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The inside dimensions of a box that is cubic is 12.7 cm on each edge with an uncertainty of 0.3 cm.

  1. What is the volume of the box?
  2. What do you estimate to be the uncertainty in the calculated volume?

The first part is simple enough. To obtain volume of the box: $(\pu{12.7 cm})^3 = \pu{2.05E3 cm^3}$

Now for the second part, "estimate the uncertainty in the calculated volume?" – I would imagine that because the uncertainty is $\rm\pm0.3\ cm$ on each side the uncertainty in the volume of the box would be $\rm\pm(0.3)^3 = \pm0.027\ cm^3$? The textbook gives $\rm\pm0.60\times10^3\ cm^3$. I am (excuse the pun) uncertain of how they have reached this answer.

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    $\begingroup$ I don't think that the textbook answer is correct. The largest possible volume would be 13*13*13= 2197cm, while the smallest possible volume would be: 12.4*12.4*12.4=1907cm. The difference between the largest and the smallest volume is 290cm. The uncertainty can't be bigger than the difference between the two extremes (it should be 145cm). $\endgroup$
    – EJC
    Sep 22 '15 at 17:10
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The volume $V$ can be calculated from the length $x$, width $y$, and height $z$.

$$\begin{align} V&=x\cdot y\cdot z\tag1\\[6pt] &=12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\times 12.7\ \mathrm{cm}\tag2\\[6pt] &=2048.383\ \mathrm{cm^3}\tag3 \end{align}$$

If we assume that the uncertainties of length $x$, width $y$, and height $z$ are not correlated (which is not necessarily true when the same ruler was used to measure length, width, and height), i.e. their correlation coefficient is $\rho=0$, the relative uncertainty of the volume may be estimated as

$$\begin{align} \left(\frac{u(V)}{V}\right)^2&=\left(\frac{u(x)}{x}\right)^2+\left(\frac{u(y)}{y}\right)^2+\left(\frac{u(z)}{z}\right)^2\tag4\\[6pt] \frac{u(V)}{V}&=\sqrt{\left(\frac{u(x)}{x}\right)^2+\left(\frac{u(y)}{y}\right)^2+\left(\frac{u(z)}{z}\right)^2}\tag5 \end{align}$$

Thus, the absolute uncertainty of the volume is

$$\begin{align} u(V)&=\sqrt{\left(\frac{u(x)}{x}\right)^2+\left(\frac{u(y)}{y}\right)^2+\left(\frac{u(z)}{z}\right)^2}\cdot V\tag6\\[6pt] &=\sqrt{\left(\frac{u(x)}{x}\right)^2+\left(\frac{u(y)}{y}\right)^2+\left(\frac{u(z)}{z}\right)^2}\cdot x\cdot y\cdot z\tag7\\[6pt] &=\sqrt{\left(\frac{0.3\ \mathrm{cm}}{12.7\ \mathrm{cm}}\right)^2+\left(\frac{0.3\ \mathrm{cm}}{12.7\ \mathrm{cm}}\right)^2+\left(\frac{0.3\ \mathrm{cm}}{12.7\ \mathrm{cm}}\right)^2}\times12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\tag8\\[6pt] &=83.809\ \mathrm{cm^3}\tag9 \end{align}$$

Note that the same result may be obtained using the general formula

$$\begin{align} u(V)&=\sqrt{\left(\frac{\partial V}{\partial x}\right)^2\cdot u^2(x)+\left(\frac{\partial V}{\partial y}\right)^2\cdot u^2(y)+\left(\frac{\partial V}{\partial z}\right)^2\cdot u^2(z)}\tag{10}\\[6pt] &=\sqrt{\left(y\cdot z\right)^2\cdot u^2(x)+\left(x\cdot z\right)^2\cdot u^2(y)+\left(x\cdot y\right)^2\cdot u^2(z)}\tag{11}\\[6pt] &=\sqrt{\left(12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\right)^2\times\left(0.3\ \mathrm{cm}\right)^2+\left(12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\right)^2\times\left(0.3\ \mathrm{cm}\right)^2\\+\left(12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\right)^2\times\left(0.3\ \mathrm{cm}\right)^2}\tag{12}\\[6pt] &= 83.809\ \mathrm{cm^3}\tag{13} \end{align}$$

However, if we have to assume that the uncertainties of length $x$, width $y$, and height $z$ are correlated, i.e. their correlation coefficient is $\rho=1$, then the equation for the estimate of the relative uncertainty of the volume simplifies to

$$\frac{u(V)}{V}=\frac{u(x)}{x}+\frac{u(y)}{y}+\frac{u(z)}{z}\tag{14}$$

and the absolute uncertainty of the volume is

$$\begin{align} u(V)&=\left(\frac{u(x)}{x}+\frac{u(y)}{y}+\frac{u(z)}{z}\right)\cdot V\tag{15}\\[6pt] &=\left(\frac{u(x)}{x}+\frac{u(y)}{y}+\frac{u(z)}{z}\right)\cdot x\cdot y\cdot z\tag{16}\\[6pt] &=\left(\frac{0.3\ \mathrm{cm}}{12.7\ \mathrm{cm}}+\frac{0.3\ \mathrm{cm}}{12.7\ \mathrm{cm}}+\frac{0.3\ \mathrm{cm}}{12.7\ \mathrm{cm}}\right)\times12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\tag{17}\\[6pt] &=145.161\ \mathrm{cm^3}\tag{18} \end{align}$$

Note that the same result may be obtained using the general formula

$$\begin{align} u(V)&=\frac{\partial V}{\partial x}\cdot u(x)+\frac{\partial V}{\partial y}\cdot u(y)+\frac{\partial V}{\partial z}\cdot u(z)\tag{19}\\[6pt] &=y\cdot z\cdot u(x)+x\cdot z\cdot u(y)+ x\cdot y\cdot u(z)\tag{20}\\[6pt] &=12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\times0.3\ \mathrm{cm}+12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\times0.3\ \mathrm{cm}\\ &\hphantom{={}}+12.7\ \mathrm{cm}\times12.7\ \mathrm{cm}\times0.3\ \mathrm{cm}\tag{21}\\[6pt] &=145.161\ \mathrm{cm^3}\tag{22} \end{align}$$

The numerical values of an estimate $y$ and its standard uncertainty $u(y)$ should not be given with an excessive number of digits. It usually suffices to quote $u(y)$ to at most two significant digits, although in some cases it may be necessary to retain additional digits to avoid round-off errors in subsequent calculations. Thus, the result for the volume $V$ according to $\text{(3)}$ and its standard uncertainty $u(V)$ according to $\text{(9)}$ or $\text{(13)}$ may be expressed as

$$V=\left(2048\pm84\right)\ \mathrm{cm^3}\tag{23}$$

or

$$V=2048(84)\ \mathrm{cm^3}\tag{24}$$

Using the uncertainty according to $\text{(18)}$ or $\text{(22)}$, the result may be expressed as

$$V=\left(2.05\pm0.15\right)\times10^3\ \mathrm{cm^3}\tag{25}$$

or

$$V=2.05(15)\times10^3\ \mathrm{cm^3}\tag{26}$$

(According to various international standards, the ± format should be avoided.)

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