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Volume of water left in a tipped barrel

  1. Dec 4, 2007 #1
    1. The problem statement, all variables and given/known data
    An open barrel of radius r and height h is initally full of water. It is tilted and water pours out until the water level coincides with a diameter of the base and just touches the rim of the top. Find the volume of water left in the barrel.

    The picture is a cylinder that is tipped with a shaded region as described above. No real information is given.

    2. Relevant equations
    To get Volume, generally, you cut the picture into slabs, get an equation for the area and then integrate the area, in this case from 0 to h

    3. The attempt at a solution

    The section its in deals only with integrating the areas of washers, slabs, and disks... but I have no idea how to go about this one.
  2. jcsd
  3. Dec 4, 2007 #2


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    Just so I have the picture clear, are you saying that the water just completely covers the circular base of the barrel and extends to where it just touches the rim of the open top at one point?

    If that is so, look at the shape of the volume in the barrel that is *not* filled with water. How does it compare with the filled portion? What would that mean about the fraction of the barrel that is filled? Since the barrel is apparently a cylinder, how would you calculate the filled volume of this cylinder?
  4. Dec 5, 2007 #3
    At the full end, the water covers exactly half of the base. Initially i was going to integrate the rotational area and divide it by two.. but I'm pretty sure its more complicated than that.
  5. Dec 5, 2007 #4


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    Oh well, so much for the problem being simple...

    I'm not sure the volume of water has the proper shape to deal with it easily that way. I'm thinking that since the layer of water at the very bottom makes a semicircle, each layer above it is a section of a circle bounded by a chord parallel to the diameter, with the limit attained at the lip of the barrel, where the chord now just touches the circle at one point. You would need to find a formula for the sector of a circle above a chord at a given distance above the diameter (work it out by integration or even look it up) and find an expression for how far the chord is from the diameter as a function of height above the base of the barrel. The volume would then be the integration of all these horizontal slices which are sectors of circle, going from a semicircle at the base of the barrel to zero area at the height of the barrel, h.
  6. Dec 6, 2007 #5


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    While putting a list of topics together for my students, I ran across a problem similar to yours with the method of solution described (it's Example 9 on page 430 in the Sixth Edition of Stewart's Calculus: Early Transcendentals).

    The water in the barrel covers a semi-circle on the bottom and extends up to the top lip of the barrel at one point. If you make slices through the volume perpendicular to the diameter of the barrel marked by the semi-circle, these cross-sections are all similar triangles. (This is reasonable because the water's surface is flat and makes a particular angle to the base of the cylindrical barrel.)

    The largest of these slices has as its base the radius of the barrel perpendicular to the diameter of the covered semicircle; that triangle has base r and height h. All the other slices have the same proportions, with the base given by sqrt[(r^2)-(z^2)], where z is the distance of the slice from the largest one; each slice will have thickness dz. Since the volume of water is symmetrical about that largest slice, you can integrate from z = 0 to z = r and double the result to obtain that volume.
    Last edited: Dec 6, 2007
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