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Math Help - Hydrostatic Force Problem!!!

  1. #1
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    Hydrostatic Force Problem!!!

    A large tank is designed with vertical ends in the shape of the region between the curves y = x^2/2 and y = 13, measured in feet. Find the hydrostatic force on one end of the tank if it is filled to a depth of 8 ft with gasoline. (Assume the gasoline's density is 42.0 lb/ft^3

    Sooo, I am not sure what I'm dealing with, but I think we are integrating from 0 to 8 and pull out the 42 of the integral, of something, I think the curves are confusing me. I know that hydrostatic force is area*pressure, and I think the pressure is 42 times x. But I don't know what the area is. Please help!
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  2. #2
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    The sides are just a parabola with depth 13. Did you graph it. That helps to see.

    depth = h(y)=13-y

    length = L(y)=2\sqrt{2y}

    42\int_{0}^{8}2(13-y)(\sqrt{2y})dy

    84\int_{0}^{8}\left(13\sqrt{2y}-\sqrt{2}y^{\frac{3}{2}}\right)dy
    Last edited by galactus; June 23rd 2008 at 05:35 PM.
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  3. #3
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    Oh I see, is the square root over the 2y for both terms?
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  4. #4
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    No, just the first one.
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  5. #5
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    Okay, so I got 7239 and it said it was wrong, I probably did something wrong, once again. I integrated, from 0 to 13 (and multiplying the whole thing by 84) the integral of (13*sq rt of 2y)-((sq rt of 2)*y^3/2) dy and that got me to, after integrating, 13sqrt of 2 over 2)*y^3/2 (-) 2*sqrt of 2 over 5)*y^5/2). That's probably really hard to read! I'm sorry!
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  6. #6
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    My mistake. That is a typo. You integrate from 0 to 8.
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  7. #7
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    Oh that's okay!! I got 8870 integrating from 0 to 8, but it's still wrong! Hmm I will double check my work again.
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  8. #8
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    Oh and you did write to integrate from 0 to 8, I the one who made the mistake!! But I still have 8870.
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