The radius of a sphere is measured to be 3.0 inches. If the measurement is correct within 0.01 inches, use differentials to estimate the error in the volume of sphere.

ΔV = 0.36π   in³

Step-by-step explanation:

Given that:

The radius of a sphere = 3.0

If the measurement is correct within 0.01 inches

i.e the change in the radius Δr = 0.01

The objective is to use differentials to estimate the error in the volume of sphere.

We all know that the volume of a sphere

The differential of V with respect to r is:

dV = 4 πr² dr

which can be re-written as:

ΔV = 4 πr² Δr

ΔV = 4 × π × (3)² × 0.01

ΔV = 0.36π   in³