(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Use differentials to estimate the amount of tin in a closed tin can with diameter 8cm and height 12cm if the tin is 0.04cm thick.

2. Relevant equations

If:

[tex] z=f(x,y) [/tex]

then

[tex] dz = f_{x}(x,y)dx+f_{y}(x,y)dy [/tex]

3. The attempt at a solution

Perhaps my problem here has to do with the top and bottom of the canister not being taken into account explicitly. If this is the case I can not see it on my own however.

First the volume of a cylinder is:

[tex] V=\pi r^{2}h [/tex]

where here:

[tex] r=8cm \;\; h=12cm [/tex]

the differential of the volume should then be:

[tex] dV=(2 \pi r h) dr + (\pi r^{2}h) dh [/tex]

where in this problem:

[tex] dr=0.04cm \;\; dh=0cm [/tex]

so the second term vanishes and in the calculation we get:

[tex] dV= 2 \pi (8)(12)(0.04) \approx 24 cm^{3} [/tex]

however the answer section tells me that it should be 16cm^3 but if i do the calculation with just geometry I get the same thing, i.e.:

[tex] V_{tin}=12 \pi (8^{2}-(8-0.04)^{2}) \approx 24 cm^{3} [/tex]

If the top/bottom of the container were to be taken into account it would only add to this number from my view.

Am I doing something wrong or is the book just wrong here?

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# Homework Help: Volume of cylinder with differentials

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