Answer
$ V=\frac{1656 \pi }{5} $
Work Step by Step
{Step 1 of 5}:
First, find the points of intersection of the curves $ x=0 $ and$ x=9-y^{2} $
To do this, set $ 9-y^{2}=0 $ or $ y^{2}=9 $ $ y= \pm 3 $
The y-coordinates of the points of intersection are -3 and 3.
{Step 2 of 5}:
Sketch the curves.
{Step 3 of 5}:
Rotate this shaded region about $ x=-1 $
Sketch the solid and the washer.
{Step 4 of 5}:
The outer radius of the washer is $ \left( 9-y^{2} \right) +1=10-y^{2} $
The inner radius of the washer is 1.
The cross-sectional area of the washer
$ A \left( y \right) = \pi \left[ \left( 10-y^{2} \right) ^{2}-1^{2} \right] $
$ = \pi \left[ 100+y^{4}-20y^{2}-1 \right] $
$ = \pi \left[ 99-20y^{2}+y^{4} \right] $
{Step 5 of 5}:
The volume of the solid
$ V= \int _{-3}^{3}A \left( x \right) dy $
$ = \int _{-3}^{3} \pi \left( 99-20y^{2}+y^{4} \right) dy $
$ = \pi \int _{-3}^{3} \left( 99-20y^{2}+y^{4} \right) dy $
$ = \pi \left[ 99y-\frac{1}{3} \left( 20y^{3} \right) +\frac{1}{3}y^{5} \right] _{-3}^{3} $
[By the fundamental theorem of calculus, part 2]
$ = \pi \left[ \left( 99 \times 3-9 \times 20+\frac{243}{5} \right) - \left( -99 \times 3+9 \times 20-\frac{243}{5} \right) \right] $
$ = \pi \left[ 297-180+\frac{243}{5}+297-180+\frac{243}{5} \right] $
$ = \pi \left[ 594-360+\frac{486}{5} \right] $
$ = \pi \left[ 234+\frac{486}{5} \right] $
$ V=\frac{1656 \pi }{5} $
