A cola company requires a can in the shape of a right circular cylinder of height $10~\rm{cm}$ and radius $3~\rm{cm}$. If the manufacturer of the cans claims a percentage error of no more than $0.2\%$ in the height and no more than $0.1\%$ in the radius, what is the approximate maximum variation in the volume of the can?

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Solution The volume $V$ of a right circular cylinder of height $h~\rm{cm}$ and radius $R\;\rm{cm}$ is $V=\pi\;R^{2}h\;\rm{cm}^{3}$. We find the differential $dV.$ $$dV=\frac{\partial V}{\partial R}dR+\frac{\partial V}{\partial h}dh=2\pi\;Rh\,dR+\pi\;R^{2}dh$$

The relative error in the radius $R$ is $\dfrac{|\Delta R|}{R}=\dfrac{|dR|}{R}=0.001$, and the relative error in the height $h$ is $\dfrac{\left\vert \Delta h\right\vert }{h}=\dfrac{|dh|}{h}=0.002$. The relative error in the volume $V$ is $$\begin{eqnarray*} \frac{|\Delta V|}{V} &\approx &\frac{|dV|}{V}=\frac{\left\vert 2\pi Rh\, dR+\pi R^{2} dh\right\vert }{\pi R^{2}h}=\left\vert 2\dfrac{dR}{R}+\dfrac{dh}{h}\right\vert\\[4pt] &=&\left\vert 2\dfrac{\Delta R}{R}+\dfrac{\Delta h}{h} \right\vert \leq 2\frac{\left\vert \Delta R\right\vert }{R}+\frac{\left\vert \Delta h\right\vert }{h} \\ &=&2(0.001)+0.002=0.004 \end{eqnarray*}$$

The maximum variation in the volume is approximately $0.4\%$, so the actual volume of the container varies as follows: $$\begin{equation*} V=\pi R^{2}h\pm 0.004( \pi R^{2}h) =\pi R^2 h ( 1\pm 0.004) =90\pi ( 1\pm 0.004) \rm{cm}^{3} \end{equation*}$$

The volume $V$ is between $89.64\pi \approx 281.612\;\rm{cm}^{3}$ and $90.36\pi \approx 283.874\rm{cm}^{3}$.