Let the error in measuring the radius of a sphere be k%.

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

Hence, we have

∴ Δx = 0.01kx

The volume of a sphere of radius x is given by

On differentiating V with respect to x, we get

We know

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 = 0.01kx

⇒ ΔV = (4πx²)(0.01kx)

∴ ΔV = 0.04kπx³

The percentage error is,

⇒ Error = 0.03k × 100%

∴ Error = 3k%

Thus, the error in measuring the volume of the sphere is approximately three times the error in measuring its radius.