consider the region bounded by the curves $y=-x^2+6x-8,y=0$ rotating the region about the line $y=-1$

Question

consider the region bounded by the curves $y=-x^2+6x-8,y=0$ rotating the region about the line $y=-1$

I know the inner radius is $1$, the outer radius is $-x^2+6x-7$,

but I'm not sure that is the $2<x<4$?

Answer 1

$2 < x < 4$ is indeed correct. How do I know this? Well I used a graphing calculator. But here's a way of doing it analytically.

You know that the curve is an upside down parabola. It will be cut off by the line $y=0$. The center of the parabola is easy, it's $$-\frac{b}{2a}=3$$ at which point the curve is above the $x$ axis. You can then imagine it slanting downwards on both sides, going through the $x$ axis. Therefore the part that is above the $x$ axis, which is the same as the part that we're going to be rotating, is the space between the two roots of the polynomial. Fortunately, $$-x^2+6x-8 = -(x-2)(x-4)$$ So that's easy. Beyond these limits, we are no longer inside the area that's bounded by the two lines.