## Calculus with Applications (10th Edition)

$$\frac{2}{{45}}\left( {{{14}^{5/2}} - {6^{5/2}} - {8^{5/2}}} \right)$$
\eqalign{ & \iint\limits_R {{x^2}\sqrt {{x^3} + 2y} }dxdy;\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \leqslant x \leqslant 2,\,\,\,\,\,\,\,0 \leqslant y \leqslant 3 \cr & {\text{replacing the limits for the region }}R \cr & \int_0^3 {\int_0^2 {{x^2}\sqrt {{x^3} + 2y} dx} dy} \cr & = \int_0^3 {\left[ {\int_0^2 {{x^2}\sqrt {{x^3} + 2y} dx} } \right]} dy \cr & {\text{solve the inner integral }} \cr & \int_0^2 {{x^2}\sqrt {{x^3} + 2y} dx} \cr & {\text{write the integrand as}} \cr & = \frac{1}{3}\int_0^2 {{{\left( {{x^3} + 2y} \right)}^{1/2}}\left( {3{x^2}} \right)dx} \cr & {\text{integrate using the power rule}} \cr & = \frac{1}{3}\left[ {\frac{{{{\left( {{x^3} + 2y} \right)}^{3/2}}}}{{3/2}}} \right]_0^2 \cr & = \frac{2}{9}\left[ {{{\left( {{x^3} + 2y} \right)}^{3/2}}} \right]_0^2 \cr & {\text{evaluating the limits in the variable }}x \cr & = \frac{2}{9}\left[ {{{\left( {{2^3} + 2y} \right)}^{3/2}} - {{\left( {{0^3} + 2y} \right)}^{3/2}}} \right] \cr & = \frac{2}{9}\left[ {{{\left( {8 + 2y} \right)}^{3/2}} - {{\left( {2y} \right)}^{3/2}}} \right] \cr & {\text{then}} \cr & \cr & \int_0^3 {\int_0^2 {{x^2}\sqrt {{x^3} + 2y} dx} dy} = \int_0^3 {\frac{2}{9}\left[ {{{\left( {8 + 2y} \right)}^{3/2}} - {{\left( {2y} \right)}^{3/2}}} \right]dy} \cr & {\text{integrating}} \cr & = \frac{1}{9}\int_0^3 {\left[ {{{\left( {8 + 2y} \right)}^{3/2}}\left( 2 \right) - {2^{5/2}}{y^{3/2}}} \right]dy} \cr & {\text{use power rule}} \cr & = \frac{1}{9}\left( {\frac{{{{\left( {8 + 2y} \right)}^{5/2}}}}{{5/2}} - \frac{{{2^{5/2}}{y^{5/2}}}}{{5/2}}} \right)_0^3 \cr & = \frac{2}{{45}}\left( {{{\left( {8 + 2y} \right)}^{5/2}} - {2^{5/2}}{y^{5/2}}} \right)_0^3 \cr & {\text{evaluate}} \cr & = \frac{2}{{45}}\left( {{{\left( {8 + 2\left( 3 \right)} \right)}^{5/2}} - {2^{5/2}}{{\left( 3 \right)}^{5/2}}} \right) - \frac{2}{{45}}\left( {{{\left( {8 + 2\left( 0 \right)} \right)}^{5/2}} - {2^{5/2}}{{\left( 0 \right)}^{5/2}}} \right) \cr & {\text{simplifying}} \cr & = \frac{2}{{45}}\left( {{{14}^{5/2}} - {6^{5/2}}} \right) - \frac{2}{{45}}\left( {{8^{5/2}}} \right) \cr & = \frac{2}{{45}}\left( {{{14}^{5/2}} - {6^{5/2}} - {8^{5/2}}} \right) \cr}