An oil tank in the shape of a right circular cylinder, with height \(30\textrm{ m}\) and radius \(5\textrm{ m}\), is two-thirds full of oil. How much work is required to pump all the oil over the top of the tank?
The work required to pump the oil over the top is the product of the weight of the oil and its distance from the top of the tank. The weight of the oil equals \(\rho gV\) newtons, where \(\rho \approx 820\) \(\textrm{ kg} \textrm{/m}^{3}\) is the mass density of petroleum (mass per unit volume, a constant that depends on the type of liquid involved), \(g\approx 9.8\textrm{ m} \textrm{/s} ^{2}\) (the acceleration due to gravity), and \(V\) is the volume of the liquid to be moved. The weight of the oil is \[ \hbox{Weight }=\rho gV\approx \hbox{ }( 820) ( 9.8) V=8036V\,\textrm{N} \]
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The oil fills the tank from \(x=0\textrm{ m}\) to \(x=20\textrm{ m}\). We partition the interval \([ 0,20]\) into \(n\) subintervals, each of width \(\Delta x=\dfrac{20}{n}\). Consider the oil in the \(i\)th subinterval as a thin layer of thickness \(\Delta x\). Now choose a number \(u_{i}\) in the \(i {th}\) subinterval. Then \begin{align*} \hbox{Volume }V_{i}\hbox{ of }i{\hbox{th}}\hbox{ layer} & =(\hbox{Area of layer})(\hbox{Thickness})=\pi r^{2}\Delta x=25\pi \Delta x \\ \hbox{Weight of }i{\hbox{th}}\hbox{ layer} & =\rho gV_{i}\approx (8036) (25\pi \Delta x)=200{,}900\pi \Delta x \\ \hbox{Distance }i{\hbox{th}}\hbox{ layer is lifted} & =30-u_{i} \\ \hbox{Work }W_{i}\hbox{done in lifting }i{\hbox{th}}\hbox{ layer} & \approx ( 200{,}900\pi \Delta x) ( 30-u_{i}) \end{align*}
\[ \begin{array}{rcl} \hbox{Weight} &=& \hbox{(Mass density)(Acceleration}\\ &&\hbox{due to gravity)(Volume)} = \rho gV \end{array} \]
Mass density \(\rho\) is often shortened to density when using SI units, since kilograms, a measure of mass, is the basic unit. When using customary U.S. units, however, we use weight density since a pound is a measure of weight. (In U.S. units, the unit of mass is called a slug. \(1 \text{slug}=32.2\textrm{ lb}\).)
The layers of oil are from 0 to \(20\textrm{ m}\). So, the work \(W\) required to pump all the oil over the top is \begin{eqnarray*} W &=&\displaystyle \int_{0}^{20}200{,}900\pi ( 30-x) \,{\it dx}=200{,}900\pi \displaystyle \int_{0}^{20}( 30-x) \,{\it dx}\\[3pt] &=&( 200{,}900~\pi )\! \left[30x-\frac{x^{2}}{2} \right] _{0}^{20} \notag \\[3pt] &=&( 200{,}900~\pi ) (600-200)=80{,}360{,}000\pi \approx 2.525\times 10^{8} \textrm{ J}\ \quad \end{eqnarray*}