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I've got this problem:

The stem of a particular mushroom has a cylindrical shape. A stem with $2$ cm of height and $r$ centimeters of radius which has a volume $V=2 \pi r^2$. Use differentials to determine the approximated increase of the stem's volume where its radius grows from $0.4$cm to $0.5$cm.

So I got the $dr$ from the difference of $0.4 - 0.5$ which means its result was $0.1$. Right after that, $Dv=4 \pi r dr$. I calculated this without the value of $r$ and I think you can't do it that way, the result was $1.256r \ \text{cm}^3$ but I think there is something left.

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    $dr$ is $0.5-0.4$ (it's `final value-initial value`), not $0.4-0.5$.2011-10-22

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Given that the volume is $V=2\pi r^2$, if $r$ increases to $r+dr$ you have the new volume V'=2\pi (r + dr)^2. You are asked for the increase in volume, which is $Dv=2\pi (r + dr)^2 - 2\pi r^2$. (I am using your notation-I would suggest if the volume is $V$, the change in volume should be $DV$ or $dV$). The method of differentials says that $(dr)^2$ is negligible compared with $dr$, so you are correct in finding $Dv=4 \pi r\ dr$. Although $r$ varies in the problem, we have agreed to ignore the variation except where it is indicated specifically, so it would be consistent to use anything in the range $0.4 - 0.5$. If you care about the difference in the answers you get, you need to use a more accurate method.