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Homework Help: Volumn by cross sections - solving in terms of which axis confusion?

  1. Feb 3, 2010 #1
    1. The problem statement, all variables and given/known data

    I need to find the volume of the solid generated by revolving this region bounded by the curves about the x-axis.

    y = sqrt(x), x+y=6, y=1


    2. Relevant equations



    3. The attempt at a solution

    I find the intersections of these curves I get:

    (1,1), (4,2), (5,1)..
    Then I see that to get the area of this cross section I need to do two sections from x = 1 to x = 4, and x = 4 to x = 5 if I integrated in terms of x..

    So I see that it will be easier to integrate in terms of y:

    I will get the area to be

    [tex]
    A(y) = \int_{1}^{2} [(-y+6)-(y^2)]dy
    [/tex]

    But now I am confused as to how I can formulate the integral to find the volume of this by revolving around the x-axis?

    I know I need to use the difference of the boundary as the radius then multiply by pi.. but I don't understand how to do that if I wrote the area in terms of y?
     
  2. jcsd
  3. Feb 3, 2010 #2

    Mark44

    Staff: Mentor

    Forget the integral you have. That just gives you the area of the region, which isn't what you want.

    Your typical volume element is a shell whose volume is 2* pi*radius*length*[itex]\Delta y[/itex]. For your problem radius is y, and length is (6 - y - y2). The limits of integration are the ones you found, y = 1 and y = 2.

    BTW, you should have put this into the Calculus and Beyond section, not the Precalcus section.
     
  4. Feb 3, 2010 #3
    So can I do this to find the volume?

    [tex]

    V = \int_{1}^{2} \pi[(-y+6)^2-(y^2)^2]dy

    [/tex]
     
  5. Feb 3, 2010 #4

    Mark44

    Staff: Mentor

    Reread what I wrote in post #2.
     
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