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Cylindrical shells to find volume of a torus

  1. Mar 2, 2014 #1
    1. The problem statement, all variables and given/known data
    Use cylindrical shells to find the volume of a torus with radii r and R.
    9jZwhrS.png


    2. Relevant equations
    V= ∫[a,b] 2πxf(x)dx
    y= sqrt(r2 - (x-R)2)


    3. The attempt at a solution
    V= ∫ [R, R+r] 2πx sqrt(r2 - x2 - 2xR + R2) dx

    I feel like this isn't going in the right direction, though. I found a different post about the same question, but that person was using disks, and the volume was 2π2r2R, and I can't see where the π2 would come from in my equation, so I think I'm going about this incorrectly. Any help is much appreciated, thanks.
     
  2. jcsd
  3. Mar 2, 2014 #2

    Simon Bridge

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    It looks like your integrand only includes the top half of the shell.

    consider: the shell at radius x: R-r<x<R+r, is dx thick 2πx around and h(x) tall.
    ... so it's volume is dV= 2πxh(x)dx

    Sketch it in - label h, then label y, and see how they are related.

    Apart from that - keep going. You'll probably need a trig substitution.
    Maybe x-R = r.sinθ ?
     
    Last edited: Mar 2, 2014
  4. Mar 3, 2014 #3
    Would h be 2y, since the height of the cylinder depends on the curve of the torus? So for any point on the torus, where the circular cross section lies in the xy plane, y=sqrt(r2-(x-R)2). Then the cylindrical shell would dip above and below the x axis, so the height should be 2y, right?

    LpolaIl.png
    gp7CeCg.png
    Like this, if that helps clarify at all.
     
  5. Mar 3, 2014 #4

    SammyS

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    That looks good to me. It appears to agree with Simon's comment.
     
  6. Mar 3, 2014 #5

    Simon Bridge

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    That's what I was thinking - nice pics BTW.
    There is no arguing with a decent sketch - people used to say "God is a geometer".

    So now you have the volume of the cylindrical shell radius x: R-r<x<R+r, is $$dV = 4\pi x \sqrt{r^2-(x-R)^2}\; dx = f(x)\; dx$$... now it is just a matter of adding up all the dV's.
    Since these are infinitesimal thickness shells, the summation sign is replaced by an integral over appropriate llmits. $$V=\int_V \; dV = \int_{a}^{b} f(x)\; dx$$ ... writing that out, with correct numbers for a, b, and f(x), is the next step - then you can use a substitution.
     
  7. Mar 4, 2014 #6
    Alright, I substituted u=x-R and solved far enough to find a shortcut using the area of a semicircle, and then evaluated to get 2π2Rr2. Thanks!
     
  8. Mar 5, 2014 #7

    Simon Bridge

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    I'd have gone right to x-R=r.sin(u).
    But well done anyway.

    Notice that ##A=\pi r^2## is the area of the crossection circle and ##C=2\pi R## is the circumference at the center. This means the volume of the torus turned out to be V=AC ... which is the volume of a cylinder radius r and height C. Does that make sense?
     
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