1. **State the problem:** Calculate the work required to pump water from the tank through a hole at the base of the tank, given the density of water as 62.4 lb/ft^3. 2. **Identify the geometry:** The tank is a cylinder with radius $r = 7$ ft and a water column height $h = 10$ ft. 3. **Setup the coordinate system:** Let $y$ be the vertical distance from the base of the tank upward, where $y=0$ at the ground (the hole) and $y=10$ ft at the water surface inside the tank. 4. **Calculate volume slice:** A thin horizontal slice of water at height $y$ has thickness $dy$. Volume is $$ dV = \pi r^2 dy = \pi (7)^2 dy = 49 \pi dy $$ 5. **Calculate weight of water slice:** Weight $dw = \text{density} \times dV = 62.4 \times 49 \pi dy = 3057.6 \pi dy$ lb 6. **Calculate the distance to pump:** Each slice of water at height $y$ must be lifted from $y$ ft above ground to the hole at ground level. The distance water is pumped downward is actually zero, but since the hole is at the base, the work needed equals the force times the distance water moves down to the hole, which is from height $y$ to 0: distance $= y$ ft. 7. **Set up integral for work:** Work $$ W = \int_0^{10} \text{force} \times \text{distance} = \int_0^{10} (3057.6 \pi) y \, dy $$ 8. **Evaluate integral:** $$ W = 3057.6 \pi \int_0^{10} y \, dy = 3057.6 \pi \left[ \frac{y^2}{2} \right]_0^{10} = 3057.6 \pi \times \frac{100}{2} = 3057.6 \pi \times 50 $$ $$ W = 152880 \pi $$ 9. **Calculate numerical value:** $$ W \approx 152880 \times 3.1416 = 479998.5 $$ 10. **Round to nearest whole number:** $$ W \approx 479999 $$ ft-lb **Final answer:** The work required to pump the water is approximately $479999$ ft-lb.