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Volume of Right Circular Cylinder

  1. Feb 15, 2006 #1
    I need to find the volume of this right circular cylinder as a function of the depth of the fluid. I am having trouble starting this problem.

    [​IMG]

    I realize the equation for the cross sectional view of the cylinder is

    [tex]x^2 + y^2 = 1^2 [/tex].


    Any suggestions on what to look at next?

    Should I be looking to set up an integral to find the area of that circle as a function of the depth of the fluid? Then, once I do that, find an integral to get me the volume of the fluid across the length cylinder?
     
  2. jcsd
  3. Feb 15, 2006 #2

    HallsofIvy

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    First you need to state the problem correctly. You are NOT trying to find the volume of the right circular cylinder. (That's [itex]5\pi[/itex] cubic meters.) You are trying to find the volume of the fluid in the cylinder.

    Think about a think layer of fluid, of thickness dz, at height z above the bottom of the tank. It is, of course, a thin rectangle with width 5 m and height 2x (x is the x coordinate at y= 1-z). Yes, x2+ y2= x2+ (1-z)2= 1 so
    [tex]x= \sqrt{1- (1-z)2} . The volume is the integral of the area of that rectangle dz with z going from 0 to d.
     
  4. Feb 16, 2006 #3
    How is the height 2x? What do you mean by the "x is the x coordinate at y=1-z" ?
     
  5. Feb 17, 2006 #4

    HallsofIvy

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    Each "layer of water" is a rectangle. One side (it doesn't matter if you think of it as "length", "width", "height") is that 5 m length of the tank. The other is measured from one side of the circular face to the other. If you take x to be horizontal, y vertical, so that the circular face has equation
    x2+ y2= 1, then that length is 2x. I am taking, as I said, z to be measured from the bottom of the circle up. At the bottom, y= -1 so that y= -1+z. (I said y= 1-z since by symmetry you could have taken positive y downward. Since your formula involves only y2 it doesn't matter whether you use z-1 or 1- z.)
     
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