You are wondering how inaccuracies in the radius or height of your can effect the volume of the can. Use differentials to answer the questions below.
An error of $p$ percent in the radius will cause an error of approximately percent in the volume of the can.
An error of $p$ percent in the height will cause an error of approximately percent in the volume of the can.
Note that although the sentences above contain the word "approximately", the answers are actually determined precisely by the procedure with which we compute them. For example for the dependence on the radius you consider $h$ constant, and you use the differentials $$d V = V'(r) d r.$$ The percentage of the error in $v$ is $$\frac{100 \hbox{dV}}{V}$$ and you know that $$p = \frac{100 d r}{r} \quad\hbox{and}\quad d V = \frac{d V}{d r}{d r}.$$ Your answer will depend on $p$. Proceed similarly for the dependence of the error on $h$.