Solids of revolution - What if the axis crosses the volume?

[Jani08]

Homework Statement

Here is the problem :

Find the volume of the solid generated when the area bounded between the following functions rotates along the y=4 line.

Functions: y=x^3 y=4x

Homework Equations

Functions: y=x^3 y=4x
Axis of rotation: y=4

The Attempt at a Solution

Points of intersection are x=-2,0,2
I tried making 2 integrals from -2 to 0 and 0 to 2, but I just ended up with a negative volume.
What I don't understand is that the axis of rotation crosses the area bounded these two lines make. How do I go about this? Never encountered thissituation. Maybe I am setting up the problem wrong...

 

[LCKurtz]

The curves do not intersect when x = -2. Show us your integral. Did you draw a picture?

 

[Jani08]

Oh I am sorry it is y=x^3 not y=x^2, yeah I drew a picture but haven't scanned it.

 

[LCKurtz]

If the function is in fact x³, you are correct that the axis of rotation passes through the region. I suspect a misprint in your text because of that. In fact, the problem would make perfect sense if the function was x², which is certainly a likely candidate for what the problem was supposed to be, especially given the limits and the axis of rotation you have.

 

[Quinzio]

I agree with LCK, but what if we wanted to go on with the original problem ?
The volume would be the union of the two volumes separately generated by the two areas (above and below the line).
Then I'd mirror one area over the other taking the line as pivot, then make the union of the areas and proceed with rotation