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Find the volume of the solid

  1. Jun 8, 2006 #1
    Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis:
    x2 - y2 = a2, x = a + h (where a > 0, h > 0); about the y-axis.

    I found the area in terms of y:
    [tex]A(y) = \pi(2ah + h^2 - y^2)[/tex]

    and the line x = a + h intersects hyperbola at:
    [tex] y = \pm\sqrt{2ah + h^2} [/tex]

    Thus, the volume is:
    [tex] V = 2\pi \int^{\sqrt{2ah + h^2}}_{0} (2ah + h^2 - y^2) dy [/tex]
    I simplify this to
    [tex] V = 2\pi (2ah + h^2)^{3/2}\frac{4}{3} = \frac{8}{3}\pi (2ah + h^2)^{3/2} [/tex]
    however, the answer is not 8/3 pi (2ah + h^2)^(3/2), but 4/3 pi (2ah + h^2)^(3/2). I'm not sure why I have that extra factor of 2 there.... Originally, I factored out a 2 so that I would integrate from 0 to [tex]\sqrt{2ah + h^2}[/tex], instead of [tex]-\sqrt{2ah + h^2}[/tex] to [tex]\sqrt{2ah + h^2}[/tex].
    Last edited: Jun 8, 2006
  2. jcsd
  3. Jun 8, 2006 #2


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    Homework Helper

    It looks like you made a mistake evaluating the integral. That should be a 2/3, not 4/3.
  4. Jun 8, 2006 #3
    hmm.. you're right. I thought I checked my solution over .. but I missed a sign change from - to +. thanks!
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