- #1

Coop

- 40

- 0

## Homework Statement

Hello,

I have solved the problem, but not the way the writer intended. I need help figuring out how they wanted me to do it.

Here's the problem:

The circumference of a sphere is found to be 0.98 m +/- 0.01 m. Calculate the volume and absolute error with five digits to the right of the decimal.

## Homework Equations

C = 2∏r

V = (4/3)∏r

^{3}

## The Attempt at a Solution

I used the low end (.97 m) and high end (.99 m) circumference measurement to solve for two radii. Then I plugged those in to the volume equation and got to volume values: .01541 m

^{3}and .01639 m

^{3}. Averaging those volumes I found the final volume to be .01589 m

^{3}.

To find the absolute uncertainty I merely subtracted the high end circumference value from the low end one and divided by two.

∴ volume = .01589 m

^{3}+/- .00049 m

^{3}.

But my question is, they wanted me to find absolute uncertainty using the following formulas:

Δ[constant]X = [constant]*ΔX to account for constants

(Δtotal/final value) = |n| * [(ΔX)/X] to find uncertainty values where n is an exponent on A

ΔX = [(ΔX)/X] * X to convert from relative to absolute error

...where Δ stands for uncertainty

How would I do this?

It seems a lot more confusing for no reason, but would there be any case when I wouldn't be able to use my method and would have to use these formulas?

P.S. Is there anyway to code when writing these? Like I know for some help sites you can input fractions using commands like \frac{num.}{denom.}