#### 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 ?

#### Solution

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

\[\text { Then }, \]

\[x = 9\]

\[ ∆ x = 0 . 03 m = 3cm\]

\[ \Rightarrow x + ∆ x = 9 + 3 = 12 cm\]

\[y = 4 \pi x^2 \]

\[\text { For } x = 9, \]

\[ y = 4\pi \times 9^2 = 324\pi\]

\[\frac{dy}{dx} = 8\pi x\]

\[ \Rightarrow \left( \frac{dy}{dx} \right)_{x = 9} = 72\pi\]

\[ \therefore ∆ y = dy = \frac{dy}{dx}dx = 72\pi \times 3 = 216\pi {cm}^2 \]

\[\text { Therefore, the approximate error in the surface area is} 216\pi c m^2 . \]

\[\text { Disclaimer: This solution has been created according to the question given in the book . However, the solution given in the book is incorrect } .\]