Determining the volume of a function ratating around an axis.

[beuller34]

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!

 

[cronxeh]

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