Answer
\[32\]
Work Step by Step
\[\begin{align}
& y=2x+4\to x=\frac{y}{2}-2 \\
& y={{x}^{3}}\to x=\sqrt[3]{y} \\
& \text{From the graph, the region }R\text{ is} \\
& R=\left\{ \left( x,y \right):\frac{y}{2}-2\le x\le \sqrt[3]{y},\text{ 0}\le y\le 8\text{ } \right\} \\
& \text{Then,} \\
& \iint_{R}{3{{x}^{2}}}dA=\int_{0}^{8}{\int_{\frac{y}{2}-2}^{\sqrt[3]{y}}{3{{x}^{2}}dx}dy} \\
& \text{Integrating} \\
& =\int_{0}^{8}{\left[ {{x}^{3}} \right]_{\frac{y}{2}-2}^{\sqrt[3]{y}}dy} \\
& =\int_{0}^{8}{\left[ {{\left( \sqrt[3]{y} \right)}^{3}}-{{\left( \frac{y}{2}-2 \right)}^{3}} \right]dy} \\
& =\int_{0}^{8}{\left[ y-{{\left( \frac{y}{2}-2 \right)}^{3}} \right]dy} \\
& =\left[ \frac{1}{2}{{y}^{2}}-\frac{1}{2}{{\left( \frac{y}{2}-2 \right)}^{4}} \right]_{0}^{8} \\
& =\left[ \frac{1}{2}{{\left( 8 \right)}^{2}}-\frac{1}{2}{{\left( \frac{8}{2}-2 \right)}^{4}} \right]-\left[ \frac{1}{2}{{\left( 0 \right)}^{2}}-\frac{1}{2}{{\left( \frac{0}{2}-2 \right)}^{4}} \right] \\
& =24+8 \\
& =32 \\
\end{align}\]
