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Volume: disc method

  1. Aug 19, 2009 #1
    1. The problem statement, all variables and given/known data

    1. Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the indicated lines.

    [tex] y = \sqrt{x}, y = 0, x = 4[/tex]

    the line x = 6


    2. Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line y=4.

    [tex] y = \frac{1}{x}, y = 0, x = 1, x = 4[/tex]


    2. Relevant equations



    3. The attempt at a solution

    1.

    [tex] V = \pi \int^{\sqrt{6}}_{0} ((6-y^{2})^{2} - (6-4)^{2}) dy[/tex]
    [tex] = \pi \int^{\sqrt{6}}_{0} (36 - 12y^{2} + y^{4} -4) dy [/tex]
    [tex] = \pi \int^{\sqrt{6}}_{0} (32 - 12y^{2} + y^{4}) dy [/tex]
    [tex] = \pi [32y - 4y^{3} + \frac{y^{5}}{5}]^{\sqrt{6}}_{0} [/tex]
    [tex] = \frac{76\sqrt{6}\pi}{5}[/tex]

    I am relatively sure that I integrated the expression I set up correctly, because I checked it with my ti-89. That said, that means the original integral I set up is incorrect. I'm not sure why.

    2.

    [tex] V = \pi \int^{4}_{1} (4^{2} - \frac{1}{x^{2}}) dx [/tex]
    [tex]= \pi \int^{4}_{1} (16 - \frac{1}{x^{2}}) dx [/tex]
    [tex]= \pi [16x + \frac{1}{x}]^{4}_{1} [/tex]
    [tex]= \pi [(64 + \frac{1}{4)}) - (16+1)][/tex]
    [tex] = \pi (\frac{257}{4} - 17)[/tex]
    [tex] = \frac{189\pi}{4}[/tex]

    Again, checked with my calculator and I evaluated integral correctly. Set up went wrong.
     
  2. jcsd
  3. Aug 19, 2009 #2
    For the first one, check the limits of integration. What values of y does the region cover?

    EDIT: I can't see anything wrong at first glance with the second one. What is the answer you were given?
     
  4. Aug 19, 2009 #3
    Answers for problems:

    1. [tex]\frac{192 \pi}{5} [/tex]
    2. [tex] \pi (8 ln 4 - \frac{3}{4}) = 32.49 [/tex]

    Oh, should the upper limit be 2?
     
  5. Aug 19, 2009 #4
    Yep!

    And as for the other one, it should be (4-1/x)^2, not 1/x^2.

    EDIT: As a quick LaTeX hint, type 'ln' as '\ln' as in [tex]\ln x[/tex].
     
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