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# Cylinder problem

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A cylinder with water is tillted till the remaining water forms an apex at the rim of the cylinder and the water line at the bottom of the cylinder is flush with the diameter. Question: what is the relationshop of the remaining water to the total volume of the cylinder. I encountered this problem many years ago. As far as I can remember the solution is a constant so that the measurements of the cylinder do not play any role. Can anyone help me please?

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If my memory serves me correct, the constant lies between 1/3 and 1/4. Part of my problem is deciding whether the surface of the water, from the two ends of the diameter to the apex, is a triangle or a parabola.

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I agree with Jay, it has to be half.

Simply draw a 2-D diagram. The cylidar becomes a rectangle, say for instance, and the waterline is drawn from one corner to the opposite corner. The area is split in half. This would translate to volume in 3-D.

EDIT: Forget what I just said. I've thought about it further and half can't be true because the liquid tapers to a point at the open end. :hihi:

EDIT 2: I've just conducted an impromptu experiment with my coffee mug, and now I'm back to thinking it's half again. I should probably just stay away from math threads. ##### Share on other sites

I agree that it is half, but I believe the OP is looking for the mathematical proof. I can't help with that off the top of my head.

Another interesting thing to ponder is the angle you must tilt the cylinder to where the water reaches the lip.

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It would be half if the water level touched the top of the circumference of the base of the cylinder. However, the water level is flush with the horizontal diameter.

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It would be half if the water level touched the top of the circumference of the base of the cylinder. However, the water level is flush with the horizontal diameter.

Based on the 2-D diagram, it would be 1/4 of the volume if it translates the same. I don't see why it wouldn't.

When I conducted my little experiment, the shape was a parabola. The water did not form a perfect triangular point.

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I suspect the answer is an irrational number.

What’s being asked, in calculus-y terms, is the sum of the volumes of an infinite series of infinitesimally short sections of a cylinder, from one with a half-circle cross section to one with a zero cross section, with cross sections, with the cosine of the half of the arc subtending the chord proportional to its height in the cylinder.

The area of section of a circle cut by a chord subtending an arc of $2a$ radians is the area of the pie-shaped section of the circle subtended, $a$, minus the area of the triangle formed by the center of the circle and the points where the chord cuts the circle, $\sin a \cos a$, that is, $a - \sin a \cos a$.

Because we want the expression in terms of $x = \cos a$ rather than $a$, we rewrite it $arccos x - x \sqrt{1-x^2}$.

For a cylinder with height and radius 1 (and thus volume $\pi$), then, the described volume is:

$\int_0^1 arccos x - x \sqrt{1-x^2} dx$

My calculus is too pathetically atrophied to quickly solve this, but it’s easy to approximate using a computer, giving about $0.2122 \pi$, or 0.2122 times the volume of the cylinder, a bit smaller than a similar subvolume of a rectangular box, which would have volume 0.25 times the box’s volume.

This offers a good excuse for a calculus refresher, or perhaps someone who doesn’t need one can find this integral exactly, and explain how?Please?

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As already mentioned, I encountered this problem many years ago along with the solution. My memory fails me somewhat but I think the answer was 0.26... or 0.28...

I think the calculation route got by without calculus. However, I may be mistaken. Perhaps somebody knows of a website where this is treated? Many thanks to all contributors thus far.

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As a reality check, I actually filled a small clear plastic cylinder (an old office party souvenir :)) with water as described, set it level, and measured it to fill .225 times the volume of the cylinder.

I think the deviation from my estimate of .2122 is due to the meniscus effect caused by the water’s surface tension. When inclined, the circumference of the surface is greater, allowing more water to accumulate against the edge of the vessel, and it also appeared to bead up on the lip of the cylinder when I was tilting out the excess water, not spilling as much as it should.

Perhaps somebody knows of a website where this is treated? Many thanks to all contributors thus far.
An exact solution would be satisfying, but just finding it on the internet would be less so. Better to actually solve the integral ourselves. ;)
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I originally misinterpreted the question, here is a graphical representation as I now understand it:

With equations x^2 + y^2 = 1 for the cylinder and z+y = 0 for the plane

I would use cylindrical polar co-ordinates to integrate:

x=r*cos(theta)

y=r*sin(theta)

z=z

dV=r*dr d[theta] dz

for the limits,

z 0->-y

theta 0->2pi

r 0->1

$\int^{2\pi}_0{\int^1_0{\int^{-rsin(\theta)}_0{r dzdrd\theta}}}$

which gave me 0 :hihi: so something went wrong..

but looking at the graph the symmetry seems to indicate the answer would be 1/4 the cylinders volume.

