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Asphyxiated
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Homework Statement
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.
Homework Equations
If:
[tex] z=f(x,y) [/tex]
then
[tex] dz = f_{x}(x,y)dx+f_{y}(x,y)dy [/tex]
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?