Answer
$ V=\frac{117 \pi }{5} $
Work Step by Step
{Step 1 of 6}:
First, find the points of intersection of the curves $ x=1+y^{2} $ and $ x=y+3 \left( x-3=y \right) $
$ 1+y^{2}=y+3 $
$ y^{2}-y-2=0 $
$ y^{2}-y-2=0 $
$ y \left( y-2 \right) +1 \left( y-2 \right) =0 $
$ y=2 $
or $ y=-1 $
The points of intersection are (5, 2) and (2, -1).
{Step 2 of 6}:
Sketch the curves.
{Step 3 of 6}:
Rotate this shaded region about the y-axis to produce a solid. Cutting this solid horizontally produces a washer. Sketch the solid and the washer.
{Step 4 of 6}:
The outer radius of the washer is $y + 3 $
The inner radius of the washer is $1+y^2 $
The cross-sectional area of the washer
$ A \left( y \right) = \pi \left[ \left( y+3 \right) ^{2}- \left( 1+y^{2} \right) ^{2} \right] $
$ = \pi \left[ y^{2}+9+6y-1-y^{4}+2y^{2} \right] $
$ = \pi \left[ 8+6y-y^{2}-y^{4} \right] $
{Step 5 of 6}:
Then the volume of the solid
$ V= \int _{-1}^{2}A \left( y \right) dy $
$ = \int _{-1}^{2} \pi \left( 8+6y-y^{2}-y^{4} \right) dy $
$ = \pi \int _{-1}^{2} \left( 8+6y-y^{2}-y^{4} \right) dy $
$ = \pi \left[ 8y+3y^{2}-\frac{1}{3}y^{3}-\frac{1}{5}y^{5} \right] _{-1}^{2} $
{Step 6 of 6}:
$ V= \pi \left[ \left( 16+12-\frac{8}{3}-\frac{32}{5} \right) - \left( -8+3+\frac{1}{3}+\frac{1}{5} \right) \right] $
$ = \pi \left[ 28-\frac{8}{3}-\frac{32}{5}+5-\frac{1}{3}-\frac{1}{5} \right] $
$ = \pi \left[ 33-3-\frac{33}{5} \right] $
$ = \pi \left[ 30-\frac{33}{5} \right] $
$ V=\frac{117 \pi }{5} $
