1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Homework Helper

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?