Monday

July 28, 2014

July 28, 2014

Posted by **Ryan** on Monday, March 7, 2011 at 8:04pm.

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.

- Calculus -
**MathMate**, Monday, March 7, 2011 at 9:29pmYes, 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!

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