# Chapter 6 - Review - Exercises: 8

$V=\frac{117 \pi }{5}$

#### 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}: $V= \pi \left[ \left( 16+12-\frac{8}{3}-\frac{32}{5} \right) - \left( -8+3+\frac{1}{3}+\frac{1}{5} \right) \right]$ $= \pi \left[ 28-\frac{8}{3}-\frac{32}{5}+5-\frac{1}{3}-\frac{1}{5} \right]$ $= \pi \left[ 33-3-\frac{33}{5} \right]$ $= \pi \left[ 30-\frac{33}{5} \right]$ $V=\frac{117 \pi }{5}$

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