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Math Help - find surface area of a donut

  1. #1
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    find surface area of a donut

    I have a problem where i dont even know where to begin. Usually, when i post problems here i have something to show but this time i dont. Im sorry for that.


    here is the problem
    :

    FInd the surface area of the torus (or donut) obtained by rotating the circle (x - b)2 + y2 = a2; (a < b) about the
    y-axis.


    would i do something like the Area of a surface of revolution theorem?:

     S = 2 \pi \int_a^b  f(t) \sqrt {(dx/dt)^2 + (dy/dt)^2} dt

    if so i am not even sure what to do to get it into the form to put it into that formula?

    Any help is appreciated since i am lost!!

    Thank you
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: find surface area of a donut

    Quote Originally Posted by icelated View Post
    FInd the surface area of the torus (or donut) obtained by rotating the circle (x - b)2 + y2 = a2; (a < b) about the y-axis.
    Use the First Guldin Theorem: the surface area S of a surface of revolution generated by the revolution of a curve about an external axis is equal to the product of the arc length s of the generating curve and the distance d traveled by the curve's geometric centroid \bar{x} .
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  3. #3
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    Re: find surface area of a donut

    Quote Originally Posted by FernandoRevilla View Post
    Use the First Guldin Theorem:
    Im not suppose to use pappus's theorem. When i googled First Guldin Theorem everything came back to pappus. What is First Guldin Theorem? =(
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