Answer
$V = \pi \int_{ - 1}^6 {\left[ {{{\left( {x + 10} \right)}^2} - {{\left( {x - 2} \right)}^4}} \right]} dx$
Work Step by Step
$$\eqalign{
& y = {\left( {x - 2} \right)^2},{\text{ }}y = x + 10;{\text{ about the }}x{\text{ - axis}} \cr
& {\text{Find the intersection points between the graphs}} \cr
& {\left( {x - 2} \right)^2} = x + 10 \cr
& {x^2} - 4x + 4 = x + 10 \cr
& {x^2} - 5x - 6 = 0 \cr
& \left( {x - 6} \right)\left( {x + 1} \right) = 0 \cr
& x = - 1{\text{ and }}x = 6 \cr
& {\text{Graph of the region shown below }}\left( {{\text{Using Geogebra}}} \right) \cr
& \cr
& {\text{When we slice through the axis }}x,{\text{ we get a disk with }} \cr
& {\text{radius }}r\left( x \right){\text{ which is the difference of the outer radius}} \cr
& x + 10{\text{ and the inner radius }}{\left( {x - 2} \right)^2}. \cr
& {\text{The area of this cross section is }}A = \pi {r^2}, \cr
& A\left( y \right) = \pi \left[ {{{\left( {x + 10} \right)}^2} - {{\left( {{{\left( {x - 2} \right)}^2}} \right)}^2}} \right] \cr
& A\left( y \right) = \pi \left[ {{{\left( {x + 10} \right)}^2} - {{\left( {x - 2} \right)}^4}} \right] \cr
& {\text{The solid lies on the interval }} - 1 \leqslant x \leqslant 6,{\text{ so using the volume}} \cr
& {\text{definition }}\left( {{\text{page 448}}} \right) \cr
& V = \int_a^b {A\left( x \right)} dx \cr
& {\text{The volume is given by}} \cr
& V = \int_{ - 1}^6 {\pi \left[ {{{\left( {x + 10} \right)}^2} - {{\left( {x - 2} \right)}^4}} \right]} dx \cr
& V = \pi \int_{ - 1}^6 {\left[ {{{\left( {x + 10} \right)}^2} - {{\left( {x - 2} \right)}^4}} \right]} dx \cr} $$
