With what accuracy must the radius of a sphere be
With what accuracy must the radius of a sphere be measured so that the relative error in the computed volume does not exceed $1\%$?
Explanation of Method
A cubic law, so the tolerance on the input is a third of the tolerance on the output.
Step 1: Volume and its differential
\[ V = \frac43\pi r^{3}, \qquad dV = 4\pi r^{2}\,dr. \]
Step 2: Form the relative error
Dividing the differential by the volume, the numerical factors and most powers of $r$ cancel: \[\frac{\bigl|dV\bigr|}{V} = \frac{4\pi r^{2}\bigl|dr\bigr|}{\tfrac43\pi r^{3}} = 3\,\frac{\bigl|dr\bigr|}{r}.\]
The relative error of the volume is exactly three times that of the radius.
Step 3: Impose the requirement
\[ 3\,\frac{\bigl|dr\bigr|}{r} \le 0.01 \;\Longrightarrow\; \frac{\bigl|dr\bigr|}{r} \le \frac{0.01}{3} \approx 0.00333. \]
Step 4: Verification
\[\ln V = \ln\frac{4\pi}{3}+3\ln r \;\Longrightarrow\; \frac{dV}{V} = 3\,\frac{dr}{r} \;\;\checkmark\]
\[\text{Test, } r = 10:\ V = 4188.79;\quad r = 10.0333:\ V = 4230.72,\ \text{a rise of } 1.001\% \;\;\checkmark\]
\[\text{A sphere of radius } 10\ \text{cm therefore needs its radius to } \pm 0.33\ \text{mm}\]
Result
Final answer:
\[\boxed{\,\frac{\bigl|\Delta r\bigr|}{r} \le \frac13\% \approx 0.33\%\,}\]
Remark
Remark. Taking logarithms before differentiating is the quickest route to any such rule: $y = x^{n}$ gives $\dfrac{dy}{y} = n\dfrac{dx}{x}$ at once. The demand is stringent in practice — three significant figures in the radius buy only about two in the volume, which is the everyday form of the same statement.