It is helpful to think of the depth below the surface of the crock as being the independent variable, so we let the positive \(x\)-axis point down, and the positive \(y\)-axis to the right, as pictured in the figure. Because the pump sits on the surface of the water, it makes sense to think about the pump moving the water one “slice” at a time, where it takes a thin slice from the surface, pumps it out of the tank, and then proceeds to pump the next slice below.

Each slice of water is cylindrical in shape. We see that the radius of each slice varies according to the linear function \(y = f(x)\) that passes through the points \((0,1.5)\) and \((4,0.75)\text{,}\) where \(x\) is the depth of the particular slice in the tank; it is a straightforward exercise to find that \(f(x) = 1.5 - 0.1875x\text{.}\) Now we think about the problem in several steps:

determining the volume of a typical slice;

finding the weight of a typical slice (and thus the force that must be exerted on it);

deciding the distance that a typical slice moves;

and computing the work to move a representative slice.

Once we know the work it takes to move one slice, we use a definite integral over an appropriate interval to find the total work.

Consider a representative cylindrical slice at a depth of \(x\) feet below the top of the crock. The approximate volume of that slice is given by

\begin{equation*}
V_{\text{slice} } = \pi f(x)^2 \Delta x = \pi (1.5 - 0.1875x)^2 \Delta x\text{.}
\end{equation*}

Since water weighs 62.4 lb/ft\(^3\text{,}\) the approximate weight of a representative slice is

\begin{equation*}
F_{\text{slice} } = 62.4 \cdot V_{\text{slice} } = 62.4 \pi (1.5 - 0.1875x)^2 \Delta x\text{.}
\end{equation*}

This is also the approximate force the pump must exert to move the slice.

Because the slice is located at a depth of \(x\) feet below the top of the crock, the slice being moved by the pump must move \(x\) feet to get to the level of the basement floor, and then, as stated in the problem description, another 9 feet to reach the drain at ground level. Hence, the total distance a representative slice travels is

\begin{equation*}
d_{\text{slice} } = x + 9\text{.}
\end{equation*}

Finally, the work to move a representative slice is given by

\begin{equation*}
W_{\text{slice} } = F_{\text{slice} } \cdot d_{\text{slice} } = 62.4 \pi (1.5 - 0.1875x)^2 \Delta x \cdot (x+9)\text{.}
\end{equation*}

We sum the work required to move slices throughout the tank (from \(x = 0\) to \(x = 4\)), let \(\Delta x \to 0\text{,}\) and hence

\begin{equation*}
W = \int_0^4 62.4 \pi (1.5 - 0.1875x)^2 (x+9) \, dx\text{.}
\end{equation*}

When evaluated using appropriate technology, the integral shows that the total work is \(W = 3463.2 \pi\) foot-pounds.