Method of cylindrical shells

  • Thread starter ktpr2
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  • #1
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I know one can figure the volume of a torus by the difference of two volumes, but I'm trying out the method of cylinderical shells. As far as i understand, you can often create a primitive with a calcuable volume and approximate the volume of the shape you wish by scaling the primitive along the curve it creates, adding infintely many times.

The problem is that my answer is wrong when i try to set up an integral when thinking in terms of cylinderical shells:

We have a rectangle, bent in the shape of a circle, with length [tex]2 \pi r[/tex] height [tex]\sqrt{1-x^2}[/tex] and width[tex] \Delta x[/tex], so it's volume should be all that multiplied together.

I have torus radius (this torus is just a circle, really) of R and the circle being revolved has a radius of r. So my integral is:

[tex]\int_{R-r}^{R+r} 4 \pi x \sqrt{1-x^2} \Delta x[/tex]

and the above is off by [tex]r^3 R[/tex] when i use differences of two volumes, what conceptual flaw am I making?
 

Answers and Replies

  • #2
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You should better define your intended torus, be more specific with which radius is which, and what you mean by rectangle, the length, and the height with respect to.

http://whistleralley.com/torus/torus.htm does a good explanation of the general integration of a torus.

If you wanted to try again, an easier way to explain your situation (even for yourself) is to refer everything to the coordinate plane, i.e.

A line drawn from the origin along the x axis to a point (R,0) is the center of a circle with radius r. The volume of the torus created by rotating the figure described is given by integrating the area of the torus along the path of revolution... etc etc
 
  • #3
192
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Hmm, i'll see if casting things in that light simpifies things.
 

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