# Density of cylinder, error propagation

1. Jul 8, 2014

### Zondrina

1. The problem statement, all variables and given/known data

The density of a cylinder is calculated from the following data:

$m= (2.8±0.8)g$
$d = (2.2±0.1)cm$
$h = (4.0±1.0) cm$

What is the error on the density, before rounding, in $\frac{g}{cm^3}$?

2. Relevant equations

$V = \pi r^2 h = \frac{1}{4} \pi d^2 h$

Ans : 0.072

3. The attempt at a solution

Not sure this is the right section for this. I figured I would first calculate the volume, then use $d = \frac{m}{V}$.

$V = \frac{1}{4} \pi d^2 h$
$V = \frac{1}{4} \pi (2.2±0.1)(2.2±0.1)(4.0±1.0) cm^3$
$V = \frac{1}{4} \pi (19±6) cm^3$

How would I propagate the scalars in front?

2. Jul 9, 2014

### ehild

The constants (pi and 4) have no error. What error propagation formula do you use?

ehild

3. Jul 9, 2014

### Zondrina

So we have (before rounding):

$V = \frac{1}{4} \pi (19.36 \pm 6.58)$
$V = (15.21 \pm 5.17)$

?

I don't really have a formula to use for this. I remember something about $\sqrt{ ( \frac{∂f}{∂x})^2 \sigma_x^2 + ( \frac{∂f}{∂y})^2 \sigma_y^2 + ... }$, but I don't think that will help.

4. Jul 9, 2014

### SteamKing

Staff Emeritus
Is the height of the cylinder truly not known to within 25% of its actual value (4.0+/-1.0) cm? This measurement error seems to be rather a lot.

5. Jul 9, 2014

### ehild

Yes, that it is. f is the density, and its error is $\sqrt{ ( \frac{∂f}{∂x})^2 \sigma_x^2 + ( \frac{∂f}{∂y})^2 \sigma_y^2 + ... }$ Now you have three variables, m, d, h. Replace σ-s by the given errors.

ehild

6. Jul 9, 2014

### Orodruin

Staff Emeritus
I agree with SteamKing - the errors seem very large. In fact, they seem so large that I think it is not appropriate to approximate using derivatives only and symmetric errors on the parameters are going to result in asymmetric errors in the final quantity. However, this is probably not what was intended in the question ...

7. Jul 9, 2014

### Zondrina

Sorry for the late reply. The errors are as such, as crazy as that is.

With $\rho = \frac{4}{\pi} \frac{m}{d^2 h}$ being the density. The error on $\rho$ is given by:

$\sigma_{\rho} = \sqrt{ ( \frac{∂ \rho}{∂m})^2 \sigma_m^2 + ( \frac{∂ \rho}{∂d})^2 \sigma_d^2 + ( \frac{∂ \rho}{∂ h})^2 \sigma_h^2 }$

$\frac{∂ \rho}{∂m} = \frac{4}{\pi} \frac{1}{d^2 h}$

$\frac{∂ \rho}{∂d} = - \frac{8}{\pi} \frac{m}{(d^3 h)}$

$\frac{∂ \rho}{∂h} = - \frac{4}{\pi} \frac{m}{(dh)^2}$

The errors are obvious to plug in. What should I do about the variables though i.e $d, h$ and $m$?

EDIT: As a side note, I got the answer awhile ago, but I did it using numerical methods rather than the partial derivatives.

I'd like to learn how to use this formula as I see the value in learning it.

8. Jul 9, 2014

### ehild

Substitute the given mean values. m=2.8, d=2.2, h=4.0.

You can notice that
$\frac{∂ \rho}{∂m} = \frac{\rho}{m}$
$\frac{∂ \rho}{∂d} = - 2\frac{\rho}{d}$
$\frac{∂ \rho}{∂h} = - \frac{\rho}{h}$

The error you get is the probable error, less than the maximum error, half the difference between the possible maximum and minimum densities. And the other posters are right, the error of h is too big.

ehild

Last edited: Jul 10, 2014
9. Jul 10, 2014

### Zondrina

Indeed the error on $h$ is ridiculous. The answer complements of wolfram for anyone who is interested:

http://www.wolframalpha.com/input/?i=sqrt%28+%28%284%2Fpi%29%281%2F%284%282.2%29^2%29%29%29^2++%280.8%29^2+%2B+%28+%288%2Fpi%29+2.8%2F%284*2.2^3%29%29^2+%280.1%29^2+%2B+%28+%284%2Fpi%29%282.8%29%2F%282.2*4.0%29^2+%29^2+%281.0%29^2++%29

Thank you.