Answer
\[0\]
Work Step by Step
\[\begin{align}
& \int_{-a}^{a}{\int_{-\sqrt{{{a}^{2}}-{{x}^{2}}}}^{\sqrt{{{a}^{2}}-{{x}^{2}}}}{\left( x+y \right)}}dydx \\
& \text{Integrating }\int_{-\sqrt{{{a}^{2}}-{{x}^{2}}}}^{\sqrt{{{a}^{2}}-{{x}^{2}}}}{\left( x+y \right)}dy \\
& =\left[ xy+\frac{1}{2}{{y}^{2}} \right]_{-\sqrt{{{a}^{2}}-{{x}^{2}}}}^{\sqrt{{{a}^{2}}-{{x}^{2}}}} \\
& =\left[ x\sqrt{{{a}^{2}}-{{x}^{2}}}+\frac{1}{2}{{\left( \sqrt{{{a}^{2}}-{{x}^{2}}} \right)}^{2}} \right]-\left[ -x\sqrt{{{a}^{2}}-{{x}^{2}}}+\frac{1}{2}{{\left( \sqrt{{{a}^{2}}-{{x}^{2}}} \right)}^{2}} \right] \\
& =x\sqrt{{{a}^{2}}-{{x}^{2}}}+\frac{1}{2}\left( {{a}^{2}}-{{x}^{2}} \right)+x\sqrt{{{a}^{2}}-{{x}^{2}}}-\frac{1}{2}\left( {{a}^{2}}-{{x}^{2}} \right) \\
& =2x\sqrt{{{a}^{2}}-{{x}^{2}}} \\
& \text{Then,} \\
& \int_{-a}^{a}{\int_{-\sqrt{{{a}^{2}}-{{x}^{2}}}}^{\sqrt{{{a}^{2}}-{{x}^{2}}}}{\left( x+y \right)}}dydx=\int_{-a}^{a}{2x\sqrt{{{a}^{2}}-{{x}^{2}}}}dx \\
& =-\int_{-a}^{a}{\left( -2x \right)\sqrt{{{a}^{2}}-{{x}^{2}}}}dx \\
& =-\left[ \frac{{{\left( {{a}^{2}}-{{x}^{2}} \right)}^{3/2}}}{3/2} \right]_{-a}^{a} \\
& =-\frac{2}{3}\left[ {{\left( {{a}^{2}}-{{x}^{2}} \right)}^{3/2}} \right]_{-a}^{a} \\
& =-\frac{2}{3}\left[ {{\left( {{a}^{2}}-{{\left( a \right)}^{2}} \right)}^{3/2}}-{{\left( {{a}^{2}}-{{\left( -a \right)}^{2}} \right)}^{3/2}} \right] \\
& =-\frac{2}{3}\left[ 0 \right] \\
& =0 \\
\end{align}\]
