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Volume of a function around an axis

  1. Nov 24, 2006 #1
    Given this problem:

    Find the volume and describe the shape of the object formed by the function f(x) = ½ x²+2 when the function is rotated around the x =3 axis and bounded by th region between the x = 0 and x = 4.

    I am not sure if I am thihnking about this correctly. I know in order to find the volume, I must use the formula 2π ∫Pxhx dx between x = 0 and x = 4. However, since it is rotatated around the x= 3 axis, do I use that as my starting point. Meaning, do I make 3 equal to zero and make the x boundaries as the difference from 3 to their points?
  2. jcsd
  3. Nov 24, 2006 #2


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    The axis of rotation is inside the region to be rotated so that the solid would intersect itself, did you copy the problem down right?
    Last edited: Nov 24, 2006
  4. Nov 25, 2006 #3
    Sorry! Never mind the question. I was looking at the problem wrong. The function is bounded by y = 0 and y = 4, not x. Now it is a piece of cake!
  5. Nov 26, 2006 #4
    I'm back and I've carefully checked to make sure I am reading it correctly.

    First I set function in terms of y because of its rotation around the verticle axis, which gives me
    y = sqrt(2x-2) Then I subtracted 3 because it revolves around x = 3. This is my radius, so I square it to get my area ( and multiply by Pi)

    which gives [sqrt(2x-2 - 3]^ 2 = 2x - 6sqrt(2x-2) - 13

    So my volume is π ∫ sqrt(2x-2) - 6sqrt(2x-2) - 13 dy from y = 0 to y = 4.

    I get [ x² -13x - 4sqrt(2x-2)^(3/2)

    Since the lowest point of this function is 2 on ther y axis, the volume should be the same whether I use the bounds of 2 to 4, or 0 to 4, but I keep gwetting differeent results for those two limits.
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