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Integral and volume

  1. May 16, 2006 #1
    Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer.

    y = x
    y = [tex]\sqrt{x}[\tex]
    rotate about y = 1

    [​IMG]

    =[​IMG]

    So, I am integrating with respect to x.
    Area = [tex]\int^1_{0}\pi[(f(x))^2-(g(x))^2]dx[\tex]

    I assumed that f(x) = x and g(x) = [tex]\sqrt{x}[\tex].

    However, the book gives f(x) = 1 - x and g(x) = 1 - [tex]\sqrt{x}[\tex].

    I don't understand how they got that.
     
    Last edited: May 16, 2006
  2. jcsd
  3. May 16, 2006 #2

    Curious3141

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    Homework Helper

  4. May 16, 2006 #3
    what axis are you rotating the function around and how does that affect the volumes you will get?

    A nicer working problem to go through, which might better illustrate the principals is this: Find the volume generated by rotating the area in between the two functions: y=1 and y=1/2 around the axis y=1. How do the two lines function when they are rotated?

    Note: You will be able to check your answer to this problem by comparing the answer with the volume of a cylinder.

    Curious: beat me to it I guess on the other thread.

    ~Lyuokdea
     
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