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Calculus II- Pressure and Force Question

  1. May 25, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the force acting on one end of a cylindrical drum with 3m. If the rim is submerged horizontally into water so that the bottom is 10m deep.

    2. Relevant equations
    A(x) - Area
    p(x) - Pressure = density * Depth

    g= 9.8

    A(x) =2[itex]\sqrt{9-x^{2}}[/itex]

    P(x) = 1000(7+x) * g

    dF = 2000*g*[itex]\sqrt{9-x^{2}}[/itex](7+x) dx

    F = 2000g [itex]\int^{3}_{-3}[/itex][itex]\sqrt{9-x^{2}}[/itex](7+x)dx

    Finally, my question is why is the limit goes from -3 to 3? Where does it come from? A picture is in the attachment.

    3. The attempt at a solution
    I have no clue about this. I am ok with Integrals and finding the equations for the problem. However, I am not sure on how to find the limits of the definite integral, so hints and information on how to find the limits (such as -3 to 3) is very much appreciated. Thank you for your time.
     

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    Last edited: May 25, 2013
  2. jcsd
  3. May 25, 2013 #2

    SammyS

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    How is x defined in this problem?

    Where is x = -3 ?

    Where is x = 3 ?


    By the way, I think you intended to say that the cylindrical drum has a radius of 3m .
     
  4. May 25, 2013 #3
    Yes that is what I meant the radius is 3. And x is marked in the attachment.

    And that is what I am asking: where did the prof get the -3 to 3 for the definite integral.
     
  5. May 25, 2013 #4

    SteamKing

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    Quote: And that is what I am asking: where did the prof get the -3 to 3 for the definite integral.

    Hint: the cylinder has a radius of 3 m. Doesn't that suggest why the limits are -3 to 3?

    Do you understand how the integral is set up?
     
  6. May 25, 2013 #5
    I know it suggests that it has something to do with the radius. I am unclear on how the integral is set up for this question, but I know what each number or variable is representing. The main thing I am not clear about is why -3 to 3? Why -3 to 3? Why not 0-6 or 4-10? Does it have to do with the area where we represented with x?
     
  7. May 25, 2013 #6

    SteamKing

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    The reference point for x is the center of the circle. The circle has a radius of 3 m. The depth of water at the center of the circle is 4m + 3m = 7m. The pressure of the water is constant along a horizontal line, like the one shown in the picture. The pressure acting at x from the center of the circle is (x + 7)*1000*g N/m^2.
    The equation of a circle with radius 3 is x^2+y^2 = 9. This implies that the half-width y of the circle at x is
    sqrt (9 - x^2). The force at x is P*dA and dA = 2y*dx. In order to find the total force on the end of the cylinder, you must integrate from 3 m above the center of the circle to 3 m below. Given the definition of water depth relative to the center of the circle, -3 m is at the top closest to the water surface and 3 m is at the bottom.
     
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