Volume of a Solid Revolved About X-Axis

[Zach Hughes]

I'm trying to practice for my final. The sample problem is:
"Find the volume of the solid generated when the region bounded by y = x⁴and y = x^1/3, 0<=x<=1, is revolved about the x-axis."

To start, I set the two y equations equal to each to find the points of intersection.
x⁴ = x^1/3, : raise both sides to power of 3
x¹², = x
x - x¹² = 0
So the intersection points that work are: x = 0, 1, -(-1)^5/11, and (-1)^6/11.

I believe I have now have to use the washer method to solve the problem.

With those multiple intersection points, it really confuses me on where to go next.

After looking through some examples in the text, I believe the integral I would have to evaluate to the find the correct volume is:
π ∫ (from 0 to 1) of (x^2/3 - x⁸) dx

But I don't know how to get to that from the given information and the points of intersection I found. Any tips would be greatly appreciated. Thanks.

 

[jedishrfu]

The two functions only intersect at zero and one right?

So just integrate the integral you have from zero to one.