Anyone good at calculus?

nolan16

nolan16

Diabloii.Net Member
Anyone good at calculus?

Ok, so here is the problem.. I'm trying to figure out how much liquid I have in a tank that is laying horizontally on the ground. So I figured that I could integrate with respect to hieght to find out how much liquid I have in the tank. This is what I came up with, its probably wrong, but please post if you are good at calculus and you can verify that I am correct or have found the correct solution.
Thanks

Integrating from 0 to H

=S (area*length)(hieght)dh
=S (pi*r^2*l)(h)dh
=(pi*r^2*l) * S (h)dh
=(pi*r^2*l) * [((h^2)/2)] <--don't really know how to represent this but a h up top and 0 on bottum of this chunk to the right in the [ ]

and if I plug in a value for h I will get a solution to how much liquid is in the tank, right?
 
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S

stillman

Diabloii.Net Member
Re: Anyone good at calculus?

Sadly, I forgot all my university calculus because of the intense workload I was under. You purposely forget everything to make room for the next exams in the other 4 subjects...subjects designed for majors of those subjets. Sadly, none of those subjects were my major. Also, exams rage on for months at a rate of 2-4 exams per week. It's almost exams through the whole term. Don't let the words "mid term exams" fool you. The exams go well beyond mid term.

Idk why I'm mentioning this; it's not helping at all. But it may help some who are just about to head into university. You should not be cheap and sell your textbooks. They are handly for looking up stuff like this.

Edit: when you say 'how much water', do you mean volume or mass?

I thought of something. If you know the mass of the empty tank (by weighing it on a scale when empty) you could figgure this out maybe without calculus. You can get the volume of the water.

-pour water into separate container for safekeeping.
-weigh the empty tank on bathroom scale.
-put water back into tank and weigh them both (tank and water together).

mass of water = mass of water in tank - mass of tank.

1 gram of water = 1 cm cubed = 1 ml.

So you have mass of water = X grams from the equation 2 lines above.

Use this conversion factor:

X grams of water (1 cm cubed/1 gram of water) = volume of water in cm cubed

which also equals water volume in ml.

It just so happens that the volume of the water in cm cubed is the same as it's mass in grams due to the simple conversion factor.

Basically, 1 kg of water is 1000 grams, which is going to equal exactly 1000 cm cubed of water. 1000 cm cubed is also one Litre becasue 1 cubic cm
of water also happens to be 1 ml.

So, conveniently, 1 cm cubed water = 1 gram of water = 1 ml of water.

Good thing you said water, or the conversion factor would be different!

(So just weigh the water. This value equals it's volume in cm cubed and in ml. But only if it's 'just water' with nothing mixed in it.)
 
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nolan16

nolan16

Diabloii.Net Member
Re: Anyone good at calculus?

I realized that the integral I made must be wrong, because I'm multiplying the total area by a number that is greater than 1 (h^2/2). It doesn't make sense.

@ffejrxx the tank has flat ends, I will check out the link, thanks.

@Stillman.. That would work, but I guess I didn't give all of the details.. Imagine reservoir fluids (oil) going into the tank, the density would constantly be changing (since the lighter molecules are produced first).
So I need to integrate this somehow =/ (or something).

Edit:
@ffejrxx - That link was exactly what I needed, thank you!
 
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ffejrxx

ffejrxx

Diabloii.Net Member
Re: Anyone good at calculus?

@Stillman,
if you could take the liquid and measure it in another tank that would be cheating
(not using calc to figure out volumes/mass)
and if its other liquids it would have different viscosity and densities making accurate measurements tougher

@OP, np
if your measuring an actual tank, you may want to check if it has any other features that take up volume inside it, they could offset your accuracy
(pipes, dents, bricks, ect..)
 
krischan

krischan

Europe Trade Moderator
Re: Anyone good at calculus?

You are approaching the problem the wrong way. Using the l*pi*r² formula for a cylinder won't help much because you want to have a formula for a partly filled tank as well.

Integrating means dividing up the area into parts, letting them become infinitesimally small and then adding them up. If we find a way of doing that which converges, the integral can be solved.

