Calculus Question about my work (I already did it, just want a double check it)

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The discussion revolves around calculating the work required to pump water from a circular swimming pool with a diameter of 14 m, a height of 4 m, and a water depth of 3.5 m. The formula used for the calculation is W = 9800(integralof)(153.94x)dx from 0.5 to 4, where 9800 represents the product of water density (1000 kg/m³) and gravitational acceleration (9.8 m/s²). The final calculated work is approximately 1.2×107 Joules, with an emphasis on using appropriate significant figures in future calculations.

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A circular swimming pool has a diameter of 14 m, the sides are 4 m high, and the depth of the water is 3.5 m. How much work (in Joules) is required to pump all of the water over the side? (The acceleration due to gravity is 9.8 ` m/s^2 ` and the density of water is 1000 ` "kg"/"m"^3 `.)

W = 9800(integralof)(153.94x)dx from 0.5 to 4

9800 = 1000*9.8 for density*gravity

153.94 = pi(7^2) to give the area of each slice of water

x notates how deep the water is in the pool at that particular point.

I get

11880168.24 joules
 
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I got the same answer. For whatever it's worth, use a reasonable number of significant figures in the future. In this case W = 1.2×107 J would have been sufficient. For completeness, I show the intermediate steps leading to the answer.
$$dW = dm~g~y=(\rho A dy)~g~y~\implies~W=\rho g \frac{\pi D^2}{4}\int_{y_i}^{y_f} y~dy=\rho g \frac{\pi D^2}{8}\left(y_{\!f}^2-y_{\!i}^2\right)$$where ##y_{i}=##0.5 m and ##y_{f}=##4.0 m
 
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