Use the Shell Method to compute the volume of the solid obtained by rotating the region bound by the graph $4y=4^2-x^2$ and the $x$-axis about the line $x=6$.
Hello, I've been having a lot of problems with this question. I feel fairly certain about what I am doing, but the answer is not correct. I'm not really looking for the answer, but rather help in determining where I am going wrong. Currently I'm working with the idea that the radius is $(6-x)$ and the height is $16-x^2$. I'm using the shell method so my area before integrating is $2\pi(6-x)(16-x^2)$, and the bounds being $0$ and $4$.
The answer I keep getting is $384\pi$, which is incorrect.
My guess is I'm making a mistake with the radius, but everything I'm reading points to it being correct. Any help would be appreciated.