Use the shell method to find the volume of the rotated region...

howsockgothap

1. The problem statement, all variables and given/known data
Use the shell method to find the area of the resulting shape when the region bounded by y=4x-x² and y=0 is rotated about the line x=5.


2. Relevant equations
2π ∫ (shell radius)(shell height) dx (from x=a to b)



3. The attempt at a solution
I know the region I am rotating, and I know that h=4x-x²

My problem comes when I try to define the limits of integration and the radius. I don't really understand how to incorporate that the region is being rotated around a line other than the y-axis

Mark44

Have you drawn a sketch of the cross-section of the solid you get? Part of the cross-section is above [0, 4]. What interval is the other part of the cross-section above?

What do you get for the volume of a typical shell?

howsockgothap

the volume of a typical shell is 2pi*r*h... and the cross section of this shape extends from 0 to 4 and then from 6 to 10... I don't see how that really helps. The radius confuses me for this mostly because for every y value there's two x values, so it doesn't seem to make sense to set it up as 5-x=r

Mark44

That's the area of an unrolled shell, not the volume. Also your formula is very general and not specific to your problem.

For the function in your problem, what is r? what is h? And what is the thickness of a typical shell? You don't have that in your formula.


I would use the interval [6, 10]. That way the radius is r = x - 5, which is a positive number. The cross-section view includes two intervals, but for your integration limits, you need to use only one of them. Each shell has a cross-section that is in [0, 4] and another that is in [6, 10], but it's still only one shell. Again, I would use the interval [6, 10] If you use the interval [0, 4], then r = 5 - x, which is positive for all values of x in [0, 4].

howsockgothap

Oops, yes, my mistake, the area is 2pi*r*h*thickness... if I were using the interval from 0 to 4, I guess I would use 2pi∫(5-x)(4x-x²)dx whereas for the interval from 6 to 10 I would use 2pi∫(x-5)(4x-x²)dx.... does that seem right?

Mark44

Yes, both will work.