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Volume of liquid in a partially filled sphere

  1. Apr 1, 2005 #1


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    On tuesday I tried to estimate how much liquid I had in a round bottom flask, but I realized I had no idea how to get a number. The liquid was about 2cm high, the entire flask is 6cm tall, and the volume of the flask when full is 100mL. How would I calculate/estimate the amount of liquid in the sphere?
  2. jcsd
  3. Apr 1, 2005 #2

    If the volume is 100mL, can you find the radius, R?

    The volume of the sphere can be given by:

    Can you find r in terms of x (r is different from R)?
  4. Apr 1, 2005 #3
    Since this is an irregular shape and there is no function defining any part of the flask, I don't believe the Calculus method will work. Using the integral above you get:

    [tex]\pi\int_{-R}^{R}r^2dx = r^2R - (-r^2R) = 2r^2R[/tex]

    Plugging in the known volume, you are solving [tex]100 = 2r^2R[/tex]

    You just need more information.
  5. Apr 1, 2005 #4
    Ok. I didn't read the complete question. You said you havea spherical flask that is 6cm tall. That means D = 6 and r = 3. Plugging in the formula for the area of a sphere (I say area because some people on this post will kill you if you don't) is :

    [tex]100 = \frac{4}{3}\pi{r}^3[/tex]

    Something doesn't quite add up here....
  6. Apr 1, 2005 #5
    The crudest way is to approximate the fluid as rectangular, or made up of a few rectangular blocks. Assuming you cannot simply weigh the fluid, or pour it in to another container, the next best method is calculus.


    This shows a misunderstanding of what a function is. What you mean to say is that you don't know a formula for a function that describes the shape of the flask.

    Imagine the shape of the fluid in the flask. Let us call the surface of the fluid, the x-y plane. Then the depth of the fluid at each point is a function d(x,y). That is, the depth is a number that corresponds to those two numbers. So the volume of the fluid is simply:

    [tex] \int \int d(x,y) dx dy [/tex]

    As for the problem you expressed earlier, of not having a formula for d(x,y), that is the hard part. The bottom line is, if you have a graphing calculator, you can use polynomials to approximate d(x,y).

    The formula you gave for the olume of a sphere is safe from technicality, no fear. To call it a formula of area is wrong, and confuses the issue.
  7. Apr 2, 2005 #6
    Dextercioby in many other posts has pointed out that the formula [tex]\frac{4}{3}\pi{r}^2[/tex] is the AREA of a sphere. I cannot fully explain why, but I just refer to it as an area now.

    You said you knew how to approximate the volume. Can you expand on this please?

    EDIT: I meant to put [tex]\frac{4}{3}\pi{r}^3[/tex]. My mistake...
    Last edited: Apr 2, 2005
  8. Apr 2, 2005 #7
    If the volumn is small, draw the liquid up into a long glass tube that has straight sides, then measure how high it fills up the tube.

    From there it should be simple to calculate the volumn of the straight sided tube.
  9. Apr 2, 2005 #8


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    Jameson, I think you are confused!

    I can imagine some people becoming upset if you use the word "area" when you mean "volume" (I can't quite imagine then killing you!) but not at using the word "volume" when you MEAN volume- which is the case here.

    The surface area of a sphere is [tex]4\pi r^2[/tex] (not 4/3) and the volume is
    [tex]\frac{4}{3}\pi r^3[/tex].
  10. Apr 2, 2005 #9


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    Semantically, the correct way to phrase it would be : the enclosed interior volume of the sphere is [tex]\frac{4}{3}\pi r^3[/tex]. Alternatively, one could call that the volume of the ball. This is what dextercioby is often insistent on.
  11. Apr 2, 2005 #10
    Jameson: Since this is an irregular shape and there is no function defining any part of the flask, I don't believe the Calculus method will work.

    Well, at one point Edison who was contemptuous of mathematicians (I hire them, they don't hire me) asked a similar question, and then looking at a page of figures, told the fellow, "Why don't you just pour it into another container you can easily measure?"
  12. Apr 2, 2005 #11

    This is what I was talking about. I have always said that the volume of a sphere was 4/3 pi(r)^3. I understand the difference between volume and surface area, I just never understood why dextercioby always challenged me when I refered to the volume of a sphere. Thank you for explaining.

    As for the original question, does anyone have a way to answer it? I've put in my two cents, but I'm stuck.
  13. Apr 2, 2005 #12


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    I don't think using 4 pi*r^2 will work because that would assume the shape of the liquid inside the sphere is also spherical; that is not true. The liquid is round on the bottom and flat on the top; like an orange cut straight across anywhere but the middle.

    I'm thinking this can be done as a series of stacked circles.

    [tex]V = \int \pi x^2 dh[/tex]

    The only problem is that trying to relate h to x gives more variables.

    edited to reduce confusion
    Last edited: Apr 2, 2005
  14. Apr 2, 2005 #13
    Relating [itex]h[/itex] to [itex]r[/itex] is just simple trigonometry...

    radius of the sphere is [itex]3[/itex], so if it's centered at the origin then at [itex]y = h[/itex], we find that [itex]x[/itex] takes extremal values [itex]\pm\sqrt{9 - h^2}[/itex].

    The volume of the region that you want is then

    [tex]\int_{-3}^{-1} \pi (9-y^2) \ dy [/tex]

    which is a pretty easy integral...
    Last edited: Apr 2, 2005
  15. Apr 2, 2005 #14
    You mean -1 for the top, right?
  16. Apr 2, 2005 #15
    yeah. fixed~
  17. Apr 3, 2005 #16


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    If we assume the entire flask is a perfect sphere, then the dimensions shown do not cohere. The volume of a sphere of diameter 6 cm is around 113 cubic cm = 113 ml. You stated that there's 100 mls when the flask is full. Are we supposed to assume there's a cylindrical stem at the top that accounts for the discrepancy ?

    That nothwithstanding, let's accept the "height" or diameter as the correct measurement, and say the flask is a perfect sphere. Then the required volume can be found easily by rotating the curve [tex]x^2 = R^2 - y^2[/tex] about the y-axis, using the required bounds. The interior volume of the solid of revolution is [tex]\pi \int_{y_1}^{y_2}x^2dy[/tex]

    Here the lower bound is -3 (bottom of the flask), the upper bound is -1 (the liquid level). The volume of water is :

    [tex]V = \pi \int_{y_1}^{y_2} (R^2 - y^2)dy = \pi[(3^2)(-1 + 3) - \frac{1}{3}(-1 + 27)] = 9\frac{1}{3}\pi {cm}^3 [/tex], which approximates [tex]29.3 {cm}^3[/tex], or about 26 % of the entire enclosed volume of the flask.
    Last edited: Apr 3, 2005
  18. Apr 3, 2005 #17
    It's round-bottom. I suppose we're all just assuming that the liquid only filled a portion corresponding to part of a ball :wink:

    And the answer is [itex] \frac{28}{3} \pi \ \mbox{cm}^3[/itex], though your decimal representation of it is correct~
  19. Apr 3, 2005 #18
    oh, you meant a mixed fraction! :rofl: :rofl: :rofl:

    So many years since I've seen anyone use one of those~
  20. Apr 3, 2005 #19


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    Yeah, I'm old school ! :biggrin:
  21. Apr 3, 2005 #20


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    You really want to check such calculations as these by getting a before and after weight. This is by far and away the best method of determining the volume of a small sample of water.
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