Posted by Ryan on .
Use the shell method to set up, but do not evaluate, an integral representing the volume of the solid generated by revolving the region bounded by the graphs of y=x^2 and y=4x-x^2 about the line x=6.
I had the shell radius as (6-x) and the shell height as (4x-2x^2).
My final integral was 2 pi * integral from 0 to 2 of [(6-x)(4x-2x^2)]. I was just wondering if I did this correct? Thanks.
Yes, your expressions for the radius and the height are correct. The expression to integrate as well. The shell method evaluates basically ∫2πr(x)h(x)dx. The limits have not been stated in the question, but I suppose they are from x=0 to x=2, as you put it.