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Find Volume of solid Produced

  1. Jul 7, 2008 #1
    volume of the torus, by revolving around line x = 3 inside circle x^2 + y^2 = 4

    i got x = sqrt(y^2 - 4)

    and what would

    v = ?
  2. jcsd
  3. Jul 8, 2008 #2


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    The torus is a big pain to do by either the "disk/washer" method or the "cylindrical shell" method. Either way, you want to keep in mind that you are taking your radius measurement from x = 3, so you have to modify your result from the equation for the circle appropriately.

    It's a good idea to draw a picture of what you are planning to integrate, in order to get the integrand and limits set up correctly. You can also remove one headache by integrating only the upper half of the torus and then doubling the result.

    Have you had Pappus' Theorem? That makes very short work of a toroid volume problem like this one.
  4. Jul 8, 2008 #3
    we are not looking for the area though we are looking for the volume of the solid that would be produced. would doubling your results work seeing as it is like a doughnut, doesn't the outer region contribute more to the inner region, towards the volume?
  5. Jul 8, 2008 #4


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    I mean to integrate only the volume of the portion above the x-axis, then double it. (And I was talking about the volume.) Having written up a solution to a similar problem, I would say that the disk/washer method is less trouble to set up, but the integration for either isn't too hard.

    If you do use the washer method, you will need to use both signs of the square root from x to get the outer and inner radii.
  6. Jul 8, 2008 #5
    so this was a test question today and i didn't get it luckily it was only 8 out of ten problems andi omitted that one

    but i did attempt it and put

  7. Jul 8, 2008 #6


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    And that integral is the result from integrating the volume of the "upper half" of the torus, because the square root will only provide a positive height for the cylindrical shell (except that 'Y' should also be an 'x'). The radius of the shell is 3-x , the height above the x-axis is [tex]\sqrt{4-x^2}[/tex], so the complete volume integral is

    [tex]V = 2 \int_{-2}^{2} 2\pi (3-x) \sqrt{4-x^2} dx[/tex]

    (your choice of whether to attach the '2' to the square root or put it out in front to get the full height of the shells), which becomes

    [tex]V = 4\pi \int_{-2}^{2} (3-x) \sqrt{4-x^2} dx[/tex] .

    The integral will have two terms, but since the function [tex]x \sqrt{4-x^2} [/tex] has odd symmetry and we are integrating symmetrically about the y-axis, the second integral term is zero, reducing our labor to

    [tex]V = 12\pi \int_{-2}^{2} \sqrt{4-x^2}dx [/tex] ,

    which, after a trig substitution, gives us [tex]V = 12\pi \cdot 2\pi = 24\pi^{2} [/tex] .

    The washer method looks worse to set up, but the integral reduces to something similar. We would use [tex]x =\pm \sqrt{4-y^2}[/tex] and we want both signs. The inner radius of the washer will be 3 minus that square root, the outer radius, 3 plus that radical, and -- again integrating only the "upper half" of the torus -- the volume is

    [tex]V = 2 \int_{0}^{2} \pi [(3 + \sqrt{4-y^2})^2 - (3 - \sqrt{4-y^2})^2 ] dy[/tex] ,

    which algebraically simplifies to

    [tex]V = 2\pi \int_{0}^{2} 12\sqrt{4-y^2} dy = 24\pi \cdot \pi[/tex]

    I went and looked it up and now I understand why you asked me why I was talking about surface area. My apologies: I find that I should have said "Pappus' Second Theorem" or "Pappus' Centroid Theorem". (A lot of calculus books I've come into contact with never mention the First Theorem, so I forgot about that one...) The Centroid Theorem says the volume of the solid of revolution equals the area of the revolved closed curve times the circumference of the path covered by the curve's centroid. We have a circle of radius 2 and center at the origin being carried around the line x = 3, so the volume is just

    [tex]V = (\pi \cdot 2^2) \cdot (2\pi \cdot 3) = 24\pi^{2}[/tex]
    Last edited: Jul 8, 2008
  8. Jul 10, 2008 #7
    thank you so much i realized my error of my y being a x after submitting it...oh well...

    anyways i did only integrate the top half and multiply by 2 resulting in 4pi but i didn't take into consideration the odd symmetry (damn! that would make it so much easier)

    lol thanks for your help

  9. Jul 10, 2008 #8


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    I mentioned the odd symmetry because the circumstances permit applying it and because I really hate going to the trouble of working out an indefinite integral, only to find on evaluating it that the whole #(%@! thing cancels out!
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