Answer
$256\pi$
Work Step by Step
Given $$ y=x^2+1, y=9-x^2 ; \quad \text { about } y=-1 $$
First, find the intersection points
\begin{aligned}
x^2+1&=9-x^2\\
2x^2-8&=0\\
(x-2)(x+2)&=0
\end{aligned}
Then
$$x=-2,\ \ \ \ x=2 $$
To find the volume of the solid , use washer method
\begin{aligned}
V&= \pi \int_a^b[R^2(x)-r^2(x)]dx
\end{aligned}
where $R$ is the outer radius and $r$ the inner, here
$$R= 1+ 9-x^2=10-x^2,\ \ \ r= 1+ x^2+1=x^2+2$$
Then
\begin{aligned}
V&= \pi \int_{-2}^2[(10-x^2)^2-(x^2+2)^2]dx\\
&=\pi \int_{-2}^{2}(-24x^2+96)dx\\
&=\pi \left(-8x^3+96x\right)\bigg|_{-2}^2\\
&= \pi (-128+384)\\
&= 256\pi
\end{aligned}
