Find the Volume of water in pool using double integrals

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SUMMARY

The volume of water in a swimming pool can be calculated using a double integral based on the depth function f(x,y) = 2sin(x/20 - 7) - 3cos((x-3)/5) + 8 and the side equations y(x) = 10 - (x-10)²/10 and y(6) = (x-10)²/20 - 5. To set up the integral, graph the parabolas representing the sides of the pool and determine the length of a rectangle perpendicular to the x-axis, which is the difference between the two y-values. The volume can be computed by integrating the product of the rectangle's length and depth function over the defined limits.

PREREQUISITES
  • Understanding of double integrals in calculus
  • Familiarity with trigonometric functions and their properties
  • Knowledge of parabolic equations and their graphical representations
  • Ability to perform definite integrals
NEXT STEPS
  • Learn how to graph parabolic functions and their intersections
  • Study the application of double integrals in calculating volumes
  • Explore the use of trigonometric identities in integrals
  • Practice setting up and solving definite integrals with variable limits
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Students in calculus courses, educators teaching integral calculus, and anyone interested in applying mathematical concepts to real-world problems involving volume calculations.

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Homework Statement



The depth of water in a swimming pool fits the equation f(x,y) = 2sin (x/20 - 7) - 3 cos ( x-3 /5)+8 when 0<=x<=20 and the sides of the pool fir the equations y(x) = 10-(x-10)^2/10 and y(6)= (x-10)^2/20 -5

Find the volume of the water in the pool using a double integral

I am not sure how to set up this integral. Any help in setting it up will be greatly appreciated.
 
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Start by graphing the two sides- they will be parabolas, of course. Imagine a thin rectangle perpendicular to the x-axis. The length of that rectangle will be the difference of the two y values for the side-that will, of course,depend upon x. The depth is also a function of x only so you don't really need a double integral. Just integrate the product of those two lengths (the area of the rectangle) with respect to x.

If you really do want to use a double integral, get that "length" of the rectangle by integrating \int dy with the the parabolic functions as limits.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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