Answer
$$9$$
Work Step by Step
$$\eqalign{
& \int_0^3 {\int_0^{\sqrt {9 - {y^2}} } y dxdy} \cr
& = \int_0^3 {\left[ {\int_0^{\sqrt {9 - {y^2}} } y dx} \right]dy} \cr
& {\text{solve the inner integral and treat }}y{\text{ as a constant}} \cr
& \int_0^{\sqrt {9 - {y^2}} } y dx = y\int_0^{\sqrt {9 - {y^2}} } {dx} \cr
& {\text{using the power rule }} \cr
& = y\left( x \right)_0^{\sqrt {9 - {y^2}} } \cr
& {\text{evaluating the limits in the variable }}x \cr
& = y\left( {\sqrt {9 - {y^2}} - 0} \right) \cr
& {\text{simplifying}} \cr
& = y\sqrt {9 - {y^2}} \cr
& {\text{then}} \cr
& \int_0^3 {\left[ {\int_0^{\sqrt {9 - {y^2}} } y dx} \right]dy} = \int_0^3 {y\sqrt {9 - {y^2}} } dy \cr
& = - \frac{1}{2}\int_0^3 {{{\left( {9 - {y^2}} \right)}^{1/2}}\left( { - 2y} \right)} dy \cr
& {\text{integrating by the power rule}} \cr
& = - \frac{1}{2}\left[ {\frac{{{{\left( {9 - {y^2}} \right)}^{3/2}}}}{{3/2}}} \right]_0^3 \cr
& = - \frac{1}{3}\left[ {{{\left( {9 - {y^2}} \right)}^{3/2}}} \right]_0^3 \cr
& {\text{simplifying}} \cr
& = - \frac{1}{3}\left( {{{\left( {9 - {3^2}} \right)}^{3/2}} - {{\left( {9 - {0^2}} \right)}^{3/2}}} \right) \cr
& = - \frac{1}{3}\left( {{{\left( 0 \right)}^{3/2}} - {{\left( 9 \right)}^{3/2}}} \right) \cr
& = - \frac{1}{3}\left( { - 27} \right) \cr
& = 9 \cr} $$
