(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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Volume of cylinder with differentials

**Physics Forums | Science Articles, Homework Help, Discussion**