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Determining the volume of a function ratating around an axis.

  1. Jan 25, 2010 #1
    Hi all. I am a recent Chemical Engineering graduate, and I recently was program recommended into the Naval Nuclear Engineering Program. While going over a study guide for my technical interviews coming up, I came across this problem, and my mind went blank! It goes something like this:

    1. Plot both y=x^2 and y=x^3 on the same graph, but only to 1.
    2. Determine the area inbetween both functions using integration.
    3. Imagine this space inbetween the functions was rotated about the x-axis. What will the volume of this region be.

    While both parts 1. and 2. are quite trivial, I couldn't seem to figure out the the last part of the problem! Any help would be greatly appreciated. Thanks!
     
  2. jcsd
  3. Jan 25, 2010 #2

    cronxeh

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    Gold Member

    The cross section you are interested in is a washer. Your inner radius is made by x^3, and your outer radius is made by x^2. The area is Pi*radius^2. So inner area is Pi*(x^3)^2, and outer area is Pi*(x^2)^2. The difference is the area you are interested in. A(x) = outer area - inner area = Pi*(x^4-x^6)

    The volume of rotation around x-axis is V= int(A(x)dx,x,0,1) = (2*pi)/35
     
    Last edited: Jan 25, 2010
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