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Question Number 5542 by FilupSmith last updated on 19/May/16

The surface area of a cylinder is: A=2πr^2 +2πrh As h→0, the shape becomes a 2D circle, so should lim_(h→0) A = πr^2 ??? Or is it that as h→0, it creates a circle of 3 dimensions with infintesimally small height, so it has two circles making it: lim_(h→0) A = 2πr^2 ???

$$\mathrm{The}\:\mathrm{surface}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{cylinder}\:\mathrm{is}: \\ $$$${A}=\mathrm{2}\pi{r}^{\mathrm{2}} +\mathrm{2}\pi{rh} \\ $$$$ \\ $$$$\mathrm{As}\:{h}\rightarrow\mathrm{0},\:\mathrm{the}\:\mathrm{shape}\:\mathrm{becomes}\:\mathrm{a}\:\mathrm{2D}\:\mathrm{circle}, \\ $$$$\mathrm{so}\:\mathrm{should}\:\:\:\:\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{A}\:=\:\pi{r}^{\mathrm{2}} \:\:\:??? \\ $$$$\mathrm{Or}\:\mathrm{is}\:\mathrm{it}\:\mathrm{that}\:\mathrm{as}\:{h}\rightarrow\mathrm{0},\:\mathrm{it}\:\mathrm{creates}\:\mathrm{a}\:\mathrm{circle} \\ $$$$\mathrm{of}\:\mathrm{3}\:\mathrm{dimensions}\:\mathrm{with}\:\mathrm{infintesimally}\:\mathrm{small} \\ $$$$\mathrm{height},\:\mathrm{so}\:\mathrm{it}\:\mathrm{has}\:\mathrm{two}\:\mathrm{circles}\:\mathrm{making}\:\mathrm{it}: \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{A}\:=\:\mathrm{2}\pi{r}^{\mathrm{2}} \:\:\:\:??? \\ $$

Commented by FilupSmith last updated on 19/May/16

That is also what i thought!

$$\mathrm{That}\:\mathrm{is}\:\mathrm{also}\:\mathrm{what}\:\mathrm{i}\:\mathrm{thought}! \\ $$

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