Chapter 13 - Multiple Integration - 13.2 Double Integrals over General Regions - 13.2 Exercises - Page 981: 29

\[12\]

\[\begin{align} & \iint_{R}{{{y}^{2}}}dA \\ & \text{The region }R\text{ is represented in the graph shown below} \\ & \text{Then,} \\ & R=\left\{ \left( x,y \right):-x-1\le y\le 2x+2,\text{ }-1\le x\le 1 \right\} \\ & \iint_{R}{{{y}^{2}}}dA=\int_{-1}^{1}{\int_{-x-1}^{2x+2}{{{y}^{2}}}}dydx \\ & \text{Integrate} \\ & =\int_{-1}^{1}{\left[ \frac{1}{3}{{y}^{3}} \right]}_{-x-1}^{2x+2}dx \\ & =\int_{-1}^{1}{\left[ \frac{1}{3}{{\left( 2x+2 \right)}^{3}}-\frac{1}{3}{{\left( -x-1 \right)}^{3}} \right]}dx \\ & =\frac{1}{3}\int_{-1}^{1}{{{\left( 2x+2 \right)}^{3}}}dx-\frac{1}{3}\int_{-1}^{1}{{{\left( -x-1 \right)}^{3}}}dx \\ & =\frac{1}{6}\int_{-1}^{1}{{{\left( 2x+2 \right)}^{3}}\left( 2 \right)}dx+\frac{1}{3}\int_{-1}^{1}{{{\left( -x-1 \right)}^{3}}}\left( -1 \right)dx \\ & =\frac{1}{6}\left[ \frac{{{\left( 2x+2 \right)}^{4}}}{4} \right]_{-1}^{1}+\frac{1}{3}\left[ \frac{{{\left( -x-1 \right)}^{4}}}{4} \right]_{-1}^{1} \\ & \text{Evaluating} \\ & =\frac{1}{6}\left[ \frac{{{\left( 2\left( 1 \right)+2 \right)}^{4}}}{4}-\frac{{{\left( 2\left( -1 \right)+2 \right)}^{4}}}{4} \right]+\frac{1}{3}\left[ \frac{{{\left( -1-1 \right)}^{4}}}{4}-\frac{{{\left( 1-1 \right)}^{4}}}{4} \right] \\ & =\frac{1}{6}\left[ \frac{{{\left( 4 \right)}^{4}}}{4}-\frac{{{\left( 0 \right)}^{4}}}{4} \right]+\frac{1}{3}\left[ \frac{{{\left( -2 \right)}^{4}}}{4}-\frac{{{\left( 0 \right)}^{4}}}{4} \right] \\ & =\frac{1}{6}\left( 64 \right)+\frac{1}{3}\left( 4 \right) \\ & =12 \\ \end{align}\]