Question

The error in the measurement of the radius of a sphere is 2%. What will be the error in the calculation of its volume?

Answer 1

To solve this problem we will apply the geometric concepts of the Volume based on the consideration made of the radius measurement. The Volume must be written in differential terms of the radius and from the formula of the margin of error the respective response will be obtained.

The error in radius of sphere is not exceeding 2%

$\frac{dr}{r} = \pm 0.02$

The objective is to find the percentage error in the volume.

The volume can be defined as

$V = \frac{4}{3} \pi r^3$

Differentiate with respect the radius we have,

$\frac{dV}{dr} = 4\pi r^2$

$dV = 4\pi r^2 \times dr$

$dV = 4\pi r^2 (\pm 0.02r)$

$dV = \pm 4\times 0.02 \times \pi r^3$

The percentage change in the volume is as follows

$\% change = \frac{dV}{V} \times 100$

$\% change = \frac{\pm 4 \times 0.02 \times \pi r^3 \times 3}{4\pi r^3}\times 100$

$\% change = \pm 6\%$

Therefore the percentage change in volume is $\pm 6\%$