Parabolic Water Tank Homework: Find Work W Done to Lower Water

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SUMMARY

The discussion focuses on calculating the work done to lower water from a parabolic water tank defined by the equation y = x², with dimensions of 4 feet deep, 4 feet across, and 6 feet long. The task involves setting up a definite integral to find the work W required to pump water to a height of 4 feet above the tank's top, specifically lowering it to a depth of 3 feet. The formula W = 62.5(l - y)(A(y)) is established, where A(y) represents the area of the rectangular cross-section at height y, with the integral needing to be evaluated with respect to y.

PREREQUISITES
  • Understanding of definite integrals in calculus
  • Familiarity with the concept of work in physics
  • Knowledge of the area of geometric shapes, specifically rectangles
  • Basic understanding of parabolic equations and their graphs
NEXT STEPS
  • Study the application of definite integrals in physics problems
  • Learn how to calculate the area of cross-sections for different shapes
  • Explore the concept of work done against gravity in fluid mechanics
  • Review the properties of parabolas and their applications in real-world scenarios
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Students in calculus or physics courses, particularly those tackling problems involving work and fluid dynamics, as well as educators looking for practical examples of integrating geometric shapes in real-world applications.

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Homework Statement



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.

Homework Equations



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

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.
 
Physics news on Phys.org
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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