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Homework Help: Finding Area Underneath Variable Curve

  1. Sep 23, 2010 #1
    1. The problem statement, all variables and given/known data
    The region enclosed by [PLAIN]http://latex.codecogs.com/gif.latex?y^2%20=%20kx [Broken] [Broken] and the line [URL]http://latex.codecogs.com/gif.latex?x%20=%20\frac{1}{4}k[/URL] is revolved about the line [URL]http://latex.codecogs.com/gif.latex?x%20=%20\frac{1}{2}k.[/URL] Use cylindrical shells to find the volume of the resulting solid. (Assume k > 0)



    2. Relevant equations

    [URL]http://latex.codecogs.com/gif.latex?V%20=%202\pi\int_{a}^{b}xf(x)[/URL]

    3. The attempt at a solution

    I started by trying to find the interval. I've managed to come up [URL]http://latex.codecogs.com/gif.latex?b=\frac{1}{4}k,[/URL] but I'm unsure of how to find a. In addition, because the area is being revolved around a vertical axis, I changed [PLAIN]http://latex.codecogs.com/gif.latex?y^2%20=%20kx [Broken] [Broken] to [URL]http://latex.codecogs.com/gif.latex?y=\sqrt{kx}.[/URL]

    Here is my current equation:

    [URL]http://latex.codecogs.com/gif.latex?V%20=%202\pi\int_{a}^{\frac{1}{4}k}(\frac{1}{2}k-x)(\sqrt{kx})[/URL]


    I need to find a. Also, I would appreciate if someone told me if I've taken the right steps so far in solving the problem.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 24, 2010 #2
    The domain of [tex] y^{2}=kx [/tex] is [tex] [0 , \infty] [/tex], so x=0 would have to be your lower limit.
     
  4. Sep 24, 2010 #3
    Your problem also looks correct so far, though you may consider whether they are asking about [tex]y=\sqrt{kx}[/tex] or [tex] y^{2}=kx [/tex]. The region between [tex] y^{2}=kx[/tex] and [tex]x=\frac{1}{4}k[/tex] is larger than the region used in your integral.
     
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