Optimizing Work in a Parabolic Water Tank: Finding the Area of Cross-Sections

[7yler]

The ends of a "parabolic" water tank are the shape of the region inside the graph of y = x^{2} 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 \int_2^4 62.5(6-y)(4-y^{2})dy , 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?

 

[tiny-tim]

hi 7yle!

your (6 - y) is correct, but I have no idea what your (4 - y²) is supposed to be

your slice is 5 ft long, dy high, and … wide?