Problem 1-25: Solving for the uncertainty of the volume of a box with dimensional uncertainties

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PROBLEM:

The sides of a small rectangular box are measured to be 1.80±0.01 cm, 2.05±0.02 cm, and 3.1±0.1 cm long. Calculate its volume and uncertainty in cubic centimeters.


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SOLUTION:

The average volume of the box is

\begin{align*}
\text{Volume} & =l\times w\times h \\
& =\left(1.80\:\text{cm}\right)\left(2.05\:\text{cm}\right)\left(3.1\:\text{cm}\right) \\
&= 11.4\:\text{cm}^3 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

The percent uncertainty for each of the dimensions:

\begin{align*}
1.80\pm 0.01\:\text{cm}\:\rightarrow & \:\frac{0.01\:\text{cm}}{1.80\:\text{cm}}\times 100\%=0.556\% \\
\\
2.05\pm 0.02\:\text{cm}\:\rightarrow & \:\frac{0.02\:\text{cm}}{2.05\:\text{cm}}\times 100\%=0.976\% \\
\\
3.1\pm 0.1\:\text{cm}\:\rightarrow &\:\frac{0.1\:\text{cm}}{3.1\:\text{cm}}\times 100\%=3.226\% \\
\end{align*}

The percent uncertainty in the volume of the box is calculated by adding the percent uncertainties of the dimensions.

\begin{align*}
\%\:\text{uncertainty}_{\text{volume}} & =0.556\%+0.976\%+3.226\% \\
& =4.758\%
\end{align*}

The uncertainty of the volume is

\begin{align*}
\delta _{\text{volume}} & =0.04758\times 11.4\:\text{cm}^3 \\
& =0.54\:\text{cm}^3 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Therefore, the volume is

\text{Volume}=11.4\pm 0.54\:\text{cm}^3 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

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