Answer
\[0\]
Work Step by Step
\[\begin{align}
& \iint_{R}{{{x}^{2}}y}dA \\
& \text{The region }R\text{ is represented in the graph shown below} \\
& \text{Then,} \\
& R=\left\{ \left( x,y \right):-\sqrt{16-{{x}^{2}}}\le y\le \sqrt{16-{{x}^{2}}},\text{ }0\le x\le 4 \right\} \\
& \iint_{R}{{{x}^{2}}y}dA=\int_{0}^{4}{\int_{-\sqrt{16-{{x}^{2}}}}^{\sqrt{16-{{x}^{2}}}}{{{x}^{2}}y}}dydx \\
& \text{Integrate} \\
& =\int_{0}^{4}{\left[ \frac{1}{2}{{x}^{2}}{{y}^{2}} \right]_{-\sqrt{16-{{x}^{2}}}}^{\sqrt{16-{{x}^{2}}}}}dx \\
& =\int_{0}^{4}{\left[ \frac{1}{2}{{x}^{2}}{{\left( \sqrt{16-{{x}^{2}}} \right)}^{2}}-\frac{1}{2}{{x}^{2}}{{\left( -\sqrt{16-{{x}^{2}}} \right)}^{2}} \right]}dy \\
& =\int_{0}^{4}{\left[ \frac{1}{2}{{x}^{2}}{{\left( 16-{{x}^{2}} \right)}^{2}}-\frac{1}{2}{{x}^{2}}{{\left( 16-{{x}^{2}} \right)}^{2}} \right]}dy \\
& =\int_{0}^{4}{0}dy \\
& =0 \\
\end{align}\]
