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Really hard Volume Question. !

  1. Jan 29, 2009 #1
    Really hard Volume Question. URGENT!

    Can someone help me with this question? I don't know where to begin. I don't want the answers, just hints in the right direction

    [​IMG]


    *edit* can a mod please fix the topic title? It's supposed to read 'Really hard Volume Question. URGENT!' << Done. >>

    Thanks
     
    Last edited by a moderator: Jan 29, 2009
  2. jcsd
  3. Jan 29, 2009 #2
  4. Jan 29, 2009 #3
    Re: ally hard Volume Question. URGENT!

    Call z the axis which h belongs to. You have to find an expression for the area of the rectangle obtained by intersecting the cylinder with an horizontal plane as a function of z and integrate this area from 0 to h.
     
  5. Jan 29, 2009 #4
    Re: ally hard Volume Question. URGENT!

    The area of the disc (what's inside the circle) from -r to h, where h is between -r and r can be figured out by doing half a circle at a time. The function is sqrt (r^2-x^2). Since a disc is symmetrical just find the area above the x-axis and multiply by 2. So the fun part here is finding out what the integral is.
     
  6. Jan 29, 2009 #5
    Re: ally hard Volume Question. URGENT!

    think simple ...
    I think it can be solved without integration using bunch of trigonometry and sector, triangle area formulas
     
  7. Jan 29, 2009 #6
    Re: ally hard Volume Question. URGENT!

    Sure area 2 is a fraction of pi r^2. I bet the fraction formula as a function of h is not simple. Area 1 is + when h is above the center of the circle and - when h is below the center. I think I'll stick with my integral since I've already solve a related problem: The probability of intersection between a tossed square and a target disc.
     
  8. Jan 29, 2009 #7
    Re: ally hard Volume Question. URGENT!

    This kind of integral is simple...
    What you actually need is the lenght of the segment obtained by intersecting the circle [tex] x^2 + z^2 = R^2 [\tex] with a straight line z = const. And this is given by

    [tex]
    b(z) = 2 \sqrt{R^2 - z^2}
    [\tex]

    Thus, let L be the height of the cylinder, the area of the rectangle is given by

    [tex]
    A(z) = L b(z) = L 2 \sqrt{R^2 - z^2}
    [\tex]

    And then the volume of the water

    [tex]
    V(h) = \int_{-R}^{h} A(z) dz
    [\tex]
     
  9. Jan 29, 2009 #8
    Re: Really hard Volume Question. URGENT!

    I dont know why but latex command doesn't work....
     
  10. Jan 29, 2009 #9
    Re: ally hard Volume Question. URGENT!

    I see.
    I failed to consider the other case.
     
  11. Jan 29, 2009 #10
    Re: Really hard Volume Question. URGENT!

    (1/2(pi r^2) + 1/2(pi (r-h)^2 ) * l

    would that work?
     
  12. Jan 29, 2009 #11
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