Calculus Problem
 

Hopefully someone can try and figure this out and hopefully I can remember all of it.
You have the graph of e^(-x).
Rotate it around the X-axis.
What is the center of mass? As in x-bar, y-bar, and z-bar as we call them. The letters with the lines over them. Obviously if you look at the 3d object, you can use symmetry to establish that the y and z coordinates are going to be 0. But what is the x? As in how far along the x-axis is the center of mass? Hopefully I've explained it enough.

Well if I understand what you're saying correctly, there would be no center of gravity on the x-axis because f(x)=e^(-x) will continue towards negative infinity therefore the center of gravity on the x-axis will continue towards negative infinity.

Actually, I think the original graph (of e^(-x)), goes from +infinity (for x approaches -infinity) to approaching zero (for x approaches +infinity)

(and of course the graph crosses the x=0 point at y=1)

Rotating would yield y= -infinity to y approaches zero
(for x=<-infinity, +infinity>).

I would have no idea on the "center of mass" value, I do not know the concept.

Edit: I think I misunderstood Frozen North's post, we probably share the same notion of the graph.

I really don't see how you can create a 3 dimensional object from someting like a line (1 dimensional) no matter how much you spin it. I already don't like your calc class TheClarkster. :)

I'd seriously tell the teacher that there can't be a center of mass because the object has no mass in the first place as it does not take up space and can not exist in a 3d universe.

When you have a X-axis, Y-axis and Z-axis, you have 3 dimensions it's probably posible to find the center of mass. But i don't know how to do that in this situation.

While I agree with Frozen North on the dimensional aspect somewhat, a correction is in order ;)

The graph has 2 dimensions. Namely in the xy-plane.

Rotating a 2D object can result in a 3D object if rotating means that you rotate in infinitely small increments and you 'copy' the object, and not 'move' it .
(or in otherwords: the resulting object is the collection of points that intersect some rotation of the original graph)

The plane that the line is graphed on has 2 dimensions, (length and width) but by definition the line itself has only 1 dimension (length but no width) and I can prove this by asking if anyone can find the area of e^(-x).

Ok... I am too far removed from calculus to fire this one up and figure it out really quickly, but I imagine there are a couple of things I can add to this discussion...

First off, you can make a 3-d object from a line. Just imagine taking a string covered in paint, holding one end, and smearing it around in a circle on a piece of paper (2-d now). Now take that 2-d paper and bend it in any sort of three dimensional shape... This isn't taking a 1-d object and magically transforming it to a 3-d shape.. Rather it is just describing the way to generate a 3-d shape from a mathmatical 1-d function.

And, to get to your question TheClarkster, I don't really remember how to find x-bar, etc, but do recall there being equations for these sorts of things... But I figure that one or two of the averages are going to be at the axis of rotation (so, at x=0 or y=0 or z=0)... As there would be equal "mass" of the line on all sides of the axis of rotation...

Hope... somebody else can be of more help...

vector calculus was never my strongsuit, but here is the formula set for centers of mass of 3d regions (rotating a curve creates a 3d region W):

xbar = ( tripleint[W] x p(x,y,z) dxdydz ) / mass,

ybar = ( tripleint[W] y p(x,y,z) dxdydz ) / mass,

zbar = ( tripleint[W] z p(x,y,z) dxdydz ) / mass, where

mass = tripleint[W] p(x,y,z) dxdydz, where

p(x,y,z) is the mass density. in your case, i'm not sure if e^(-x) is the mass density or its rotation the region, but i suspect it's the mass density.

this is all from Vector Calculus 4th ed by Marsden & Tromba.

Alright, I'll buy that we can create 3d region out of a curve if the spinning of the curve is just defining the surface and isn't the object itself.

i think we are just suffering from sloppy shorthand definitions, no biggie.

Or I'm just retarded and forgot to say it's from 0 to 1 on the x-axis. Ugh.

yea giving it bounds would be the only way to do this or else the graph crosses former self when you rotate.

i contend that since you're rotating about the x axis that ybar = 0. due to symmetry.

also, zbar = 0 also because of the symmetry that you get from rotating about the x axis.

so the only thing that matters is xbar which is simply the x value of the average value of e^(-x) between 0 and 1.

so the average value of e^(-x) would be 1/(1-0) * int[0,1][e^(-x)dx] and you get -1/e - (-1) = 1-1/e or (e-1)/e.

to get the x value of that you say that e^(-x) = (e-1)/e, take the ln() of both sides and multiply by -1 to get x.

that leaves -ln[(e-1)/e] or ln[e/(e-1)]
which becomes 1- ln[e-1]. what ever that value is should be your xbar.

xbar = 1-ln[e-1].
ybar = 0
zbar = 0

answer to this problem
 

DUDE I have the same problem! WTF!

Anyways. Center of Mass, or X bar is found many ways. The way I use is da best.

Center of mass is described as total moment divided by total mass. So to find total moment, take the integral of moment (moment being mass times displacement) and that gives you total moment. now divide by total mass (In this case, mass is described as the integral of the function.)

Since you are rotating this around the X axis, center of mass for y doesn't matter. It's symetrical. X will be less than 0.5, if i recall correctly. I don't wanna give the answer away though. lol

Cool!