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.
Good work!
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.
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