Calculating Work to Empty a Water-Filled Trough

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SUMMARY

The discussion focuses on calculating the work required to empty a water-filled trough with a vertical cross-section shaped like the graph of x^6, spanning from x=-1 to x=1. The trough measures 2 feet in length and 1 foot in height, with water density at 62 pounds per cubic foot. To find the work, participants emphasize the need to determine the volume of the cross-section, which is essential for integrating the weight over the distance the water must be pumped. The area of the cross-section is identified as a rectangle, with the length being 2 feet and the width derived from the equation x^6=y.

PREREQUISITES
  • Understanding of calculus, specifically integration techniques.
  • Familiarity with the concept of work in physics, defined as weight times distance.
  • Knowledge of volume calculation for irregular shapes, particularly using cross-sectional areas.
  • Basic understanding of the properties of water, including density.
NEXT STEPS
  • Calculate the volume of the cross-section using the integral of the area from 0 to 1 with respect to y.
  • Explore the application of the Fundamental Theorem of Calculus to solve for work in similar problems.
  • Research methods for calculating the area of irregular shapes, particularly using functions like x^6.
  • Study the principles of fluid mechanics related to work and energy in pumping systems.
USEFUL FOR

Students and professionals in physics, engineering, and mathematics who are involved in fluid dynamics, particularly those interested in calculating work done in pumping fluids from containers.

chimbooze
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A trough is 2 feet long and 1 foot high. The vertical cross-section of the trough parallel to an end is shaped like the graph of x^6 from x=-1 to x=1. The trough is full of water. Find the amount of work in foot-pounds required to empty the trough by pumping the water over the top. Note: The density of water is 62 pounds per cubic foot.

Work = weight x distance traveled.

Weight = volume x density
Weight =

distance traveled = 1 - y

I'm just having issue with coming up with the volume for the cross section of the trough. If I know that, I can take the integral from 0 to 1 with respect to y.
 
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The volume is the area of the cross section*dy. The area is a rectangle. The length is 2 and the width is the distance from -x to x, where x^6=y. Sound right? What's the area?
 

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