volume of revolution

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radu 101 Sep 23, 2005 at 13:41

Hi,

Simple question but I don’t seem to find a formula for confirmation.

Suppose we have a curve y = f(x). The volume of revolution of this curve along the x axis is V = PI * Integral (y*y)dx = PI * Integral (f(x)*f(x))dx

If we use parametric coordinates, x = f(t), y = g(t), we end up with

V = PI * Integral( g(t) * g(t) * f ‘ (t))dt

Ok, all fine so far.

What about for a curve defined x = f(t), y = g(t), z = h(t). What is the volume of revolution of this curve along the x axis ? I cannot seem to find a formula. My guess is

V = PI * Integral ((g(t)*g(t) + h(t)*h(t))* f ‘ (t) dt

but I need this confirmed (book, web pagem anything)

thanks
radu

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Reedbeta 167 Sep 23, 2005 at 22:35

Your formula looks right to me, if your intent is basically to draw a disc perpendicular to the x-axis for each point on the curve and then add up the area of all those discs to get the volume. I’m not sure, though, why this would be useful - when would you have a 3D (non-planar) curve that you wanted to revolve?

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NomadRock 101 Sep 24, 2005 at 00:48

I dont think this would even be totally well defined. With parametric equations like that, you can start looping back and tying knots and all sorts of strange things, if you took that curve and rotated it, deciding which parts should be solid and which should not is not a trivial task.

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Reedbeta 167 Sep 24, 2005 at 21:38

Definitely true. The equation posted will work, but if the curve is a strange one you may not get the results you expect.

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geon 101 Sep 25, 2005 at 20:22

You can “project” your curve onto the revolution plane, giving a 2D curve. (This might not be reasonable for you specific case, but anyway.)

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SigKILL 101 Sep 26, 2005 at 09:28

You would generally not get the right result trying to find the volume of a parameterized curve using the formula described above. However, by the implicit function theorem, you can divide it up in intervals and create implicit functions (i.e. y = f(x) sort).

In general the curve x = f(t), y = g(t), z = h(t) will not ensure the revolting it around an axis actually will generate something with a volume at all. All you know is that it generates a surface…

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Einheri 101 Sep 26, 2005 at 09:40

@geon

You can “project” your curve onto the revolution plane, giving a 2D curve. (This might not be reasonable for you specific case, but anyway.)

This would most likely be the correct approach. Because multivariable calculus can often is a pretty hairy subject for things like this, you probably want to get a 2 dimensional curve by taking your 3 dimensional one and doing something best described as ‘looking at it from one side’; basically, if you want the rotation around the x axis, then ignore the z parameter entirely and perform the integration on the 2d curve that you’re left with.