\mathrm{\pi }{x}^3 \\ \mathrm{Let}∆x\mathrm{be}\mathrm{the}\mathrm{error}\mathrm{in}\mathrm{the}\mathrm{radius}\mathrm{and}∆y\mathrm{be}\mathrm{the}\mathrm{error}\mathrm{in}\mathrm{the}\mathrm{volume}. \\ \mathrm{Then},\frac{∆x}{x}\times 100=0.1 \\ \Rightarrow \frac{dx}{x}=\frac{1}{1000} \\ \mathrm{Now},y=\frac{4}{3}\mathrm{\pi }{x}^3 \\ \Rightarrow \frac{dy}{dx}=4{\mathrm{\pi x}}^2 \\ \Rightarrow \mathrm{dy}=4{\mathrm{\pi x}}^2\mathrm{dx} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{y}}=\frac{4{\mathrm{\pi x}}^2\mathrm{dx}}{{\displaystyle \frac{4}{3}}\mathrm{\pi }{\mathrm{x}}^3}=\frac{3}{\mathrm{x}}\mathrm{dx} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{y}}=\frac{3}{1000} \\ \Rightarrow \frac{∆\mathrm{y}}{\mathrm{y}}\times 100=0.3 \end{gathered}$$

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