## Calculus: Early Transcendentals 8th Edition

$V=\frac{1656 \pi }{5}$
{Step 1 of 5}: First, find the points of intersection of the curves $x=0$ and$x=9-x^{2}$ To do this, set $9-x^{2}=0$ 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 $A \left( y \right) = \pi \left[ \left( 10-y^{2} \right) ^{1}-1^{2} \right]$ $= \pi \left[ 100+y^{4}-20y^{2}-1 \right]$ $= \pi \left[ 99-20y^{2}+y^{4} \right]$ {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] $= \pi \left[ \left( 99 \times 3-9 \times 20+\frac{243}{5} \right) - \left( -99 \times 3+9 \times 20-\frac{243}{5} \right) \right]$ $= \pi \left[ 297-180+\frac{243}{5}+297-180+\frac{243}{5} \right]$ $= \pi \left[ 594-360+\frac{486}{5} \right]$ $= \pi \left[ 234+\frac{486}{5} \right]$ $V=\frac{1656 \pi }{5}$