# Approximation of Volume using Differential

• Frank69
In summary, the differential equation is solved to give the approximate volume of a can of tin with a radius of 4 cm., height of 12 cm., and thickness of 0.04 cm.
Frank69

## Homework Statement

Use the differential to estimate the volume of tin a can of radius 4 cm., height 12 cm.
and thickness 0.04 cm is made of.

## Homework Equations

(differential)

df = f'(x,y)dx + f'(x,y)dy

The approximated volume should be 14cm^3. However, I need the procedure to get to the result (since it was given).

## The Attempt at a Solution

I know how to compute the differential. In order to do so I need the two derivatives in x and y. then the formula is straight forward: df = f'(x,y)dx + f'(x,y)dy
Unfortunately I have no clue how to apply this concept to solve the problem.
Any help would be appreciated.

For a cylinder the volume is V=πr2h.

So in your formula it would be

dV= (∂V/∂r)dr + (∂V/∂h)dh

So when you put in ∂V/∂r and ∂V/∂h into the formula, 'dr' is the change in radius (your thickness) and 'dh' is your change in height (which is zero).

Frank69 said:

## Homework Statement

Use the differential to estimate the volume of tin a can of radius 4 cm., height 12 cm.
and thickness 0.04 cm is made of.

## Homework Equations

(differential)

df = f'(x,y)dx + f'(x,y)dy

The approximated volume should be 14cm^3. However, I need the procedure to get to the result (since it was given).

## The Attempt at a Solution

I know how to compute the differential. In order to do so I need the two derivatives in x and y. then the formula is straight forward: df = f'(x,y)dx + f'(x,y)dy
Unfortunately I have no clue how to apply this concept to solve the problem.
Any help would be appreciated.

You're given height (h), radius (r), and thickness (dr). You have all the info you need. Make the necessary substitutions and you'll have it.

If you get stuck, post what you have and we can provide further hints (if we've determined you're actually putting some effort into understanding the solution).

Thanks to everyone. Geezer, you say that dh is zero. I instead set it to 0.04 (I considered also the bases).

V(r,h) = πr2h.

(∂V/∂r) = 2πrh

(∂V/∂h) = πr²

Then I plugged r = 4 and h = 12 in dV

dV = (∂V/∂r)(4,12)dr + (∂V/∂h)(4,12)dh

And I got: dV = 301.44*0.04 + 50.24 *0.04 = 12.0576 + 2.0096 = 14.0672

14 cm^3 was the result expected.

If instead I used dh = 0 as you suggest I would get:

dV = 301.44*0.04 + 50.24 *0 = 12.0576 + 0 = 12.0576

which is not the right answer.

Where's the catch?

Frank69 said:
Thanks to everyone. Geezer, you say that dh is zero. I instead set it to 0.04 (I considered also the bases).

I didn't say dh=0. In fact, I never commented on dh (dh=dr since the thickness of the can is uniform).

rock.freak667 said:
'dh' is your change in height (which is zero)

Sorry Gezeer, I meant what was written by rock.freak667 in the previous post.

Thanks to all of you anyway :)

I don't think your answer should be 14 cm3 since if you calculate

dV as dV = πh(r22-r12) where r2=4.04, r1= 4 and h =12, you would get around 12.1 or so.

Frank69 said:
Sorry Gezeer, I meant what was written by rock.freak667 in the previous post.

Thanks to all of you anyway :)

No biggie.

## 1. What is an approximation of volume using differential?

An approximation of volume using differential is a mathematical technique used to estimate the volume of a three-dimensional object by breaking it into smaller, simpler parts and calculating their volumes. This method uses the concept of differentials, which represent small changes in a variable, to approximate the volume of the entire object.

## 2. How does the approximation of volume using differential work?

The approximation of volume using differential works by dividing a three-dimensional object into smaller, simpler parts. The volume of each part is then calculated using the differential volume formula, and the volumes are added together to get an estimate of the total volume of the object.

## 3. What is the differential volume formula?

The differential volume formula is a mathematical formula used to calculate the volume of a small part of a three-dimensional object. It is given by dV = A(x)dx, where dV is the differential volume, A(x) is the cross-sectional area of the part at a specific value of x, and dx is the small change in x. This formula is used in the approximation of volume using differential.

## 4. What are the advantages of using the approximation of volume using differential?

The approximation of volume using differential has several advantages, including its ability to provide a quick and easy estimate of the volume of a three-dimensional object, its versatility in handling complex shapes, and its accuracy when the parts used are small enough.

## 5. Are there any limitations to the approximation of volume using differential?

Yes, the approximation of volume using differential has some limitations. It can only provide an estimate of the volume, which may not be exact. The accuracy of the approximation also depends on the size of the parts used, with smaller parts leading to a more accurate estimate. Additionally, this method may not work well for irregular or non-geometric shapes.

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