Work to pump water from a circular pool 3.5 m deep

[the7joker7]

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

 

[kuruman]

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