#### Question

If the radius of a sphere is measured as 7 m with an error of 0.02 m, find the approximate error in calculating its volume ?

#### Solution

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

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

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

\[x = 7\]

\[ ∆ x = 0 . 02\]

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

\[ \Rightarrow \left( \frac{dy}{dx} \right)_{x = 7} = 196\pi\]

\[ \therefore ∆ y = dy = \frac{dy}{dx}dx = 196\pi \times 0 . 02 = 3 . 92\pi \]

\[\text { Hence, the approximate error in calculating the volume of the sphere is } 3 . 92\pi m^3 .\]

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Solution If the Radius of a Sphere is Measured as 7 M with an Error of 0.02 M, Find the Approximate Error in Calculating Its Volume ? Concept: Approximations.