What is the height of liquid in a horizontal cylinder with 1/3 volume filled?

[BrettJimison]

Good day all,

someone gave me this problem to try to figure out (for fun) and I'm having a bit of a hard time:

A closed cylinder of radius R and length L is filled with liquid. The volume of the liquid is equal to 1/3 the total volume of the cylinder. If the cylinder is layed flat (horizontal - long ways), how high up the sides does the liquid go?
The answer involves an inverse trig function. The answer is not in terms of elementary functions - find an approximate solution graphically. Well,
I I am not sure my approach is correct, but I tried integrating "slices of water" from y= -R up to X - with X being the height of water along the side in terms of R (i think the bounds on my integral are wrong).

I ended up with :
$$\frac{13\pi }{12}= \frac{x\sqrt{R^{2}+x^{2}}}{2R^{2}}+sin^{-1}(\frac{x}{R})(\frac{1}{2})$$

...which seems horribly wrong. Any help on how to approach this problem?

Thanks!

 

[Matterwave]

Well, the length part is easy, you just have to multiply by L, so the only hard part of the problem is figuring out the area of a cut-off circle. You can do this by integration as you suggest. What we have is an arc of $x^2+y^2=R^2$ from -R up to X where $X>-R$, that's correct. We can look at just the top half of the arc, and then the bottom half of the arc is symmetric so the area as a function of X is:

$$A(X)=2\int_{-R}^X \sqrt{R^2-x^2} dx$$

Is this the integral you had? Wolfram is giving me a function with an inverse tangent in it when I integrate this...

 

[BrettJimison]

ahh...well my integral looked a little different . I was integrating volume so I was integrating slices of water...but this looks better. so I integrate this (sin trig sub - note: invese tan comes from just integrating "1" with d(theta) so a "theta" comes out in the answer which converts back invese trig function - I chose sin inverse, but I guess wolfram likes tan inverse. Let me work with this a little bit.

So I will compute the integral, multiply by L and set it equal to...? $$\pi R^{2} L (1/3)$$?

I guess I am also a little confused on how to make my volume i equal to one third to total volume of the cylinder while only having X in terms of R..

 

[Matterwave]

So, once you have $A(X)$ you can get a volume simply by multiplying by $L$, the volume of water will be $A(X)L$ and you have to set that equal to $\frac{1}{3}\pi R^2 L$ and the $L$ will cancel on both sides, so what you really have to do is simply find when $A(X)$ is equal to $\pi R^2/3$.

To check if your integration is correct, make sure you get $A(R)=\pi R^2$.

 

[BrettJimison]

Ah ha! Thanks!

 

[ehild]

The volume of the water in the " horizontal" cylinder is that of a cylinder of length L but with base a segment of the circle of height X. No need to integral.The area of the segment is area of sector - area of triangle.

ehild

 

[Matterwave]

If that's what your integral looked like, then we are just off by the factor of 2, and you are doing exactly the same thing I was doing. Notice the arc $y=\sqrt{R^2-x^2}$ is only the top half of the circle, we multiply by 2 to get the bottom half of the arc. In other words, the multiplication by 2 is so we don't have to calculate the integral of $y=-\sqrt{R^2-x^2}$ which is the other valid root in the equation $y^2=R^2-x^2$

 

[Matterwave]

The volume of the water in the " horizontal" cylinder is that of a cylinder of length L but with base a segment of the circle of height X. No need to integral.

ehild

View attachment 74344

Do you know of a nice formula for the area of the base "a segment of the circle of height X"? I saw no way to calculate that except with an integral.

 

[Matterwave]

one more thing: where did the 2 come from?

Answered in the post right above your post here.

 

[ehild]

Do you know of a nice formula for the area of the base "a segment of the circle of height X"? I saw no way to calculate that except with an integral.

The area of the sector is $R^2\theta / 2$. The area of the triangle is $R^2\sin(\theta/2)cos(\theta/2)=R^2sin(\theta)/2$. The area of the segment is the difference $0.5R^2(\theta-sin(\theta))$.
And $X=R(1-\sin(\theta/2))$.

 

[BrettJimison]

The area of the sector is $R^2\theta / 2$. The area of the triangle is $R^2\sin(\theta/2)cos(\theta/2)=R^2sin(\theta)/2$. The area of the segment is the difference $0.5R^2(\theta-sin(\theta))$.
And $X=R(1-\sin(\theta/2))$.

hello ehild,

the answer must be in terms of R and x... with your method, is there any way to get rid of theta?

