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Homework Help: Volume of a Cylinder/Tank

  1. Feb 19, 2010 #1
    1. The problem statement, all variables and given/known data
    You have a cylinder standing upright. Now, imagine a plane cutting the cylinder in half diagonaly from bottom left to top right.

    Develop a function for the volume of the half-cylinder (the top left piece) as a function of its height. Volume would be liquid rising from the bottom of the piece up to the top.

    The only variables I have are radius of the cylinder, and height of the cylinder.

    This isn't a homework problem... it's an oddly shaped tank I need to develop a curve for, but I figured this is the best place to post.

    2. Relevant equations

    V = Pi*R^2*H -not very usefull

    A = 1/2 * R^2 * (Theta - Sin(theta)) -This is the area of the segment of a circle made by a chord, where theta is the angle formed by drawing two radii from the center of the circle to each end of the chord.

    3. The attempt at a solution

    For a solution I'd really like a simple equation and/or something I can reference from a good source.

    The way I did it is the following (I'd like something more simple though if it's possible):

    Imagine the top view of the half-cylinder tank. As water rises from the bottom, you will see a cross sectional area fill that is equal to an area segment created by a chord. You can take that area and multiply it by a differential height then integrate over the height of the tank.

    So I take the equation for the area segment: A = (1/2) * R^2 * (Theta - Sin(Theta))

    Theta would start at 0 and increase to 2Pi as the tank fills. However, I need to turn this equation into a function of height (y):

    Theta = 2*ArcCos[1-(2y/H)] where H = total height of the tank and y = height

    Plug in the new value for theta, and simplify:

    A = (1/2) * R^2 * ((2*ArcCos[1-(2y/H)]) - ((1-(2y/H))*(1-(1-(2y/H)^2)^.5))

    To make this easier to read I'll consider Z = 1-(2y/H) and substitute it in (I can't use latex on this browser, sorry):

    A = (1/2) * R^2 * ((2*ArcCos[Z]) - (Z*(1-Z^2)^.5))

    This equation can then be integrated from 0 up to y (where y would be the height of interest).

    Solving that integral analytically is very hard (impossible?). I used Excel to numerically integrate, but it is kind of sketchy. I'd like a more simple equation that gives an exact answer if it exists. Thanks for any help.
  2. jcsd
  3. Feb 20, 2010 #2
    equality of the circle:
    [tex]r^{2} = x^{2} + y^{2}[/tex]
    solving it for y, choosing the positive one gives the function for the positive half:
    [tex] f(x) = \sqrt{r^{2} - x^{2}}[/tex]
    integrating it from -r, and doubling is should give us the area of a cross section. Now you should only find the function between h and x, and integrate it again to get the volume.
  4. Feb 20, 2010 #3
    You can just look up the part with ArcCos[Z].
    The other term can be done with substituting Z = sin U and then integration by parts
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