The width w of a cylinder with the radius of r at a height of h (with h=0 being at the center of it) can be calculated by r² = h² + (w/2)², so w = 2*sqrt(r² - h²). Divide up the area into rectangles with an infinitesimal height dh and a width of w and add them up (H is the fill height, counted from the bottom):

V(H) = l*Integral(-r,H-r,2*sqrt(r² - h²)*dh) = 2*l*r*Integral(-r,H-r,2*sqrt(1 - (h/r)²)*dh)

You can solve that with the substitution h/r = sin a and then applying Pythagoras to the square root, producing an integral of cos² a * da which can be solved by partial integration IIRC (I forgot the proper English term yet again).
 
Kijya

Kijya

Diabloii.Net Member
Re: Anyone good at calculus?

I'd utilize the "segment of circle" formula and multiply by the length of the cylinder.

Assuming you've got the values for the water height inside the tank and radius of the cylinder, it's just some basic trigonometry to get the angle you need for it.

Or have I understood your problem incorrectly?
 
krischan

krischan

Europe Trade Moderator
Re: Anyone good at calculus?

Kijya said:
I'd utilize the "segment of circle" formula and multiply by the length of the cylinder.

Assuming you've got the values for the water height inside the tank and radius of the cylinder, it's just some basic trigonometry to get the angle you need for it.

Or have I understood your problem incorrectly?
Click to expand...
The question probably isn't about finding the book in which the solution is printed. There are different ways of finding a solution, of course. You can find out the result by adding/subtracting the area of a triangle to/from the area of a 0-360° pie slice as well, but I guess that won't find the consent of the tutor because it's probably about using integral calculus to solve it. I wouldn't be surprised if there is yet another way of solving it, with linear algebra :azn:



 
Kijya

Kijya

Diabloii.Net Member
Re: Anyone good at calculus?

krischan said:
The question probably isn't about finding the book in which the solution is printed. There are different ways of finding a solution, of course. You can find out the result by adding/subtracting the area of a triangle to/from the area of a 0-360° pie slice as well, but I guess that won't find the consent of the tutor because it's probably about using integral calculus to solve it. I wouldn't be surprised if there is yet another way of solving it, with linear algebra :azn:
Click to expand...
Hey, that was printed in the bible (physics handbook) so it must be accepted by the tutor! Otherwise it's heresy! :yes:

One could make it a triple integral in cylindrical coordinates of course ...



 
AeroJonesy

AeroJonesy

Diabloii.Net Member
Re: Anyone good at calculus?

Ah, so it's a cylinder? Seems like a lot of spilled ink over calculating the volume of a cylinder.

What's wrong with just using V=pi*r^2*h?
 
Kijya

Kijya

Diabloii.Net Member
Re: Anyone good at calculus?

AeroJonesy said:
Ah, so it's a cylinder? Seems like a lot of spilled ink over calculating the volume of a cylinder.

What's wrong with just using V=pi*r^2*h?
Click to expand...
As I understood it the cylinder is lying horizontally, and not 100% full :wink:



 
nolan16

nolan16

Diabloii.Net Member
Re: Anyone good at calculus?

Yes something like that..

@Kiyja- yes that would work too =P

@krischan- I got lost at r^2=h^2 hehe 8o.. but thanks anyways =)

Thanks for the help all! =)
 
krischan

krischan

Europe Trade Moderator
Re: Anyone good at calculus?

More confusion for you:

krischan said:
Integrating means dividing up the area into parts, letting them become infinitesimally small and then adding them up. If we find a way of doing that which converges, the integral can be solved.
Click to expand...
Nobody noticed that this is wrong :smug:. You actually have to do it with two sets of infinitesimally small areas, one summing up to more than the integral and one summing up to less and if you find a pair which converges to the same value, the integral can be solved.



 
jakotaco

jakotaco

Diabloii.Net Member
Re: Anyone good at calculus?

krischan said:
More confusion for you:



Nobody noticed that this is wrong :smug:. You actually have to do it with two sets of infinitesimally small areas, one summing up to more than the integral and one summing up to less and if you find a pair which converges to the same value, the integral can be solved.
Click to expand...
even more confusion, this definition is for the Riemann way to integrate, there are other ways aswell. :crazyeyes:



 
krischan

krischan

Europe Trade Moderator
Re: Anyone good at calculus?

I was going to write that as well, but I hoped that I can lure in Dondrei :whistling:
 
SnickerSnack

SnickerSnack

Diabloii.Net Member
Re: Anyone good at calculus?

If something is Riemann integrable, then it will also be integrable by most other methods (looking at lebesgue measure here), and the integrals will be equal. The nice part about the lebesgue integral is that you only have to do it from one side.
 
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