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Area of a Surface of Revolution

  1. Jul 26, 2006 #1
    I'm having more Calculus troubles here. Here's the problem:

    Write the definite integral that represents the area of the surface formed by revolving the graph of [tex]f(x) = 81 - x^2[/tex] on the interval [tex][0, 9][/tex] about the y axis; evaluate the integral to determine the surface area.

    By knowing that [tex]f(x) = 81 - x^2[/tex], [tex]f'(x) = -2x[/tex]. I then set up the integral for surface area, and I get this:

    [tex]2\pi \int_{0}^{9} (81 - x^2)\sqrt{1 + (-2x)^2} dx[/tex]

    Assuming I did that correctly, I can't figure out for the life of me how to evaluate the integral. Any ideas?

    Edit: I'm pretty sure I set up the integral wrong. I think it should be:

    [tex]2\pi \int_{0}^{9} (x)\sqrt{1 + (-2x)^2} dx[/tex]

    If so, I think I found the answer, and it comes out to be about 3068.
    Last edited: Jul 26, 2006
  2. jcsd
  3. Jul 26, 2006 #2
    Your second integral is correct.

    Do you know why the first is not?
  4. Jul 27, 2006 #3
    Well, the forumula for surface area, as far as I know, is

    [tex]2\pi \int_{a}^{b} r(x)\sqrt{1 + f'(x)} dx[/tex]

    where r(x) is the radius of the ring at the given x. 81 - x^2 isn't the radius; x is. Is that the correct reasoning?
  5. Jul 27, 2006 #4
    Yep, that's the reason. :)
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