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Integral and volume

  1. May 16, 2006 #1
    Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer.

    y = x
    y = [tex]\sqrt{x}[/tex]
    rotate about y = 1

    [​IMG]

    =[​IMG]

    So, I am integrating with respect to x.
    Area = [tex]\int^1_{0}[(f(x))^2-(g(x))^2]dx[/tex]

    I can't figure out how to get f(x) and g(x). I would think that they are simply f(x) = x and g(x) = [tex]\sqrt{x}[/tex].

    The book gives f(x) = 1 - x and g(x) = 1 - [tex]\sqrt{x}[/tex]

    I don't understand how that works.
     
    Last edited: May 16, 2006
  2. jcsd
  3. May 16, 2006 #2

    Curious3141

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    I found the easiest way to do it is to demarcate the element of the volume of revolution.

    In this case, one element is like a small piece of a cylindrical annulus (which consists of a cylinder with a centered cylindrical hole burrowed within it).

    The cylindrical annulus has outer radius given by [tex](1 - y_1) = 1 - x[/tex] and inner radius given by [tex](1 - y_2) = 1 - \sqrt{x}[/tex]. The radii have the "one minus" because you are rotating about y = 1 and not the x-axis. From geometry you know that the volume of a cylindrical annulus is [tex]Volume = \pi(r_o^2 - r_i^2)h[/tex] where [tex]r_o[/tex] is external radius, [tex]r_i[/tex] is internal radius and [tex]h[/tex] is height. Apply the same logic to the element of volume here and you'll get the integrand below.

    The complete volume can be evaluated as the integral

    [tex]V = \int_0^1 \pi ((1 - y_1)^2 - (1 - y_2)^2) dx = \int_0^1 \pi ((1 - x)^2 - (1 - \sqrt{x})^2) dx[/tex]. The bounds are of course derived from seeing where [tex]x = \sqrt{x}[/tex]

    That is in essence the same thing your book has given, in this case, [tex]f(x) = (1-x)[/tex] and [tex]g(x) = (1 - \sqrt{x})[/tex]
     
    Last edited: May 16, 2006
  4. May 16, 2006 #3
    Thanks,

    I think I'm starting to understand why you get the 1 - x and 1 - [tex]\sqrt{x}[/tex]. Is it that you have to treat the radius as a function? Why can't you pretend that you are rotating over the origin and take the radius from there?

    I also don't understand why you have to subtract the function...in another example where you rotate over y = -1, you add 1 to the function.

    I don't understand the difference.
     
    Last edited: May 16, 2006
  5. May 16, 2006 #4

    Curious3141

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    You can't "pretend" you're rotating over the x-axis (you meant this when you said "origin" didn't you?), because the axis of rotation makes a big deal to the calculation. It's not a simple question of rotating about the x-axis then translating the volume. Just cut the required area at x = 1 and see how much the "curvey part" you're rotating differs from the "curvey part" you're rotating when you use the x-axis to rotate the object. There's no easy way to get from the solid of revolution formed from rotating about the x-axis to the one formed by rotating about y = 1.

    Here we're taking (1 - y) as the radius. Now when rotating about y = -1, you replace the "1" with "-1" giving you (-1-y).

    That seems different from (1+y) doesn't it? But when you square it to do the volume, you get the same thing : (-1-y)^2 = (1+y)^2. Simple, no?:smile:
     
  6. May 17, 2006 #5
    Ok, but what if you moved the function down 1 so that it is over the x-axis and then rotated it?

    letting f(x) = x - 1 and g(x) = [tex]\sqrt{x}[/tex] - 1.
     
  7. May 17, 2006 #6

    Curious3141

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    Yes, that would be fine. Verify that it works out the same way when you square it, e.g. (x-1)^2 = (1-x)^2.
     
  8. May 17, 2006 #7
    Thanks, I understand now!
     
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