TITLE: Calculus 2 Shell Method
QUESTION [8 upvotes]: I have tried to use the standard formula for shell method and when I used this formula, the answer that I produced was incorrect. I do not know if it was a mathematical error or a formulaic error. If anybody could help with this problem, it would be greatly appreciated.

REPLY [2 votes]: Draw a picture. The curves meet at $x=2$ (and $x=-2$, but that is irrelevant). 
Look at a slice of width "$dx$" going from $x$ to $x+dx$. This is roughly at distance $x$ from the $y$-axis. So the radius of the cylindrical shell is $x$.
The height of the cylindrical shell is $(8-x^2)-x^2$, so the volume of the shell is approximately $2\pi x(8-2x^2)\,dx$. "Add up" (integrate) from $x=1$ to $x=2$. Our volume is
$$\int_1^2 2\pi x(8-2x^2)\,dx.$$
The rest is mechanical. Simplify the integrand to $(4\pi)(4x-x^3)$.