Answer
$$\frac{{81}}{5}$$
Work Step by Step
$$\eqalign{
& \int_0^3 {\int_0^{\sqrt {9 - {z^2}} } {\int_0^x {xy} dydxdz} } \cr
& = \int_0^3 {\int_0^{\sqrt {9 - {z^2}} } {\left[ {\int_0^x {xy} dy} \right]dxdz} } \cr
& {\text{Solve the inner integral}}{\text{, integrate with respect to }}y,{\text{ treat }}z{\text{ and }}x{\text{ as constants}} \cr
& \int_0^x {xy} dy = \left[ {\frac{{x{y^2}}}{2}} \right]_0^x \cr
& {\text{Evaluate the limits in the variable }}y \cr
& = \frac{{x{{\left( x \right)}^2}}}{2} - \frac{{x{{\left( 0 \right)}^2}}}{2} \cr
& = \frac{{{x^3}}}{2} \cr
& {\text{Then}}{\text{,}} \cr
& \int_0^3 {\int_0^{\sqrt {9 - {z^2}} } {\left[ {\int_0^x {xy} dy} \right]dxdz} } = \int_0^3 {\int_0^{\sqrt {9 - {z^2}} } {\left( {\frac{{{x^3}}}{3}} \right)dxdz} } \cr
& = \int_0^3 {\left[ {\int_0^{\sqrt {9 - {z^2}} } {\left( {\frac{{{x^3}}}{2}} \right)dx} } \right]} dz \cr
& {\text{Integrate with respect to }}x,{\text{ treat }}y{\text{ as a constant}} \cr
& \int_0^{\sqrt {9 - {z^2}} } {\left( {\frac{{{x^3}}}{2}} \right)dx} = \left( {\frac{{{x^4}}}{8}} \right)_0^{\sqrt {9 - {z^2}} } \cr
& = \frac{{{{\left( {\sqrt {9 - {z^2}} } \right)}^4}}}{8} - \frac{{{{\left( 0 \right)}^4}}}{8} \cr
& = \frac{1}{8}{\left( {9 - {z^2}} \right)^2} \cr
& = \frac{1}{8}\left( {81 - 18{z^2} + {z^4}} \right) \cr
& \int_0^3 {\left[ {\int_0^{\sqrt {9 - {z^2}} } {\left( {\frac{{{x^3}}}{3}} \right)dx} } \right]} dz = \frac{1}{8}\int_0^3 {\left( {81 - 18{z^2} + {z^4}} \right)} dz \cr
& {\text{Integrate}} \cr
& = \frac{1}{8}\left( {81z - 6{z^3} + \frac{{{z^5}}}{5}} \right)_0^3 \cr
& {\text{Evaluate}} \cr
& = \frac{1}{8}\left( {81\left( 3 \right) - 6{{\left( 3 \right)}^3} + \frac{{{{\left( 3 \right)}^5}}}{5}} \right) - \frac{1}{{12}}\left( 0 \right) \cr
& = \frac{1}{8}\left( {\frac{{648}}{5}} \right) \cr
& = \frac{{81}}{5} \cr} $$
