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Cylindrical Rotation Volume Help

  1. Jan 2, 2008 #1
    1. The problem statement, all variables and given/known data
    The region R enclosed by the coordinate axes and the graph of y = k(x-2)^2 is shown above. When this region is revolved around the y - axis, the solid formed has a volume of 8*pi cubic units. What is the value of k?

    A) 1
    B) 4/3
    C) 3/pi
    D) 2
    E) 3


    2. Relevant equations

    V = 2*pi f a -> b x [ f(x) - g(x) ] dx


    3. The attempt at a solution

    Ok so I pluged in 8 pi in for volume and the function for f(x) and o for g(x) and tried solving it.
    After simplifying I got 4 = 8/3k
    and for k I got 12/8

    But that is not an answer choice.

    Can I get some help, I would really appreciate it!
    Thank You very much!!
     
  2. jcsd
  3. Jan 2, 2008 #2
    Using the shell method, volume is

    [tex]V = 2\pi\int_a^b x * f(x) dx[/tex]

    plugging in y=f(x) and the bounds determined by the x and y intercepts
    [tex]= 2\pi k \int_0^2 x * (x-2)^2 dx [/tex]

    [tex]= 2\pi k \int_0^2 x^3 - 4x^2 + 4x dx [/tex]

    [tex]= 2\pi k \left[ \frac{x^4}{4} - \frac{4x^3}{3} + 2x^2 \right]_0^2 [/tex]

    [tex]= 2\pi k \left[ 4 - \frac{32}{3} + 8 \right][/tex]

    [tex]= \frac{8}{3}\pi k[/tex]

    since we know the volume is [tex]8\pi[/tex]

    [tex]8\pi = \frac{8}{3}\pi k[/tex]

    [tex]k = 3[/tex]
    bam
     
  4. Jan 2, 2008 #3
    Ohh, thanks, yeah I made a calculation error, Thanks!!!
     
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