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Homework Help: Volumes question, volume of a torus

  1. Oct 23, 2011 #1
    Problem solved

    I'm not sure how easy this will be to understand without a diagram, but I don't know how to upload one :(

    1. The problem statement, all variables and given/known data
    Let a and b be constants, with a > b > 0.A torus is formed by rotating the
    circle (x - a)^2 + y^2 = b^2 about the y-axis.

    The cross-section at y = h, where –b ≤ h ≤ b, is an annulus. The annulus has
    inner radius x1 and outer radius x2 where x1 and x2 are the roots of
    (x - a)^2 = b^2 + y^2

    (i) Find x1 and x2 in terms of h.

    (ii) Find the area of the cross-section at height h, in terms of h.

    (iii) Find the volume of the torus.

    2. Relevant equations
    I think they're all up there.

    3. The attempt at a solution

    I did part (i) and got x1 = a - [itex]\sqrt{b^2 - h^2}[/itex]
    and x2 = a + [itex]\sqrt{b^2 - h^2}[/itex], which is correct according to the answers.

    I did part (ii) and got A = [itex]\pi[/itex]a[itex]\sqrt{b^2 - h^2}[/itex], which again is correct.

    But for part 3 I thought V = [itex]\pi[/itex]a[itex]\int^{b}_{-b}[itex]\sqrt{b^2 - h^2}dh[/itex] (sorry I don't know how to get rid of that itex in the integral)

    But in the answers it is V = [itex]\pi[/itex]a[itex]\int^{a}_{-a}[itex]\sqrt{b^2 - h^2}dh[/itex], and im not sure what it's between -a and a. The answers are from a different source then the question so I'm not sure if they have made a mistake or not.

    Any help will be much appreciated, thanks.
    Last edited: Oct 23, 2011
  2. jcsd
  3. Oct 23, 2011 #2
    I just remembered an easy way to find the volume of a torus and my answers was right, don't need help anymore.
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