1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    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:


    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook