Just a Simple* Math Question

**Sasaki Kojiro** · 05-01-2007, 08:42

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.

**naut** · 05-01-2007, 08:49

Thanks for clearing that up for me.

**Sasaki Kojiro** · 05-01-2007, 08:52

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.

**Rodion Romanovich** · 05-01-2007, 09:14

Look at http://en.wikipedia.org/wiki/Volumes_of_revolution

**naut** · 05-01-2007, 09:19

Edit: I know how to integrate and find volume using integration.

**Ironside** · 05-01-2007, 10:13

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).

**doc_bean** · 05-01-2007, 10:17

truncated cone

They still subject people to these ?

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 ...

**Rodion Romanovich** · 05-01-2007, 11:17

Originally Posted by Rythmic

Edit: I know how to integrate and find volume using integration.

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:
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

**Kralizec** · 05-01-2007, 12:15

Gah!

**The_Mark** · 05-01-2007, 12:16

Originally Posted by LegioXXXUlpiaVictrix

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:
pi * INTEGRAL((3+0.1x)^2)dx
with the integral limits being 0 and 5, and 0 and 10, respectively.

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.

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