# 1.12: Squared and Cubic Units

- Page ID
- 213139

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\id}{\mathrm{id}}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\kernel}{\mathrm{null}\,}\)

\( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\)

\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\)

\( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)

\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)

\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vectorC}[1]{\textbf{#1}} \)

\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

Units for areas and volumes often include exponents, due to the calculations required to determine them. For example, consider the following area calculation.

\( {\text {Area}} = {\text {length}} \times {\text {width}}\)

If an object has a length of 17.2 inches and a width of 6.9 inches, its area would be calculated by applying this formula, as follows.

\( {\text {Area}} = {\text {17.2 in}} \times {\text {6.9 in}}\)

\( {\text {Area}} = {\text {120 in}^{2}}\)

Since no unit cancelation occurs in the calculation, the units are multiplied, just as the numbers are. Multiplying a unit by itself (\( {\text {in}} \times {\text {in}}\)) yields a final unit of \(\text{in}^2\), which is read as "square inches" or "inches squared." However, what if this area needed to be expressed in square feet? Much like in the previous sections, the six-step dimensional analysis process must be applied to execute a unit conversion, and both Step 3 *and* Step 4 will need to be considered more thoughtfully, in order to achieve the desired unit cancelation.

- Determine the "given" unit in the problem. The given number is 120, and its unit is square inches (\(\text{in}^2\)).

- Determine the "desired" unit in the problem. The given quantity should be changed
*to*square feet (\(\text{ft}^2\)), making "square feet" the desired unit.

- Determine which equality or equalities relate the given and desired units. Section 1.6 does not reference an equality that directly relates square inches to square feet. However, a measured equality that relates inches and feet
*does*exist; specifically, \( { \text{1 ft}} = { \text{12 in}}\).

- Use the appropriate conversion factor derived from this equality to achieve unit cancelation. Using the methodology developed in the previous sections, the conversion factor should be set up such that the unit "in" is written in the denominator.
\( { \text{120 in}^{2}} \times \dfrac{ \text{1 ft}}{\text{12 in}} \)

However, remember that the unit \(\text{in}^2\) really means \( {\text {in}} \times {\text {in}}\). If the area unit is written out to show its true meaning, it can be seen that only partial unit cancelation occurs when a single conversion factor is applied.\( { \text{120 in} \; \times \cancel{\rm{in}}} \times \dfrac{1 \; \rm{ft}}{12 \; \cancel{\rm{in}}}\)

Therefore, in order to cancel out the second "in" within the initial \(\text{in}^2\) unit, a second conversion factor must be applied. This second conversion factor must be set up identically to the first, such that the unit "in" is again written in the denominator.\( { \text{120} \; \cancel{\rm{in}} \times \cancel{\rm{in}}} \times \dfrac{1 \; \rm{ft}}{12 \; \cancel{\rm{in}}} \times \dfrac{1 \; \rm{ft}}{12 \; \cancel{\rm{in}}}\)

Note that the conversion factor is effectively multiplied by itself to achieve the required unit cancelation. Therefore, the conversion factor could itself be raised to a power, as shown below.\( { \text{120} \; \cancel{\rm{in^{2}}}} \times \left( \dfrac{1 \; \rm{ft}}{12 \; \cancel{\rm{in}}}\right)^{2}\)

While the last two notations are equivalent, correctly using the second representation requires a solid understanding of its meaning. Note that the power must be written*outside*of the parentheses, as the*entire conversion factor*is being multiplied by itself. Therefore, when performing the correct calculation, the*numbers and the units*must both be squared. Explicitly writing out multiple conversion factors is recommended, as doing so usually results in fewer student errors, relative to using the second (condensed) representation.

- Perform the calculation that remains once the units have been canceled. When using a calculator, the conversion factors should be entered in parentheses
the "=" key should be used after*or*division. In this case,*each*\( {120} \times {\text {(1 ft ÷ 12)}} \times {\text {(1 ft ÷ 12)}} = {\text {0.83333333... in}} \; \times {\text {in}} = {\text {0.83333333... in}^{2}}\)

- Apply the correct number of significant figures to the calculated quantity. Since the conversions used are based on prefix modifier equalities, neither the "1" nor the "12" are considered when determining the number of significant figures that the answer should have. The given number has 2 significant figures, so the final answer must have 2 significant figures. Since the number of digits that the answer actually has is more than the number of digits that it should have, the calculated answer must be rounded to 0.83 in
^{2}.

Using only conversion factors based on the equalities from Section 1.6, convert 18,900,000 cubic centimeters to cubic meters.

**Solution**

The given unit is "cubic centimeters (cm^{3})," and the desired unit is "cubic meters (m^{3})." Section 1.6 does not reference an equality that directly relates cubic centimeters to cubic meters. However, a prefix modifier equality that relates centimeters and meters does exist: \( { \text{1 m}} = { \text{100 cm}}\). However, in order to achieve complete unit cancelation, this conversion factor must be applied *three* times, as \(\text{cm}^3\) really means \( {\text {cm}} \times {\text {cm}} \times {\text {cm}}\). Therefore,

\( { \text{18,900,000} \; \cancel{\rm{cm}} \times \cancel{\rm{cm}} \times \cancel{\rm{cm}}} \times \dfrac{1 \; \rm{m}}{100 \; \cancel{\rm{cm}}} \times \dfrac{1 \; \rm{m}}{100 \; \cancel{\rm{cm}}} \times \dfrac{1 \; \rm{m}}{100 \; \cancel{\rm{cm}}} = {\text {18.9 m}^{3}}\)

Each of the numbers found within a prefix modifier equality is exact. The given number has 3 significant figures, so the final answer must have 3 significant figures. Since the number of digits that the answer actually has is exactly equal to the number of digits that it should have, no adjustments need to be made, and the calculated answer is the final answer.

Using only conversion factors based on the equalities from Section 1.6, convert 0.196 square miles to square yards.

**Answer**- The given unit is "square miles (mi
^{2})," and the desired unit is "square yards (yd^{2})." Section 1.6 does not reference an equality that directly relates square miles to square yards, nor does it include an equality that directly relates miles to yards. Therefore, multiple equalities will be necessary to achieve the required unit change. Specifically, the measured equalities \( { \text{1 mi}} = { \text{5,280 ft}}\) and \( { \text{1 yd}} = { \text{3 ft}}\) will*each*need to be applied*twice*in order to achieve complete unit cancelation. Therefore,\( { \text{0.196} \; \cancel{\rm{mi}} \times \cancel{\rm{mi}}} \times \dfrac{5,280 \; \bcancel{\rm{ft}}}{1 \; \cancel{\rm{mi}}} \times \dfrac{5,280 \; \bcancel{\rm{ft}}}{1 \; \cancel{\rm{mi}}} \times \dfrac{1 \; \rm{yd}}{3\; \bcancel{\rm{ft}}} \times \dfrac{1 \; \rm{yd}}{3\; \bcancel{\rm{ft}}} = {\text {607,129.6 yd}^{2}}\)

Each of the numbers found within a measured equality is exact. The given number has 3 significant figures, so the final answer must have 3 significant figures. Since the number of digits that the answer actually has is more than the number of digits that it should have, the calculated answer must be rounded to 607,000 yd^{2}.