A sphere was measured and its radius was found to be 21 cm with a possible error of no more than 0.05 cm. Estimate the maximum possible error in the volume if we use this value of the radius. equation to use:$${\mathtt{v}} = \left({\frac{{\mathtt{4}}}{{\mathtt{3}}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\right){\mathtt{\,\times\,}}{{\mathtt{r}}}^{{\mathtt{3}}}$$.
$$\\V=\frac{4}{3}\pi*r^3\\
estimated\;r\;=21cm\qquad max\;r\;=21.05cm\\
$biggest error = max volume - estimated volume$\\
max\; error=\frac{4}{3}\pi*21.05^3-\frac{4}{3}\pi*21^3\\\\
max\; error=\frac{4}{3}\pi*(21.05^3-21^3)\\\\$$
$${\frac{{\mathtt{4}}}{{\mathtt{3}}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}\left({{\mathtt{21.05}}}^{{\mathtt{3}}}{\mathtt{\,-\,}}{{\mathtt{21}}}^{{\mathtt{3}}}\right) = {\mathtt{277.748\: \!730\: \!102\: \!649\: \!218\: \!5}}$$
max possible error approx 278 cm^3
$$\\V=\frac{4}{3}\pi*r^3\\
estimated\;r\;=21cm\qquad max\;r\;=21.05cm\\
$biggest error = max volume - estimated volume$\\
max\; error=\frac{4}{3}\pi*21.05^3-\frac{4}{3}\pi*21^3\\\\
max\; error=\frac{4}{3}\pi*(21.05^3-21^3)\\\\$$
$${\frac{{\mathtt{4}}}{{\mathtt{3}}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}\left({{\mathtt{21.05}}}^{{\mathtt{3}}}{\mathtt{\,-\,}}{{\mathtt{21}}}^{{\mathtt{3}}}\right) = {\mathtt{277.748\: \!730\: \!102\: \!649\: \!218\: \!5}}$$
max possible error approx 278 cm^3