A cylindrical water tank has a radius of 2 feet and a height of 6.0 feet. Compute the work done to pump the water out of a filled tank through the top. [The density of water is 62.4 lbs/ft³.] A) 14,115 ft-lbs B) 2352 ft-lbs C) 1123 ft-lbs D) 7390 ft-lbs

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The work done to pump water out of the tank can be calculated using the formula for work, $ W = \rho g h A d $, where $ \rho $ is the density of water, $ g $ is the acceleration due to gravity (which is often included in the density value for these types of problems, making $ \rho g = 62.4 $ lbs/ft$^3$ as given), $ h $ is the height at which the water is being lifted, $ A $ is the cross-sectional area of the water being lifted, and $ d $ is the depth of the water. Given a cylindrical tank, the volume of water $ V $ is $ \pi r^2 h $, where $ r $ is the radius and $ h $ is the height. The cross-sectional area $ A $ of the cylinder is $ \pi r^2 $. To find the work done to pump all the water out, we integrate from the bottom of the tank to the top, considering that the distance the water needs to be lifted changes as we move up the tank. The work done to lift a thin slice of water at a height $ y $ from the bottom to the top of the tank is $ dW = \rho g A dy $, where $ dy $ is the thickness of the slice and $ A $ is the cross-sectional area of the water. The total work is the integral of $ dW $ from $ 0 $ to $ h $, the height of the tank.Given $ \rho g = 62.4 $ lbs/ft$^3$, $ r = 2 $ feet, and $ h = 6.0 $ feet, the cross-sectional area $ A = \pi r^2 = \pi (2)^2 = 4\pi $ ft$^2$. The work done to pump the water out of the tank is:$$ W = \int_0^h \rho g A (h - y) dy $$$$ W = 62.4 \times 4\pi \int_0^6 (6 - y) dy $$$$ W = 249.6\pi \left[ 6y - \frac{y^2}{2} \right]_0^6 $$$$ W = 249.6\pi \left[ 36 - \frac{36}{2} \right] $$$$ W = 249.6\pi \times 18 $$$$ W = 4492.8\pi $$$$ W \approx 14,115 $$ ft-lbs.Therefore, the correct answer is 14,115 ft-lbs.

A Cylindrical Water Tank Has a Radius of 2 Feet