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*Simple is subjective ~D
I've got this question for maths; but I can't get the equation which I have to integrate:
A drinking glass has a shape of a truncated cone. If the internal radii of the base and the top are 3cm and 4cm respectively and the depth is 10cm, find by integration, its capacity. If the glass is filled with water to a depth of 5cm, find the volume of the water.
Anybody know how to deduce an equation from this question?
Well I remember those problems were annoying if that's any consolation. Let's see, by integrating what you are doing is taking a really small slice with the area of the base, and than another and so on infinitely many times which gives you an approximation of the volume. Soooo I don't remember the details. You'll probably have 3 and 4 on the little squiggly thing.
Thanks for clearing that up for me. :laugh4:
Maybe c is 10?
Maybe you take the formula for volume of a cone and do some calculus thing to it and put 3 and 4 and 10 in their somehow? Man I don't remember calculus.
Edit: I know how to integrate and find volume using integration.
Create the whole cone and remove the non-existant part.
4^2*pi*h(40 unless I'm mistaken)/3-3^2*pi*h(30)/3
3,5^2*pi*h(35)/3-3^2*pi*h(30)/3
Considering what I just wrote, it seems that it's either a double integral or a triple.
int: int: int: r/3 drad/dt dr/dt, dh/dt or something like that (i don't think this one handles the cone part correctly).
They still subject people to these ?Quote:
truncated cone
We saw them in the alst year of HS, never needed to know anything about them again (and I'm an engineer).
What i think:
first a formula for the area of a cirkel:
2 pi int(r dr)
now integrate it over the height
2 pi int(int(r dr) dx)
with dr/dx =0.1 -> dx=10dr
20 pi int(int(r)dr dr )
You now have a double integral over r, solving it you get
20/6 pi r^3 between r=3 and r=4
calculate that and you should get the volume.
I checked using the basic cone formulas, it should be correct, i hope this isn't for credit ... :bow:
Equation is 3+0.1x for the line of the side of the cone. Integrate this equation by using the disc solid of revolution method from 0 to 5 for the water-filled 5 cm stuff, and from 0 to 10 for the full capacity. The full equation would be something like:Quote:
Originally Posted by Rythmic
pi * INTEGRAL((3+0.1x)^2)dx
with the integral limits being 0 and 5, and 0 and 10, respectively.
This should become:
pi * INTEGRAL0to5 (9 + 0.6x + 0.01x^2) dx =
pi * (9*5 + 0.3 * 5^2 + 0.01/3* 5^3) =
pi * (45 + 7.5 + 0.01/3* 125) =
pi * (52,5 + 1.25/3) =
pi * 635/12 = approx. 166 cm^3
I hope I got the calculation right, but I'm not sure
Gah!
Yes. This is how I'd do it. I'll just explain how to get the line of the side of the cone, as it would seem to be your problem with this thing.Quote:
Originally Posted by LegioXXXUlpiaVictrix
Just think of the cone and take a cross-section of it along the axis and you'll see the line on the cross-section plane. Then, if you think of the cross-section plane as your basic (x,y) plane you can simply regard the question as the usual analytic geometry - find the equation of the line, and after that it's the integration described above. Try drawing a picture of it, pictures are nice. Pictures are your friend.
Edit. Ahh, it feels good to integrate for a while, even if only schematically. *leaves to integrate happily into the night*
Edit2. Yes, it's simple. I'd be damned if it was for credit. Where are you studying? We might as well compare educational systems while at it, as well :beam:
Thanks for the responses. :bow:
I did that using the 2-point formula, and it seems I got the right answer for f(y).Quote:
Originally Posted by The_Mark
Picture? This is what I understand it as being (Note just realised it should be (f(y))^2 in the picture):
https://img354.imageshack.us/img354/511/mathsmk7.jpg
Ok I got that using the 2-point formula; I also got the step before integration right. So that must mean my integration was wrong.Quote:
Originally Posted by LegioXXXUlpiaVictrix
That's the answer. ~:cheers:Quote:
Originally Posted by LegioXXXUlpiaVictrix
Quote:
Originally Posted by The_Mark
I'm a high-school student. What's credit?Quote:
Originally Posted by doc_bean
Edit:Seconded.Quote:
Originally Posted by Fenring
An american terminologyQuote:
I'm a high-school student. What's credit?
*walks into thread*
*dies*
@ Stig: :laugh4:
Thanks for the working LegioXXXUlpiaVictrix, I know exactly what I did wrong now. When integrating I divided the constant and the index wrongly. (:wall:)
I take it you do this stuff at Uni then? I've got to do it at school, quite irritating really. And questions like these are pretty common in our assessments. The hardest one I've ever got was:Quote:
Originally Posted by The_Mark
A hemispherical bowl of radius a units is filled with water to a depth of a/2 units. Find, by integration, the volume of the water.
The teacher couldn't even do it, I'm proud to say that it took me under 5 minutes to do it.
This thread is enough to fry my dyslexic brain into oblivion... :wall:
I am glad I'm not studying what Rythmic is, in my day it was:
"A man has five camels, he trades two of these for another man's 45 year old spinster sister, besides being totally ripped off, how many camels does he have left?"
:bow:
I took all the advanced courses I could in High School and I just realized that all of that information is just gone. What's pie? :stupido2:
Here in NSW we have the following levels of Math:Quote:
Originally Posted by Vladimir
General - Basic, such as algebra, probability and statistics
2 Unit (which is what I'm doing) - Relatively difficult, the hardest stuff we do is application of calculus and the trigonometric functions
3 Unit (which I used to do) - Basically they do the 2 Unit course but very quickly and then do 5-6 more topics on top of that.
4 Unit - Simply put death.
Naw, I'm just about to graduate from the Finnish equivalent of high school; finished my matriculation exams this March and I'm hoping for a sizeable scholarship for a full 60/60 score from maths :beam:Quote:
Originally Posted by Rythmic
The bowl assignment doesn't sound that difficult.I'm just screwing something obvious up right now with it. Gah. Lack of routine is not your friend :shame:
I don't know, it's American and probably something about getting bonus points from doing some voluntary and/or hard assignments.Quote:
I'm a high-school student. What's credit?
Forcing you to do it by integration is evil.
Easy way out is find the volume of the cone and subtract the cut off top bit.
I hope the you get the results you wanted. I've got half a year left of my schooling.Quote:
Originally Posted by The_Mark
It's just a bit confusing.Quote:
Originally Posted by The_Mark
V = π∫ a/2 to a (a^2 - y^2) dy
= π [(a)^3/3 - (y)^3/3] a/2 to a
= π/3 {a^3 - (3(a)^3)/8 - a^3 + (8(a)^3)/8}
= π/3 {(5(a)^3)/8}
= (5π(a)^3)/24
But strangely satisfying if you get it right.Quote:
Originally Posted by Avicenna
4 if you count the sister https://img76.imageshack.us/img76/65...awesomepm9.gifQuote:
Originally Posted by Cambyses II
Credit is what you get for doing work. Like in the definition of the word ~;)
Yep, thanks, I got quite what I wanted; I can pretty much walk into whatever physics/maths uni in Finland :2thumbsup:Quote:
Originally Posted by Rythmic
Good luck with your last days, enjoy them :beam:
That's true. The wonders of maths...Quote:
But strangely satisfying if you get it right.