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anonymous
 5 years ago
a small dam is constructed across a stream. a vertical cross section of the stream is y=2x^2. The dam is 4 ft tall. set up an integral that estimates the hydrostatic force on the dam when the water is all the way at the top. Force=Pressure*Area. water weighs 62.5 lb/cubic foot
anonymous
 5 years ago
a small dam is constructed across a stream. a vertical cross section of the stream is y=2x^2. The dam is 4 ft tall. set up an integral that estimates the hydrostatic force on the dam when the water is all the way at the top. Force=Pressure*Area. water weighs 62.5 lb/cubic foot

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so far i have \[\int\limits_{0}^{4}62.5 \] i need help with coming up with the Area part of the integral to be evaluated

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so far i have \[\int\limits_{0}^{4}62.5 \] i need help with coming up with the Area part of the integral to be evaluated

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think you need to find the area of the horizontal cross section since you're integrating from 0 to 4. get x in terms of y > x = sqrt(y/2) do we know how wide the stream is?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats what i thought too but i l looked up the definition of vert. cross section and it said from teh side so it makes sense. i think the width of the stream is dx since it changes with the depth of the water, looking at the vertical cross section.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ill take a picture and attach what ive drawn

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok the area of the inside of the parabola is sum of 2*x distance * dy >x = sqrt(y/2) >area = integral 2sqrt(y/2)dy so add that to integral above,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i got \[\int\limits\limits_{0}^{4}62.5*2x^{2} dx\] using dx as the width of the estimating rectangle and \[y=2x^{2}\] as the height of the estimating rectangle but you're saying it's \[\int\limits_{0}^{4}62.5*2\sqrt{y/2}dx \] ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0correct except its dy not dx i get an answer of 471.39

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because we are integrating from 0 to 4 which is the height or y value imagine that 2sqrt(y/2) is width of a rectangle and dy is the height

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea im just having trouble getting the equation from the problem. F=Pressure*Area and I got the pressure because it was given and im not understanding what makes up the Area part of the equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh ok. x=width and dy is the height because its changing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its backwards a little bit. ok your welcome
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