Answer
$\frac{1656}{5} \pi$
Work Step by Step
Given
$$ x=0, \quad x=9-y^2 ; \quad \text { about } x=-1$$
find the intersection points
\begin{aligned}
9-y^2&=0\\
(3-y)(3+y)&=0
\end{aligned}
then $$y=-3,\ \ \ y=3$$
To find the volume of the solid when the bounded region rotate about $ x=-1$.
\begin{aligned}
V&= \pi\int_a^b[R^2(y)- r^2(y)]dy
\end{aligned}
Here
$$R= 9-y^2+1= 10-y^2,\ \ \ r= 1 $$
Hence
\begin{aligned}
V&=\pi \int_a^b[R^2(y)- r^2(y)]dy \\
& =2 \pi \int_0^3\left[\left(10-y^2\right)^2-1\right] d y\\
&=2 \pi \int_0^3\left(100-20 y^2+y^4-1\right) d y \\
& =2 \pi \int_0^3\left(99-20 y^2+y^4\right) d y\\
&=2 \pi\left[99 y-\frac{20}{3} y^3+\frac{1}{5} y^5\right]_0^3 \\
& =2 \pi\left(297-180+\frac{243}{5}\right)\\
&=\frac{1656}{5} \pi
\end{aligned}
