Answer:
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
[tex]V = \dfrac{4}{3} \pi r^3[/tex]
The differential of V with respect to r is:
[tex]\dfrac{dV}{dr }= 4 \pi r^2[/tex]
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
