Chapter 13 - Multiple Integration - 13.2 Double Integrals over General Regions - 13.2 Exercises - Page 981: 52

\[\frac{232}{15}\]

\[\begin{align} & y=\sqrt{x}\to x={{y}^{2}} \\ & y=x-2\to x=y+2 \\ & \text{From the graph, the region }R\text{ is} \\ & R=\left\{ \left( x,y \right):{{y}^{2}}\le x\le y+2,\text{ 0}\le y\le 2\text{ } \right\} \\ & \text{Then,} \\ & \iint_{R}{3{{x}^{2}}}dA=\int_{0}^{2}{\int_{{{y}^{2}}}^{y+2}{{{x}^{2}}ydx}dy} \\ & \text{Integrating} \\ & =\int_{0}^{2}{\int_{{{y}^{2}}}^{y+2}{{{x}^{2}}ydx}dy} \\ & =\int_{0}^{2}{\left[ \frac{1}{3}{{x}^{3}}y \right]_{{{y}^{2}}}^{y+2}dy} \\ & =\frac{1}{3}\int_{0}^{2}{\left[ {{x}^{3}}y \right]_{{{y}^{2}}}^{y+2}dy} \\ & =\frac{1}{3}\int_{0}^{2}{\left[ {{\left( y+2 \right)}^{3}}y-{{\left( {{y}^{2}} \right)}^{3}}y \right]dy} \\ & =\frac{1}{3}\int_{0}^{2}{\left[ \left( {{y}^{3}}+6{{y}^{2}}+12y+8 \right)y-{{y}^{7}} \right]dy} \\ & =\frac{1}{3}\int_{0}^{2}{\left( {{y}^{4}}+6{{y}^{3}}+12{{y}^{2}}+8y-{{y}^{7}} \right)dy} \\ & =\frac{1}{3}\left[ \frac{1}{5}{{y}^{5}}+\frac{3}{2}{{y}^{4}}+4{{y}^{3}}+4{{y}^{2}}-\frac{1}{8}{{y}^{8}} \right]_{0}^{2} \\ & =\frac{1}{3}\left[ \frac{1}{5}{{\left( 2 \right)}^{5}}+\frac{3}{2}{{\left( 2 \right)}^{4}}+4{{\left( 2 \right)}^{3}}+4{{\left( 2 \right)}^{2}}-\frac{1}{8}{{\left( 2 \right)}^{8}} \right]-\frac{1}{3}\left[ 0 \right] \\ & =\frac{1}{3}\left( \frac{232}{5} \right) \\ & =\frac{232}{15} \\ \end{align}\]