What is the mass of a column of air above a one-meter square table top? Imagine a one meter square vertical column, $C,$ of air starting at sea level and extending upwards to 40,000 meters. Shown in Figure 9.1 .12 are 'standard' values of atmospheric density as a function of altitude up to 40,000 meters (U.S. Standard Atmospheres $1976,$ National Oceanic and Atmospheric Administration, NASA, U.S. Air Force, Washington, D.C. October 1976 ). The density is not constant throughout $C,$ and the mass of $C$ may be approximated by partitioning $C$ into regions of 'small' volume, assuming that the density is constant within each region, and approximating the mass within each of the regions. Then the object will approximately have $$\text { Mass } \quad=\delta_{1} \times v_{1}+\delta_{2} \times v_{2}+\cdots+\delta_{n} \times v_{n}$$
where $\delta_{k} \mathrm{Kg} / \mathrm{m}^{3}$ and $v_{k} \mathrm{~m}^{3}$ are respectively the density and volume in the $k \underline{\text { th }}$ region.
a. Partition the altitude interval from 0 to 40,000 meters into at least 6 subintervals (they need not be all the same length). For each interval, find the volume of the region of $C$ corresponding to the interval, find a density of the air within that region, and approximate the mass of the region. Approximate the mass in each region and add the results to approximate the mass of air in $C$.
b. Does the size of your answer surprise you? Show that it is consistent with the standard pressure of one atmosphere being $760 \mathrm{~mm}$ of mercury? The density of mercury is $1.36 \times 10^{4} \mathrm{Kg} / \mathrm{m}^{3}$.