Significant figures and error rounding for volume calculation

[Ian Mount]

Hello,

I have a question asking me to find the volume of a rectangular prism. The dimensions are as follows:

x = 20 ± 0.2 cm, y = 30 ± 0.2 cm, z = 70 ± 0.4 cm

I am asked to report the answer with the correct number of significant figures and include the error.

What I have so far:

V = xyz = 42000 ± 558.93 cm³

So I'm stuck on determining how many significant figures the answer should have and how to round the error.

I'm leaning towards 3 significant figures because 20 ± 0.2 is anything between 19.8 and 20.2, which have 3 significant figures. The same can be said for the other given dimensions. Therefore the volume would be V = 4.20*10⁴ cm³.

Also, I'm not sure how to round errors. Do I round to the same number of significant figures or do I round to the same precision as the answer? If it's the same precision, assuming three significant figures, the answer would be 4.20*10⁴ ± 6*10² cm³. This feels correct, if you know what I mean.

Any help would be greatly appreciated. Thanks!

 

[mfb]

I moved the thread to our homework section.

With a proper uncertainty estimate the concept of significant figures is not that relevant any more. Typically two digits of the uncertainty are reported, but as your input uncertainties have only one significant figure rounding to one digit of the uncertainty should be fine as well.

 

[Ian Mount]

I moved the thread to our homework section.

With a proper uncertainty estimate the concept of significant figures is not that relevant any more. Typically two digits of the uncertainty are reported, but as your input uncertainties have only one significant figure rounding to one digit of the uncertainty should be fine as well.

So would the answer, 42000 cm³, be reported with only one significant figure as well? Leading to a final answer of 40000 ± 600 cm³? What you're saying makes sense to me it just feels so wrong, as 20*30*70 is clearly 42000, or were you only referring to the uncertainties, with regards to the number of significant figures?

 

[mfb]

No, the central value always keeps the same absolute precision as the uncertainty: (420 +- 6) * 10^2

 

[Ian Mount]

No, the central value always keeps the same absolute precision as the uncertainty: (420 +- 6) * 10^2

Oooh okay, that makes sense. Thank you so much! I was having a ridiculously hard time finding a definitive answer on google.

 

[HallsofIvy]

There is an engineer's "rule of thumb" that "when quantities add, their errors add, when quantities multiply their relative errors add".