 

[BrettJimison]

Hello guys:

so I ended up with an eq:

$$-\frac{\pi}{3}= \frac{x}{R^{2}}\sqrt{R^{2}-x^{2}}-arcsin(-\frac{x}{R})$$

...when my friend sent this to me he said:
You should be able to find a formula involving inverse trig functions that can be solved
for the result, but it does not have solutions in terms of elementary functions. Find
an approximate solution graphically.

Not even sure what to do...

 

[OmCheeto]

Hello guys:

so I ended up with an eq:

$$-\frac{\pi}{3}= \frac{x}{R^{2}}\sqrt{R^{2}-x^{2}}-arcsin(-\frac{x}{R})$$

...when my friend sent this to me he said:
You should be able to find a formula involving inverse trig functions that can be solved
for the result, but it does not have solutions in terms of elementary functions. Find
an approximate solution graphically.

Not even sure what to do...

What your friend meant was, that there is no explicit solution.
You need to solve for volume, in terms of R, graph the function, and then divide the graph into thirds.
The answer is the point where volume is 1/3, obviously.

 

[ehild]

hello ehild,

the answer must be in terms of R and x... with your method, is there any way to get rid of theta?

You get rid of theta if you solve the equation volume of segment-based cylinder = 1/3 volume of cylinder.
Expressed with theta, this means
$$0.5R^2L(\theta-sin(\theta))=LR^2\pi/3$$
simplified: $$\theta-sin(\theta)=2\pi/3$$ The approximate solution is about 2.6. And X=R(1-cos(theta/2)). (My previous X was wrong)

 

[BrettJimison]

You get rid of theta if you solve the equation volume of segment-based cylinder = 1/3 volume of cylinder.
Expressed with theta, this means
$$0.5R^2L(\theta-sin(\theta))=LR^2\pi/3$$
simplified: $$\theta-sin(\theta)=2\pi/3$$ The approximate solution is about 2.6. And X=R(1-cos(theta/2)). (My previous X was wrong)

Ehild,

Sorry I am a little confused, I understand getting rid of theta, but 2.6 confuses me. 2.6 what?

 

[BrettJimison]

using a graphical method, I got x = 9/16R (which seems a little small..)

 

[ehild]

Ehild,

Sorry I am a little confused, I understand getting rid of theta, but 2.6 confuses me. 2.6 what?

theta=2.6 radian.

 

[ehild]

Are you sure your x is what you think is?

 

[Matterwave]

The area of the sector is $R^2\theta / 2$. The area of the triangle is $R^2\sin(\theta/2)cos(\theta/2)=R^2sin(\theta)/2$. The area of the segment is the difference $0.5R^2(\theta-sin(\theta))$.
And $X=R(1-\sin(\theta/2))$.

Ah, that's quite ingenious. I like this solution! :)

 

[ehild]

Ah, that's quite ingenious. I like this solution! :)

Note that X = R (1-cos(theta/2)) I made a mistake in that post.

 

[Matterwave]

Note that X = R (1-cos(theta/2)) I made a mistake in that post.

Yes, I just read through the method, I did not bother to follow through with the calculation haha.

 

[BrettJimison]

ahh.. perfect sense!

Nice approach ehild! Thanks again!

 

[SteamKing]

What you were missing is knowing how to calculate the area of a circular segment:

http://en.wikipedia.org/wiki/Circular_segment

This type of problem is encountered on a regular basis. Engineers often must prepare tables showing the depth of fluid in a tank and the corresponding amount of fluid at various intervals. At service stations, for example, you'll sometimes see an attendant put a long measuring stick into the underground fuel tanks thru the fill lines, so that he can gauge how much fuel is remaining in the tank.

 

[OmCheeto]

What your friend meant was, that there is no explicit solution.
You need to solve for volume, in terms of R, graph the function, and then divide the graph into thirds.
The answer is the point where volume is 1/3, obviously.

I would like to thank everyone for ignoring my obviously incorrect post.
My sincerest apologies.

 

[A.T.]

At service stations, for example, you'll sometimes see an attendant put a long measuring stick into the underground fuel tanks thru the fill lines, so that he can gauge how much fuel is remaining in the tank.

Even more crucial for estimating when you will run out of beer in a barrel.