Answer
$$36$$
Work Step by Step
$$\eqalign{
& \int_0^3 {\int_0^{\sqrt {9 - {x^2}} } {4x} } dydx \cr
& = \int_0^3 {\left[ {\int_x^{\sqrt {9 - {x^2}} } {4x} dy} \right]} dx \cr
& {\text{Integrate with respect to }}y \cr
& \int_x^{\sqrt {9 - {x^2}} } {4x} dy = \left[ {4xy} \right]_0^{\sqrt {9 - {x^2}} } \cr
& = 4x\left( {\sqrt {9 - {x^2}} } \right) - 4x\left( 0 \right) \cr
& = 4x\sqrt {9 - {x^2}} \cr
& \int_0^3 {\left[ {\int_x^{\sqrt {9 - {x^2}} } {4x} dy} \right]} dx = \int_0^3 {4x\sqrt {9 - {x^2}} } dx \cr
& = \int_0^3 {4x\sqrt {9 - {x^2}} } dx \cr
& = - 2\int_0^3 {\left( { - 2x} \right)\sqrt {9 - {x^2}} } dx \cr
& {\text{Integrating}} \cr
& = - 2\left[ {\frac{{{{\left( {9 - {x^2}} \right)}^{3/2}}}}{{3/2}}} \right]_0^3 \cr
& {\text{Simplifying}} \cr
& = - \frac{4}{3}\left[ {{{\left( {9 - {x^2}} \right)}^{3/2}}} \right]_0^3 \cr
& = - \frac{4}{3}\left[ {{{\left( {9 - {3^2}} \right)}^{3/2}} - {{\left( {9 - 0} \right)}^{3/2}}} \right] \cr
& = - \frac{4}{3}\left( { - 27} \right) \cr
& = 36 \cr} $$
