Given the radius of a sphere is measured as 7 m with an error of 0.02 m.

Let x be the radius of the sphere and Δx be the error in measuring the value of x.

Hence, we have x = 7 and Δx = 0.02

The volume of a sphere of radius x is given by

On differentiating V with respect to x, we get

We know

When x = 7, we have.

Recall that if y = f(x) and Δx is a small increment in x, then the corresponding increment in y, Δy = f(x + Δx) – f(x), is approximately given as

Here, and Δx = 0.02

⇒ ΔV = (196π)(0.02)

∴ ΔV = 3.92π

Thus, the approximate error in calculating the volume of the sphere is 3.92π m³.