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Volume using integral

  1. Dec 22, 2011 #1
    1. The problem statement, all variables and given/known data
    The planar region bounded by [itex]y = x, y = \sqrt{x}[/itex] is rotated about the line x = -2.
    Find the Volume.

    2. Relevant equations
    [itex]V = 2\pi\int_{0}^{4} R dA[/itex]


    3. The attempt at a solution

    Solution:
    y = -2
    (-2)^2 = x
    x = 4
    y^2 = +- 2
    the point of intersection should be (0,0) and (4,2)

    now im using the cylindrical method in obtaining the volume
    [itex]V = 2\pi\int_{0}^{4} R dA[/itex]
    where dA = (y2-y1)dx
    for
    y2 = sqrt(x)
    y1 = x
    and
    radius be r = x+2 since the revolved line is at the 2nd quadrant and x in the 1st quadrant and it needs to be added
    [tex]V = 2\pi\int_{0}^{4} (x+2)(y_{2}-y_{1}) dx [/tex]

    [tex]V = 2\pi\int_{0}^{4} (x+2)(\sqrt{x}-{x}) dx [/tex]


    V = -(416/15)pi ????
     
    Last edited: Dec 22, 2011
  2. jcsd
  3. Dec 22, 2011 #2

    Mark44

    Staff: Mentor

    Your description is not as clear as it could be. I'm assuming that you mean the region in the first quadrant that lies between the two curves.
    (0, 0) yes, but (4, 2) no. That point is not on the line.

    Why did you start off with y = -2? The line the region is being rotated around is x = -2.
    The answer should be a positive number.
     
  4. Dec 22, 2011 #3
    [itex]V = 2\pi\int_{0}^{1} (x+2)(\sqrt{x}-x)dx[/itex]

    so the answer is [itex]\frac{4\pi}{5}[/itex]?
     
  5. Dec 22, 2011 #4

    Mark44

    Staff: Mentor

    That's what I get.
     
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