Volume of cylinder with differentials

In summary, the conversation is about using differentials to estimate the amount of tin in a closed tin can with given dimensions and thickness. The volume of the can is calculated using the formula for a cylinder and the differential of the volume is found. The discrepancy in the final answer is discussed and resolved, with the correct answer being 16cm^3 if the top and bottom of the can are taken into account.
  • #1
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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?
 
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  • #2
The diameter of the can is 8 cm; the radius is 4 cm.
 
  • #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:

1. What is the formula for finding the volume of a cylinder with differentials?

The formula for finding the volume of a cylinder with differentials is V = πr^2h, where V is the volume, π is the constant pi, r is the radius of the cylinder, and h is the height of the cylinder.

2. How do you calculate the volume of a cylinder with differentials?

To calculate the volume of a cylinder with differentials, you can use the formula V = πr^2h, where V is the volume, π is the constant pi, r is the radius of the cylinder, and h is the height of the cylinder. Simply plug in the values for each variable and solve for V.

3. What units are used for the volume of a cylinder with differentials?

The units for the volume of a cylinder with differentials are typically cubic units, such as cubic inches, cubic feet, or cubic meters. This is because volume is a measure of three-dimensional space.

4. Can the volume of a cylinder with differentials be negative?

No, the volume of a cylinder with differentials cannot be negative. The volume represents the amount of space occupied by the cylinder, and space cannot have a negative value.

5. How does changing the radius or height of a cylinder affect its volume with differentials?

Changing the radius or height of a cylinder will directly affect its volume with differentials. The volume is directly proportional to both the radius and height, meaning that as either one increases, the volume will also increase. This can be seen in the formula V = πr^2h, where changing the values of r or h will result in a different volume.

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