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for the limits,

theta 0->2pi

which gave me 0 :hihi: so something went wrong.

Try Theta from 0 to pi, not to 2pi. I think you’re getting 0 because you’re including points both with z>0 and z<0 – a wedge-shaped volume above the “table top”, and one below it with identical negative volume.

The axes in the diagram appear mislabeled, with a green Y axis where your formulae use Z, though I’m not sure if this is causing any problem.

I think my integration by “partial disks” is more intuitive, though I don’t see why your approach shouldn’t give the same result.

What do your definite integrals equal, exactly? Are you using some sort of algebra/calculus software? Is it integrating symbolically, or approximating, as I did?

PS: Here’s the little MUMPS program (just command lines) I used to calculate mine:

s I=10000,B=0 f X=0:1/I:1 s A=arccos(X)-(1-(X**2)**.5*X),B=A+B
s B=B/I
w B/pi

It’s output is: .2122315913181805995

As with any MUMPS keyword beginning with z, $zarccos and$zpi (for some wierd reason, I had to omit the "\$z" to prevent the code block stripping out the whole function name) aren’t standard MUMPS, but common implementor-specific function for the arccosine (inverse cosine) and pi. As in FORTRAN, ** is the exponentiation operator, everything else is pretty much like every other language, except that there's no operator precidence, hence the need for extra parenthesis.

I is the number of iterations used for the approximation. Increasing I to 100000 results in an output of .2122090907944941077, 1000000 of .2122068407892468055.

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As a reality check, I actually filled a small clear plastic cylinder (an old office party souvenir ) with water as described, set it level, and measured it to fill .225 times the volume of the cylinder.

I think the deviation from my estimate of .2122 is due to the meniscus effect caused by the water’s surface tension. When inclined, the circumference of the surface is greater, allowing more water to accumulate against the edge of the vessel, and it also appeared to bead up on the lip of the cylinder when I was tilting out the excess water, not spilling as much as it should. An exact solution would be satisfying, but just finding it on the internet would be less so. Better to actually solve the integral ourselves. :hyper:

It appears that the material of the cylinder is a factor in an exact answer which includes the meniscus. Is the cylinder plastic, or glass? :hihi: Meniscus curve: glass vs plastic beaker

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I come from an architectural and engineering background.

Please consider the following diagram, and extrapolate three dimensionally.

Wouldn't this be a proper representation of the conditions?

Assuming the cylinder is perfectly round, why wouldn't the liquid represent exactly 1/4 of the total volume?

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Craig, I was using good 'ol pen and paper :hihi:

I made the graphics only as a guide and yes the axes are wrong I just found it easier to draw that way. But alas I think that has caused me to make an error.

So im trying again..

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$\int^1_0{\int^{\pi}_0{\int^r_0{r}}}dzd\theta dr$

eventually gives 2/3 units^3

Or in terms of the overall volume pi: (2/3)/pi = 0.2122065907892...

:)

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$int_0^1 arccos x - x sqrt{1-x^2} dx$

I didn't really read carefully enough to know if the integral is what needs to be calculated, but it seems (ACTUALLY seemed before really trying it) an easy one to solve for me:

First term (see any table):

$\int_0^1 arccos(x)=x\cdot arccos x -\sqrt{1-x^2}=1$

Second term (by substitution x=cos(y)-->dx=-sin(y)dy):

$-\int_0^1 x \sqrt{1-x^2}dx=\int_{\cos(0)}^{\cos(1)} (\cos (y) \sin^2(y))dy=[\frac{\sin^3(y)}{3}]_{\cos(0)}^{\cos(1)}=-0.153237658$

So the total would be 0.846762342 ... so there seems to be a mistake somewhere, but where?

EDIT: it has been years that I haven't done integration by substitution any more... if I replace y again with arccos(x) I get -1/3 which is what a maple like program gives me...what do I do wrong with the upper and lower limits?

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