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The ends of a "parabolic" water tank are the shape of the region inside the graph of y = [tex]x^{2}[/tex] for 0 ≤ y ≤ 4; the cross sections parallel to the top of the tank (and the ground) are rectangles. At its center the tank is 4 feet deep and 4 feet across. It is 5 feet long. Rain has filled the tank and water is removed by pumping it up to a pipe 2 feet above the top of the tank. Set up an iterated integral to find the work W that is done to lower the water to a depth of 2 feet and then find the work. [Hint: You will need to integrate with respect to y.]

So I came up with [tex]\int_2^4 62.5(6-y)(4-y^{2})dy[/tex] , and I'm not sure why this integral isn't correct. The water is being pumped to 6 feet, so it's 6-y, right?

So I came up with [tex]\int_2^4 62.5(6-y)(4-y^{2})dy[/tex] , and I'm not sure why this integral isn't correct. The water is being pumped to 6 feet, so it's 6-y, right?

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