Answer
a) $1.2$%
b) $1.8$%
Work Step by Step
a)
For a sphere of radius $r$
the circumference is $C$ = $2{\pi}r$
the surface area is $S$ = $4{\pi}r^{2}$
so
$r$ = $\frac{C}{2\pi}$
$S$ = $4\pi(\frac{C}{2\pi})^{2}$ = $\frac{C^{2}}{\pi}$
$dS$ = $\frac{2}{\pi}CdC$
When $C$ = $84$ and $dC$ - $0.5$
$dS$ = $\frac{2}{\pi}(84)(0.5)$ = $\frac{84}{\pi}$
so the maximum error is about $\frac{84}{\pi}$ $\approx$ $27$ $cm^{2}$
Relative error $\approx$ = $\frac{\frac{84}{\pi}}{\frac{84^{2}}{\pi}}$ = $\frac{1}{84}$ $\approx$ $0.012$ = $1.2$%
b)
$V$ = $\frac{4}{3}{\pi}r^{3}$ = $\frac{4}{3}{\pi}(\frac{C}{2\pi})^{3}$ = $\frac{C^{3}}{6{\pi}^{2}}$
$dV$ = $\frac{1}{2{\pi}^{2}}C^{2}dC$
When $C$ = $84$ and $dC$ = $0.5$
$dV$ = $\frac{1}{2{\pi}^{2}}(84)^{2}(0.5)$ = $\frac{1764}{{\pi}^{2}}$ $\approx$ $179$ $cm^{3}$
The relative error is approximately
$\frac{dV}{V}$ = $\frac{\frac{1764}{{\pi}^{2}}}{\frac{(84)^{3}}{6{\pi}^{2}}}$ $\approx$ $0.018$ = $1.8$%
