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I Surface of revolution of a donuts

  1. Nov 15, 2016 #1
    HELP I can't find the surface of revolution!! By donuts I mean a circle that doesn't touch the axes (tore in french)

    y^2+(x-4)^2=2^2 is my function ( y^2+x^2=r^2) and the axe of rotation is y

    so y= sqrt(r^2-x^2)
    the formula I know :
    2* pi (Integral from a to b (F(x)*sqrt( 1+ (f``(x))^2))

    • 1) what are the bornes of the integral and how did you find them
    • 2)where do you go from there to have the result of 32pi^2
  2. jcsd
  3. Nov 15, 2016 #2
    It is necessary to state what you are trying to find. People can guess, of course, but it would be better for you to tell us. The phrase "surface of revolution" refers to a subset of 3-dimensional space R3.

    But maybe you are trying to determine the area of that surface?

    Also, when using notation like f(x) and F(x) it is important to say what these functions are supposed to represent, and what their domain is.
  4. Nov 15, 2016 #3
    yes the area of that surface! hahah sorry I though it was the right way to say it in English sorry..!!!

    and there is no further info in the problem the question is:
    calculate the area of the surface if you rotate a circle y^2+(x-4)^2=2^2 around the vertical axe (x=0)

    and again sorry I meant f(x) and f ' (x)!!!!!!!!!!!!!

    2* pi (Integral from a to b (f(x)*sqrt( 1+ (f `(x))^2))
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