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Registered Member #95
Joined: Thu Feb 09 2006, 04:57PM
Location: Norway
Posts: 1308
I've tried my best to not bring my homework here, but I need to ask a quick question. I have to find the volume of an object rotated around both the X and Y axises (axi?). I have the function expressed as a function of X, and I know that I have to integrate the function, and multiply by pi get the volume for the X-axis object. At least if I remember last year's math correctly. Anyway, for the Y-axis object I'm stumped. I thought I could just try to express the function by Y instead of X, but it's a third degree function and I can't solve it.
Then I considered the problem some more, in both cases the area of the graph (F) when seen in two dimensions is the same, it's just rotated around either the X-axis or the Y-axis. SOOO- Won't the volume be the same in both cases?
Registered Member #1739
Joined: Fri Oct 03 2008, 10:05AM
Location: Moscow, Russia
Posts: 261
>and multiply by pi Wrong! PI is a scaleless constant, so [m^2]*pi is NOT equal to something in [m^3].
If I were you, I'd integrate the circular area (eg if you have an X-axis lathe object with a given envelope Y=f(x), you just integrate the Pi * f(x)^2 (or /2, can't recall that basic formula lol)).
>Won't the volume be the same in both cases? Wrong! A simple counter-example - you have a rectangle with dimensions of WxH. The X-lathe is a cylinder with the volume of Pi*H^2*W. The Y-lathe has the volume of Pi*W^2*H.
Registered Member #29
Joined: Fri Feb 03 2006, 09:00AM
Location: Hasselt, Belgium
Posts: 500
In cylindrical coordinates, volume V=int(int(int(r.dr).dphi). dx) in this case where r = f(x). Doing the first two inner integrals with limits 0->f(x) and 0->2pi, we get
Rotating about the x axis, V=pi * int(f(x)^2. dx) from 0->1, as the Lord of Lithium correctly points out!
To rotate around the y axis, the brute-force method would be to do a similar operation using the inverse of f
V=pi * int(inv(f(x))^2. dx) from 0->f(1)
However, a more elegant way would be to make 0<r<x and 0<y<f(x) to get
V= pi * int(x^2 * f(x). dx)
if I am not mistaken.
Geometrically speaking, rotating around the x-axis, you "add up" the volume of x-centered "disks" of differential thickness dx and radius f(x) along the x axis.
In the case of rotation about the y-axis, you sum up the volumes of annular rings of differential thickness dx, of radius x and height f(x). Does that make sense?
Try to look at these problems in a geometric way rather than a pure "maths" way. I find it helps alot!
Registered Member #95
Joined: Thu Feb 09 2006, 04:57PM
Location: Norway
Posts: 1308
We had a lecture about it today. It appears V = 2*pi*int(x*f(x)dx) when rotating around the Y-axis. Summing of cylinder volumes in other words, with r = x and the height being f(x).
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