Answer
$ V=256 \pi $
Work Step by Step
{Step 1 of 5}:
First, find the points of intersection of the curves $ y=x^{2}+1 $
and $ y=9-x^{2} $
$ x^{2}+1=9-x^{2} $
$ 2x^{2}=8 $
$ x^{2}=4 $
$ x= \pm 2 $
The points of intersection are (-2, 5) and (2, 5).
{Step 2 of 5}:
Sketch the curves.
{Step 3 of 5}:
Rotate this shaded region about $ y=-1 $
Sketch the solid obtained after rotation and a washer.
{Step 4 of 5}:
The outer radius of the washer $ 1+ \left( 9-x^{2} \right) =10-x^{2} $
The inner radius of the washer $ 1+ \left( 1+x^{2} \right) =2+x^{2} $
The cross-sectional area of the washer
$ A \left( x \right) = \pi \left[ \left( 10-x^{2} \right) ^{2}- \left( 2+x^{2} \right) ^{2} \right] $
$ = \pi \left[ 100+x^{4}-20x^{2}-4-x^{4}-4x^{2} \right] $
$ = \pi \left[ 96-24x^{2} \right] $
{Step 5 of 5}:
The volume of solid
$ V= \int _{-2}^{2} \pi \left[ 96-24x^{2} \right] dx $
$ = \pi \int _{-2}^{2} \left( 96-24x^{2} \right) dx $
$ = \pi \left[ 96x-\frac{24}{3}x^{3} \right] _{-2}^{2} $
[By the fundamental theorem of calculus, part 2]
$ = \pi \left[ 96 \times 2-8 \times 8+96 \times 2-8 \times 8 \right] $
$ = \pi \left[ 384-128 \right] $
$ V=256 \pi $
