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(a) First, we construct a picture of the tank as described. Below we see a sketch of $f(x) = 3\cos \left( \frac{x^3}{4} \right)$ in the setting where the positive $x$-axis points down; note that this function first crosses the $x$-axis at about $1.84527$. We reflect the curve from the first quadrant across the positive $x$-axis to generate a solid of revolution. Cross-sections perpendicular to the $x$-axis have thickness $\triangle x$ and are circular disks. We now suppose that the resulting tank is filled to a depth of $1.5$ feet with water weighing $62.4$ pounds per cubic foot. Since the bottom of the tank is at $1.84527$, being full to a depth of $1.5$ feet means that there is water from $x = 0.34527$ to $x = 1.84527$. To find the work done to pump the water until $0.5$ feet remain in the tank (i.e. to a level where $x = 1.34527$), we have to first find the weight of a typical slice, which requires us to know the volume of such a slice. Observe that for a slice at depth $x$, $$V_{\text{slice}} \approx \pi \left( 3\cos \left( \frac{x^3}{4} \right) \right)^2 \triangle x\,.$$ Since water weighs $62.4$ pounds per cubic foot, the weight (and thus force to be exerted) of such a slice is $$F_{\text{slice}} \approx 62.4 \cdot 9 \cdot \pi \cos^2 \left( \frac{x^3}{4} \right) \triangle x\,.$$ Finally, the work done to pump such a slice to the top of the tank requires the slice to move $x$ units, and thus work for that slice is $$W_{\text{slice}} \approx x \cdot 561.6 \pi \cos^2 \left( \frac{x^3}{4} \right) \triangle x\,.$$ To find the total work accomplished, we use a definite integral to sum the work done on each slice from $x = 0.34527$ (the top of the water when the tank is initially full to $1.84527$ feet) to $x = 1.34527$ (the height of the water after $1$ foot has been emptied). Thus, $$W = \int_{0.34527}^{1.34527} 561.6 \pi \cdot x \cos^2 \left( \frac{x^3}{4} \right) \, dx\,.$$ Evaluating the integral, we find $W \approx 1353.55$ foot-pounds. (b) If instead we envision the tank not as a solid of revolution but as a trough, where the two-dimensional figure above forms one end of a tank that is 10 feet long, we now have a similar problem but with slices that are thin rectangular slabs, rather than thin cylindrical disks. Note that each such slice is $2 f(x)$ wide, $10$ feet long, and $\triangle x$ thick. Thus, for a slice at depth $x$, $$V_{\text{slice}} \approx 2 \left( 3\cos \left( \frac{x^3}{4} \right) \right) \cdot 10 \cdot \triangle x\,.$$ Since water weighs $62.4$ pounds per cubic foot, the weight (and thus force to be exerted) of such a slice is $$F_{\text{slice}} \approx 62.4 \cdot 60 \cos \left( \frac{x^3}{4} \right) \triangle x\,.$$ Finally, the work done to pump such a slice to the top of the tank requires the slice to move $x$ units, and thus work for that slice is $$W_{\text{slice}} \approx x \cdot 3744 \cos \left( \frac{x^3}{4} \right) \triangle x\,.$$ Assuming that the tank is completely full, to lower the tank by $h$ feet, we consider slices from $x = 0$ to $x = h$, and thus using a definite integral to sum the work required for each slice, we find that the total work is $$W = \int_0^h 3744 x \cos \left( \frac{x^3}{4} \right) \, dx\,.$$ (c) For the trough described in (b), we now seek to find the total hydrostatic force on one of the ends of the tank when it is completely full. First, we note that a typical slice at location $x$ will be at a depth of $x$ feet below the water and have area $$A_{\text{slice}} \approx 2 f(x) \triangle x = 2 \left( 3\cos \left( \frac{x^3}{4} \right) \right) \triangle x\,.$$ At any point on this slice, which again lies at depth $x$, the pressure exerted is $$P_{\text{slice}} \approx 62.4 x$$ since pressure is the weight density of water times depth. Finally, since $F = P \cdot A$, the total force water is exerting on this slice is $$F_{\text{slice}} = P_{\text{slice}} \cdot A_{\text{slice}} \approx 62.4 x \cdot 6 \cos \left( \frac{x^3}{4} \right) \triangle x\,.$$ Integrating to sum the total force, we find that $$F = \int_0^{1.84527} 374.4 x \cos \left( \frac{x^3}{4} \right) \, dx \approx 462.637$$ pounds.