# Parabolic Water Tank

1. Feb 15, 2012

### amw2829

1. The problem statement, all variables and given/known data

The ends of a "parabolic" water tank are the shape of the region inside the graph of
y = x2
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. The tank is 6 feet long. Rain has filled the tank and water is removed by pumping it up to a spout that is 4 feet above the top of the tank. Set up a definite integral to find the work W that is done to lower the water to a depth of 3 feet and then find the work. [Hint: You will need to integrate with respect to y.

2. Relevant equations

W=62.5(l-x)(A(x)) dx

3. The attempt at a solution

Since you need to integrate with respect to y, I would assume the integral would be: 62.5(6-y)(A(y)). However, I'm not sure what A(y) would be or how I would identify the bounds.

2. Feb 16, 2012

### HallsofIvy

A(y) is the area of the rectangular cross section of the tank at height y. One side of that rectangle has length 6. The other side has length 2x where $y= x^2$.

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