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Integral to find the volume of a torus

  1. Aug 25, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the volume of the torus formed when the circle of radius 2 centered at (3,0) is revolved about the y-axis. Use geometry to evaluate the integral.


    2. Relevant equations
    Formula for the semi-circle: [itex]y=\sqrt{4-x^2}[/itex]
    Solving for x give [itex]x=\pm\sqrt{4-y^2}[/itex]


    3. The attempt at a solution
    I was thinking of solving the integral from -2 to 2 using both the positive and negative sides, then evaluate the integral and then multiply by 2, but I'm not so sure about my equation.

    [tex]2\int_{-2}^2\pi\big( (3-\sqrt{4-y^2})^2 - (3+\sqrt{4-y^2})^2 \big) \mathrm d y[/tex]

    I just seem lost on this one :(
     
  2. jcsd
  3. Aug 25, 2013 #2
    I believe the equation is:

    [tex]V=\int_{-2}^2\pi\big( (3+\sqrt{4-y^2})^2 - (3-\sqrt{4-y^2})^2 \big) \mathrm d y[/tex]

    [tex]V=\int_{-2}^2\pi\cdot 12(\sqrt{4-y^2}) \mathrm d y[/tex]

    then trigonometric substitution...
     
  4. Aug 25, 2013 #3

    Zondrina

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    Edit : Got beat to it.
     
  5. Aug 25, 2013 #4

    mathwonk

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    the way to do this is to use pappus' theorem. the product of the distance traveled by the center of the circle by the area of the circle.
     
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