or a bell?

all i have is the height and the radius


Im thinking the answer is simple

im just too spent to think of it

3/28/2007 10:10:05 AM

Volume of a disc? For just a cylindrical solid, it's just [h*pi*(r)^2].

If it's an uneven disc, you could always do slices and integrate.

3/28/2007 10:51:15 AM

water displacement

3/28/2007 10:52:59 AM

^^she said what i would have said, only better

3/28/2007 10:54:06 AM

Quote :

"water displacement"

cheater!

3/28/2007 2:17:31 PM

Say you'll give him a nice new sextant if he tells you the height of the building.

3/28/2007 9:59:45 PM

"du -sk"

3/29/2007 2:09:44 AM

Quote :

"bell"

what's a bell shape, mathematically speaking?

and disk = cylinder. use formula above. here is the rationale (and how you remember it):

you can look at a cylinder as made up of a number of flat circles stacked together. so you find the area of one circle (which is the area of one of the circular faces), and then multiply by the height of the cylinder (which in effect gives you how many of those circles are stacked together).

hence, V = (pi*r^2)*h

3/29/2007 2:16:21 AM

I assume you mean a bell curve?

something like y=e^(-x^2)

Using 0EPII1's rational, you can think of this as a stack of disks of increasing size,

at height y, each has radius -ln(y)^(1/2). Again, you'll want to add the area of each disk to get the volume, so we integrate pi*(ln(y)^(1/2))^2 where y goes from 0 to 1.

= pi * int ln(y) =pi*[ -y ln(y) + y] evaluated from 0 to 1 (note, youll have to take the limit as y goes to zero, since ln y is undefined at 0) = pi

Also, I just pulled that out of my ass, so don't trust anything i said.

[Edited on March 29, 2007 at 2:40 AM. Reason : its late]

3/29/2007 2:40:07 AM

oh is that what he meant, a bell curve?

he said "volume", so i had the picture of a 3d bell in my head.

we still don't know what he meant.

3/29/2007 3:13:35 AM

I figured it out

It's something like 4/3pi * radius^2 * length.

[Edited on March 29, 2007 at 4:18 AM. Reason : the volume of a parabola]

3/29/2007 4:18:28 AM

that sounds like volume of a sphere

3/29/2007 4:18:58 AM

Quote :

"the volume of a parabola"

dude, you are CONFUSED.

how can a parabola have volume?

Quote :

"4/3pi * radius^2 * length."

that is indeed the volume of something, as it has 3 dimensions, but i don't know what of. length of what? radius of what?

(^ sphere V = 4/3 * pi * r^3)

really, i am curious now. what solid shape are you talking about?

AHA... are you talking about a PARABOLOID? (like a 3-dimensional parabola)

i think so, because you mentioned "parabola" now and "bell" at the beginning.

but your original question was how to find volume of [b]a disc OR a bell[/i], and i don't see how they are related. that's like asking, how do you find the volume of a sphere or a cone?

3/29/2007 5:51:30 PM

(pi*r^2)*h

everyone else here is either wrong or talking too much and trying to sound smart.

3/29/2007 6:03:58 PM

KEVIN STOP BEING A FAGGOT AND TELLING TWOOZLES YOU THINK I STILL HAVE THE PICTURES...I TOLD YOU I FUCKING DELETED THEM

3/29/2007 6:06:04 PM

lets see,

if it is z=1-x^2-y^2 bounded by z=0 below ( a paraboloid that opens down with z-intercept 1) then the volume can be found easily by switching to cylindrical coordinates where

0 < theta < 2pi
0 < r < 1
0 < z < 1 - x^2 - y^2 = 1- r^2

Then V = Int(dV) = I I I ( r d(theta) dz dr )
= I I (2*pi r dz dr)
= I (2*pi*r (1-r^2) dr )
= I (2*pi*[r -r^3] dr )
= 2*pi*[1/2 - 1/4]
= pi / 2.

On the other hand if you have in mind z = exp( - x^2 - y^2 ) bounded by z=0 below, well then it only approaches z=0 in the limit of infinite radius, again a change of coordinates makes it much easier to integrate ( I mean can you integrate exp(-x^2) directly? ),

V = Int(dV) = I I I ( r d(theta) dz dr )
= I I (2*pi r dz dr)
= I (2pi r exp( - r^2 ) dr)
= -pi*exp( - r^2 ) + pi (and then take r to infinity)
= pi.

3/29/2007 6:15:01 PM

^^you mean like the first reply in the thread where they said:

Quote :

"[h*pi*(r)^2]."

3/29/2007 6:20:39 PM

i enjoy this thread

3/29/2007 9:51:35 PM