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Homework Help: I can't believe I am not getting this, shell method

  1. Aug 14, 2011 #1
    1. The problem statement, all variables and given/known data

    Let's say I have the area bounded by [tex]y_1 = \sqrt{x}[/tex] and [tex]y_2 = x^2[/tex] in (0,1). Rotate that about the x-axis, find that volume of solid.

    3. The attempt at a solution

    Which one is right?

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

    [tex]\pi \int_{0}^{1} x^4 - x dx[/tex]

    I think the second integral is right because it uses washeres?

    If the second integral is wrong, what is the meaning of the first integral?
  2. jcsd
  3. Aug 14, 2011 #2
    The second is the correct one.
    The first maybe be.... something like a "cone" shaped solid whose radious is [itex]y_1(x)-y_2(x)[/itex]
  4. Aug 14, 2011 #3
    I think you have the signs of the terms in the second integral reversed. The square root of x is larger than x squared on the interval from 0 to 1. If you evaluate your integral, you get a negative volume.
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