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Surface Area

  1. Mar 16, 2004 #1
    i've been stuck on this problem for an hour now. how do you find the surface area for y=(x)^1/2 when 0_<x_<2, rotating about the x axis?
  2. jcsd
  3. Mar 16, 2004 #2
    You have a nice little formula for this kind of thing.

    [tex]SA = \int_{x_i}^{x_f} 2\pi y(x) \sqrt{1 + \Big(\frac{dy}{dx}\Big)^2}\,dx[/tex]

    You'll recognize that this is [itex]2\pi y(x)[/itex] times the formula for the arc length. Basically what you're doing is adding up a bunch of little rings of radius y and length an infintesimal piece of the arc length. Because the radius is y, and we know that circumference = [itex]2\pi r[/itex], that's where the [itex]2\pi y(x)[/itex] comes from. The "width" of the ring depends on y and x and can be made into a triangle of base 1 and height dy/dx.

    Last edited: Mar 16, 2004
  4. Mar 16, 2004 #3
    but how do you figure out what the dx part is? that's the part that i don't understand.
  5. Mar 16, 2004 #4
    dx? dx is dx. It's necessary to evaluate the integral.

    Do you mean dy/dx? You have to integrate y with respect to x to get dy/dx.

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