Question

If the radius of a sphere is measured as 9 cm with an error of 0.03 m, find the approximate error in calculating its surface area.

Answer 1

Let x be the radius and y be the surface area of the sphere.

$$\begin{gathered} \mathrm{Then}, \\ x=9 \\ ∆x=0.03\mathrm{m}=3\mathrm{cm} \\ \Rightarrow x+∆x=9+3=12\mathrm{cm} \\ y=4{\mathrm{\pi x}}^2 \\ \mathrm{For}\mathrm{x}=9, \\ \mathrm{y}=4\mathrm{\pi }\times 9^2=324\mathrm{\pi } \\ \frac{\mathrm{dy}}{\mathrm{dx}}=8\mathrm{\pi x} \\ \Rightarrow {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\mathrm{x}=9}=72\mathrm{\pi } \\ \therefore ∆\mathrm{y}=\mathrm{dy}=\frac{\mathrm{dy}}{\mathrm{dx}}\mathrm{dx}=72\mathrm{\pi }\times 3=216\mathrm{\pi }{\mathrm{cm}}^2 \\ \mathrm{Therefore},\mathrm{the}\mathrm{approximate}\mathrm{error}\mathrm{in}\mathrm{the}\mathrm{surface}\mathrm{area}\mathrm{is}216\mathrm{\pi }c{\mathrm{m}}^2. \\ \mathrm{Disclaimer}:\mathrm{This}\mathrm{solution}\mathrm{has}\mathrm{been}\mathrm{created}\mathrm{according}\mathrm{to}\mathrm{the}\mathrm{question}\mathrm{given}\mathrm{in}\mathrm{the}\mathrm{book}.\mathrm{However},\mathrm{the}\mathrm{solution}\mathrm{given}\mathrm{in}\mathrm{the}\mathrm{book}\mathrm{is}\mathrm{incorrect}. \end{gathered}$$