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Volume of cylinder with differentials

  1. Aug 28, 2011 #1
    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

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


    [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?
  2. jcsd
  3. Aug 28, 2011 #2


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    The diameter of the can is 8 cm; the radius is 4 cm.
  4. Aug 28, 2011 #3
    ha, ok... for some reason I always do that when I haven't dealt with radius vs diameter in awhile, sometimes I need the obvious pointed out :)

    thanks vela

    edit: so then its 12....

    if i take the top and bottom its 16!

    thanks again
    Last edited: Aug 28, 2011
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