$x = 0, x = 9 - y^2$ rotated about $x = -1$
I'm having a lot of trouble deciding whether to use the disc method or the shell method. Intuitively, it makes sense that the shell method would be simpler when you are rotating horizontally, like around the y-axis or x = -1.
I know that the shell method is:
$[circumference][height][width]$, where $C = 2\pi r$, $h = f(x)$, and $w = \Delta x$
I must find the radius of the solid about the line $x = -1$.
So, I see that the radius must be $y - (-1) \rightarrow r = y + 1$
Therefore, $C = 2\pi (y + 1)$, right?
The height varies with $f(y) = 9 - y^2$.
Here is where things get really muddled for me.