Volume of water left in a tipped barrel

In summary, the water in the barrel completely covers a semicircle on the bottom of the barrel and extends up to the top lip of the barrel. The largest of these slices has the same proportions as the water in the barrel and the volume of water is symmetrical about the largest slice.
  • #1
raddishgirl07
2
0

Homework Statement


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.


Homework 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


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.
 
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  • #2
raddishgirl07 said:

Homework Statement


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.

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?
 
  • #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.
 
  • #4
raddishgirl07 said:
At the full end, the water covers exactly half of the base.

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

Initially i was going to integrate the rotational area and divide it by two.. but I'm pretty sure its more complicated than that.

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.
 
  • #5
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:

1. What is the formula for calculating the volume of water left in a tipped barrel?

The formula for calculating the volume of water left in a tipped barrel is: V = πr²h, where V is the volume, π is the mathematical constant pi, r is the radius of the barrel, and h is the height of the water remaining in the barrel.

2. How do you measure the radius and height of a tipped barrel to calculate the volume of water?

To measure the radius, place a ruler or measuring tape from the center of the barrel to the edge. To measure the height, place a ruler or measuring tape from the bottom of the barrel to the waterline. Make sure to take accurate measurements in the same units (e.g. inches, centimeters) for an accurate calculation.

3. How does the shape of the barrel affect the volume of water remaining after it has tipped?

The shape of the barrel does not directly affect the volume of water remaining after it has tipped. As long as the barrel is round and the same dimensions are used in the formula, the calculation will be accurate regardless of the shape.

4. Is there a specific unit of measurement for the volume of water in a tipped barrel?

The unit of measurement for the volume of water in a tipped barrel can vary depending on the units used for the radius and height measurements. Common units include liters, gallons, and cubic meters.

5. Can the volume of water in a tipped barrel be negative?

No, the volume of water in a tipped barrel cannot be negative. If the calculation yields a negative value, it means that the height of the water remaining in the barrel is less than the radius, and there is no water left in the barrel.

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