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 Bully For You (Posted on 2004-12-02)

Take a regular torus (doughnut) shaped object and cut a vertical slice through it at line A-A1(Fig1).

Now look at the cross section formed(Fig2). Is it possible to calculate the volume of the original Torus? Prove your results.

 No Solution Yet Submitted by Juggler Rating: 3.6667 (3 votes)

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 possible solution | Comment 3 of 5 |
first, I am assuming that a regular torus is a circle of radius r rotated about an axis with the center of the circle a distance R from the axis.

From this assumption, the basic circle is known to have a height of 12, therefore a radius of 6 (r = 6).  The equation of the circle is therfore (assuming this circle to be centered on an arbitrary x-y axis) x^2+y^2=36.  We know at a certain distance from the center of the circle we have a point which y = 1 (for the vertical distance of 2).  Solving at this point for x we get ±√35. The distance from the center of the torus to this point on the x axis would be R-√35.

Now, creating an isoceles triangle with long length 50, height of R-√35, and the two other legs of length R+r.  Using pathag. we can solve for R...25^2+(R-√35)^2=(R+r)^2

Solving for R I got ≈ 26.183

The volume of a torus is 2π^2r^2R ≈ 18606 mm^3

I think this is correct, a little hard to explain without pictures...

Edited on December 2, 2004, 7:25 pm
 Posted by Christian on 2004-12-02 18:35:03

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