# Volume of cylinder with differentials

1. Aug 28, 2011

### Asphyxiated

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:

$$z=f(x,y)$$

then

$$dz = f_{x}(x,y)dx+f_{y}(x,y)dy$$

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:

$$V=\pi r^{2}h$$

where here:

$$r=8cm \;\; h=12cm$$

the differential of the volume should then be:

$$dV=(2 \pi r h) dr + (\pi r^{2}h) dh$$

where in this problem:

$$dr=0.04cm \;\; dh=0cm$$

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

$$dV= 2 \pi (8)(12)(0.04) \approx 24 cm^{3}$$

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.:

$$V_{tin}=12 \pi (8^{2}-(8-0.04)^{2}) \approx 24 cm^{3}$$

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. Aug 28, 2011

### vela

Staff Emeritus
The diameter of the can is 8 cm; the radius is 4 cm.

3. Aug 28, 2011

### Asphyxiated

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