Calculating Density of Rock from Mars w/ Uncertainty

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Homework Statement



the mass of a particular sample of rock retrived from Mars is m=772.2+or-0.2g. The sample is in the shape of cylinder of height h=10+or-0.1 cm and base radius r=2.5 +or-0.1cm. calculate the density of the rock sample in units of kg/m^3. You must state your result in the correct units with its associated uncertainty. The volume of a cylinder is V=pir^2h.

Homework Equations


p=m/v



The Attempt at a Solution



I can get the density but I cannot get the correct uncertainty.

The density I get is

(772.2g/196.35cm^3)((1 kg/m^3)/10^-3g/cm^3)=3932.77 kg/m^3

I believe the uncertainty is supposed to be +or-400 but I end up with four.

(delta p)/(3.9g/cm^3) = .2/772.2 + .14/196.35

delta p = .004 g/cm^3

.004 g/cm^3 ((1kg/m^3)/(10^-3 g/cm^3) = 4 kg/m^3

Im new to this propagation of errors stuff. Using sig figs other people in my class got either 4000 +or- 400(or +or- 100) I can't remember.
 
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I tried to find the uncertainty for volume again. I am probably calculating the error wrong.

(delta v/196.35) = 2pi(.1/2.5)+(.1/10)

delta v = 51 cm^3

(delta p)/(3.9g/cm^3) = .2/772.2 + 51/196.35 = 1.014 g/cm^3


1.014 g/cm^3 ((1kg/m^3)/(10^-3 g/cm^3) = 1014 kg/m^3

this seems way to high. I was given an online manuel on how to calculate propagation of error but my calculations always seem off.

To Dr Transport I do not believe have never seen of been taught that method.
 
shade585 said:
I tried to find the uncertainty for volume again. I am probably calculating the error wrong.

(delta v/196.35) = 2pi(.1/2.5)+(.1/10)

I think this is the mistake:

looking at the quantity without pi, call it x = r^2*h:

(delta x/62.25) = 2(.1/2.5)+(.1/10) = 0.09

delta x = 0.09*62.5 = 5.625

So the error in pi*r^2*h is 3.14*deltax = 17.67

so delta v = 17.67