Answer
$V = \pi \int_0^2 {\left[ {{{\left( {6 - {y^2}} \right)}^2} - 4} \right]} dy$
Work Step by Step
$$\eqalign{
& y = \sqrt x ,{\text{ }}y = 0,{\text{ }}x = 4,{\text{ about }}x = 6 \cr
& y = \sqrt x \Rightarrow {x^2} = y \cr
& {\text{Graph of the region shown below }}\left( {{\text{Using Geogebra}}} \right) \cr
& \cr
& {\text{For this exercise the inner radius is given by:}} \cr
& {r_i} = 6 - f\left( y \right) \cr
& {r_i} = 6 - 4 \cr
& {r_i} = 2 \cr
& {\text{And the outer radius is given by:}} \cr
& {r_o} = 6 - f\left( y \right) \cr
& {r_o} = 6 - {y^2} \cr
& {\text{The cross - sectional area is}} \cr
& A\left( y \right) = \pi {\left( {{r_o}} \right)^2} - \pi {\left( {{r_i}} \right)^2} \cr
& A\left( y \right) = \pi {\left( {6 - {y^2}} \right)^2} - \pi {\left( 2 \right)^2} \cr
& {\text{The volume is given by}} \cr
& V = \int_c^d {A\left( y \right)} dy \cr
& V = \int_0^2 {\left[ {\pi {{\left( {6 - {y^2}} \right)}^2} - \pi {{\left( 2 \right)}^2}} \right]} dy \cr
& V = \int_0^2 {\left[ {\pi {{\left( {6 - {y^2}} \right)}^2} - 4\pi } \right]} dy \cr
& V = \pi \int_0^2 {\left[ {{{\left( {6 - {y^2}} \right)}^2} - 4} \right]} dy \cr} $$
