1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Area of a Surface Revolution

  1. Aug 1, 2006 #1
    Hello

    can someone please give me some advice on applying the area of a surface revolution. I never seem to plugin the right formula.

    In the formula S(from a to b) 2pi f(x)**sqrt(1 + f'(x)^2) I never know what to put for f(x)** some times the answers have just x and sometime it has the function of f(x).[/quote]
    I don't know what you mean by "some times the answers have just x and sometime it has the function of f(x)." What does the problem itself have? In order to find the surface area of a "surface of revolution", you have to know what is being revolved! If you are given a function y= f(x), that is then revolved around the x-axis to form a 3 dimensional figure (the "surface of revolution"), then the surface area is given by that formula:
    [tex]2\pi \int_a^b f(x)\sqrt{1+ f'^2(x)}dx[/tex].

    The "f(x)" has to be given as a description of the curve, y= f(x), being rotated around the x-axis.

    You probably know that the surface area of a cylinder, of radius r and length L is given by [itex]2\pi r L[/itex]: that's really the circumference of the circle times the length.

    If y= f(x) is rotated around the x-axis, then each point describes a circle of radius f(x) and so circumference [itex]2\pi f(x)[/itex]. The length of a tiny ("infinitesmal") cylinder is given by the arclength formula: [itex]L= ds= \sqrt{1+ f'^2(x)}dx[/itex]. So the differential of area is given by [itex]dA= 2\pi f(x)\sqrt{1+ f'^2(x)}dx[/itex]. You integrate that to get the surface area: [itex]A= \int_a^b2\pi f(x)\sqrt{1+ f'^2(x)}dx[/itex].

    HOWEVER, and this may be what is causing your confusion, if the graph of y= f(x) is rotated around the y-axis rather than the x-axis, each point follows a circle of radius x, not f(x). That circle has radius x, the distance from the point to the y-axis, not f(x), the distance from the point to the x-axis.

    If you are given y= f(x), its graph being rotated around the x-axis, then the surface area is given by
    [tex]2\pi\int_a^b f(x)\sqrt{1+ f '^2(x)}dx[/tex]

    If you are given y= f(x), its' graph being rotated around the y-axis, then the surface area is given by
    [tex]2\pi\int_a^b x \sqrt{1+ f '^2(x)}dx[/tex]
     
    Last edited by a moderator: Aug 1, 2006
  2. jcsd
  3. Aug 1, 2006 #2

    siddharth

    User Avatar
    Homework Helper
    Gold Member

    This post is confusing. Haven't you answered your own question?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Area of a Surface Revolution
Loading...