Question

How does one find the percent uncertainty in the volume of a sphere whose radius is r = 1.56 +/- .05?

Answer 1

Doesn't seem to be working. The actual values are r=2.86m and the uncertainty is +/- 0.08m. The answer is supposed to be 9%. I cannot seem to get there.

Answer 2

Shouldn't there be 3 volumes, the difference V 1.56 the original, and then the V 1.61- V 1.51?

Answer 3

delta V/ V original times 100?

Answer 4

1/2 delta V/ V1.56. Half delta V because of the +, _

Answer 5

I don't get 9% either but these are based on your second set of numbers:

Radius of a sphere is:
\[\frac{4}{3}\pi r^2\]
Low end volume:
\[\frac{4}{3}\pi (2.86-0.08)^2\approx32.37\]
Normal volume:
\[\frac{4}{3}\pi (2.86)^2\approx34.26\]
To find the percentage difference:
\[\frac{32.37-34.26}{34.26}100 \% =-5.5\%\]

The upper end will be the same so the percent of uncertainy for a sphere of radius 2.86m with an uncertainty of +/- 0.08m is:
\[\pm 5.5\%\]

Answer 6

Hmm, boobooed. Radius of sphere is r^3, so Low end is 89.996. Normal is 97.991. Final uncertainty is +/- 8.2%.

Answer 7

For the numbers on top plugging them in I get +/- 9.3% which I believe is what you are looking for:
Low end volume:
\[\frac{4}{3}\pi (1.56-0.5)^3\approx14.42\]
Normal volume:
\[\frac{4}{3}\pi (1.56)^3\approx15.90\]

Findinding percentage difference:
\[\frac{14.42-15.90}{15.90}100\%=-9.3\%\]
So the result is \[\pm9.3\%\]