The ends of a \"parabolic\" water tank are the shape of the region inside the gr

Question

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 5 feet long. Rain has filled the tank and water is removed by pumping it up to a spout that is 3 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.]

Explanation / Answer

Your bounds will be from 3 to 4.
You will integrate (water density)(cross-sectional area)(the distance) .
You have the bounds for integration,
water density is the constant rho = 62.5,
Cross-sectional area = 2bh or 2lw (area of a rectangle * 2) where length = 5ft and width= the given equation in terms of y,

Distance is the height of the water pumped out. (l - y) where l = 4-y (they didn't tell you where the water is raised to so we'll assume it's to the top.)

therefore,

Work=integral(3,4) of (62.5(10sqrt(y))(4-y)) dy

=625 integral (3,4) of (sqrt(y))(4-y) dy

=625 integral (3,4) of ((4 sqrt(y))-y^3/2 ) dy

=250/3(64-33sqrt(3))

=570.193

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