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Surface of revolution

  1. Nov 25, 2005 #1

    I am trying to find the surface area of a torus. I know there are equations
    for finding this but in this problem I must use a surface of revolution.

    Given in the problem is the distance to the center of the torus is 12 and
    the radius of the torus is 2.

    By using the equation for a circle I come up with the following
    equation for my surface of revolution,

    [tex] \[(x - h)^2 + (y - k)^2 = r^2\] [/tex]

    I replace x with 0 since I suppose it doesn't matter where on the
    x axis I revolve this surface. The equation becomes,

    [tex] \[x^2 + (y - 12)^2 = 2^2\] [/tex]


    [tex] \[x^2 + y^2 - 24y + 140\] [/tex]

    Solving for y,

    [tex] \[y = \pm \left( {\sqrt {4 - x^2 } - 12} \right)\] [/tex]

    A quick plot in Maple to show my surface of revolution,


    I am confused on where I go next because y is not a function of x,
    nor is x a function of y. Do I need have two different surfaces of
    revolution? The top half of the circle and the bottom half of the
    Last edited: Nov 25, 2005
  2. jcsd
  3. Nov 25, 2005 #2


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    Staff Emeritus
    Science Advisor

    Yes, use just what you have.
    But it is not
    [tex] \[y = \pm \left( {\sqrt {4 - x^2 } - 12} \right)\] [/tex]
    it is
    [tex] y = 12 \pm\sqrt {4 - x^2 } [/tex]
    One function is
    [tex] y = 12+ \sqrt {4 - x^2 } [/tex]
    and the other is
    [tex] y = 12-\sqrt {4 - x^2 }[/tex]
    Last edited: Nov 25, 2005
  4. Nov 25, 2005 #3


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    Homework Helper

    Fifty Ways To Find The Volume Of A Torus

    There is a perfect answer to your exact question: check this out,Fifty Ways To... Find The Volume Of A Torus, it starts on the bottom of pg 2, under the heading (3) Slip out the back, Jack.


    Vol(torus) = Vol(upper semi-circle rotated about the x-axis) - Vol(lower semi-circle rotated about the x-axis).

    Have fun :rofl: .
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