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Integral and cylindrical shell

  1. May 17, 2006 #1
    Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the y-axis.

    y = [tex]x^2[/tex]
    y = 4
    x = 0

    0 <= x <= 2

    So I drew these:

    I don't know how to determine which region to use (above or below
    y = x^2 ). Either seems like it will work.
    Last edited: May 17, 2006
  2. jcsd
  3. May 17, 2006 #2


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    Think (as the question advised) in terms of cylindrical shells (elements). You are rotating about the y-axis. What is the radius of the cylinder? What is the height? Hence determine the volume of that element, and set up the integral.
  4. May 17, 2006 #3
    The region above y=x^2 is sort of in the shape of a bullet pointed straight down the y-axis, and the region below y=x^2 forms a sort of dog dish shaped object. Notice that x=2 is not a bound for the curve. Look at what does bound the curve and that is what you need to find the volume of.
  5. May 17, 2006 #4
    Actually, x = 2 is a bound because 0 <= x <= 2. Yeah, the book gives the area for the top part...but I don't see why it can't be the lower part.

    If it was the lower part, you still could use cylindrical shells, right?
  6. May 17, 2006 #5
    I think I see where you are getting confused...there is a difference between what x varies over, and what the bounds are.

    In this case x varies from 0 to 2. That is different from a bound. Suppose the curve y=x^2 was bounded by y=0, and x=1, but x varied from 0 to a billion? It doesn't much matter how x varies so long as the problem makes sense. In this case, x=2 is not a bound, in which case you have the curve y=x^2, y=4 and x=0 (y-axis is x=0) as bounds. Draw that region, then rotate it about the y-axis and you have your volume. Set up the integral and evaluate.
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