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Homework Help: Surfaces of Revolution

  1. Oct 11, 2007 #1
    Consider the ellipse:

    [tex] (\frac{x}{2})^2 + y^2 = 1 [/tex]

    We rotate this ellipse about the x-axis to form a surface known as ellipsoid. Determine the area of this surface.

    Start off by solving for y.

    [tex] y = \sqrt{1-\frac{x^2}{4}} [/tex]

    Then find the derivative.

    [tex]y' = \frac{-x}{2\sqrt{4-x^2}} [/tex]

    Then plug into the formula for surface of revolution.

    [tex] S = \int 2\pi y \sqrt{1+(\frac{dy}{dx})^2} dx [/tex]

    [tex] 2 \pi \int \sqrt{1-\frac{x^2}{4}} \sqrt{1+(\frac{-x}{2\sqrt{4-x^2}})^2} dx [/tex]

    Plenty of simplifications later yields

    [tex] \frac{\pi}{2} \int \sqrt{16-3x^2} dx [/tex]

    Now I haven't found that anti-derivative yet but just looking at it tells you its going to be extremely ugly. But all I remember my professor showing us in class was beautiful little problems that come out to [tex]\int 4x dx [/tex] or something simple like that. So that makes me think I'm wrong.

    Any help?
  2. jcsd
  3. Oct 11, 2007 #2


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    Well I found a formula for it but I am not really sure how to show the working for it. But here is the formula

    [tex]\int \sqrt{a^2-x^2} dx =\frac{1}{2}(x\sqrt{a^2-x^2}+a^2sin^{-1}(\frac{x}{a}))+ c[/tex]
  4. Oct 11, 2007 #3
    I trust you that it works but we haven't been taught to use that formula yet, so I'm pretty sure he wouldn't have us use it.

    I'm thinking I got something wrong in getting from the initial plug in to the end simplification.

    Thanks tho =P
  5. Oct 11, 2007 #4


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    AH seems that I should actually work out the problems more...when I did it...the end result was less complicated than what you had...you did some wrong algebra in there
  6. Oct 11, 2007 #5
    Hrmm...any hints? Been through it a few times and all I have changed is a positive to a negative and a negative to a positive. Haven't found anything to make it easier.
  7. Oct 11, 2007 #6


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    Well it seems i made a mistake..twice...i keep getting what you initially had..the only help i can suggest is

    [tex]\frac{\pi}{x}\int \sqrt{16-(\sqrt{3}x)^2} dx[/tex]

    Let [tex](\sqrt{3})x=4sin\theta[/tex] and work from there....but since it is a surface area of revolution I would expect some limits so then it would be easy from here
  8. Oct 11, 2007 #7
    So after playing with it and using that substitution I get to
    [tex]8\pi + \frac{4\pi(sin^-^1(3/2)-sin^-^1(-3/2))}{3}[/tex]

    Can't help but think I got something wrong in there.
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