Find the Volume of water in pool using double integrals

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The discussion focuses on finding the volume of water in a swimming pool using double integrals based on the given depth function and side equations. The depth of water is defined by f(x,y) = 2sin(x/20 - 7) - 3cos((x-3)/5) + 8, while the sides of the pool are represented by parabolic equations. Participants suggest that a double integral may not be necessary, as the volume can be calculated by integrating the product of the lengths of the rectangles formed by the side equations with respect to x. For those who prefer using a double integral, it is recommended to integrate dy with the parabolic functions as limits to find the rectangle's length. The conversation emphasizes understanding the setup of the integral for accurate volume calculation.
mridgwa
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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.
 
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