calculate the work required to pump water into the tank shown below through a hole at the base of the tank…

Explanation:

Step1: Define coordinate system and thin water layer

Let ground level be ( y = 0 ). A horizontal water layer at height ( y ) (2 ≤ ( y ) ≤ 8 ft, since tank base is at 2 ft and height is 6 ft) has thickness ( dy ), radius 10 ft. Volume ( dV = \pi(10)^2 dy = 100\pi dy , \text{ft}^3 ).

Step2: Calculate weight of thin layer

Weight ( dF = \text{density} \times dV = 62.4 \times 100\pi dy = 6240\pi dy , \text{lb} ).

Step3: Determine lifting distance for thin layer

Water is pumped from ground (( y = 0 )) to height ( y ), so lifting distance is ( y , \text{ft} ).

Step4: Set up work integral for thin layer

Work for layer: ( dW = dF \times \text{distance} = 6240\pi y dy , \text{ft-lb} ). Total work: ( W = \int_{2}^{8} 6240\pi y , dy ).

Step5: Evaluate the integral

[ W = 6240\pi \int_{2}^{8} y , dy = 6240\pi \left[ \frac{y^2}{2} \right]_{2}^{8} = 3120\pi \left( 8^2 - 2^2 \right) = 3120\pi (64 - 4) = 3120\pi \times 60 = 187200\pi ]

Step6: Compute numerical value and round

( 187200\pi \approx 187200 \times 3.1416 \approx 587174.4 ), rounded to nearest whole number.

Answer:

587174