Answer
\[4\pi \]
Work Step by Step
\[\begin{align}
& \int_{-2}^{2}{\int_{-\sqrt{4-{{x}^{2}}}}^{\sqrt{4-{{x}^{2}}}}{dy}dx} \\
& \text{Find the integral} \\
& =\int_{-2}^{2}{\left[ y \right]_{-\sqrt{4-{{x}^{2}}}}^{\sqrt{4-{{x}^{2}}}}dx} \\
& =\int_{-2}^{2}{\left[ \sqrt{4-{{x}^{2}}}+\sqrt{4-{{x}^{2}}} \right]dx} \\
& =\int_{-2}^{2}{2\sqrt{4-{{x}^{2}}}dx} \\
& \text{By tables }\int{\sqrt{{{a}^{2}}-{{u}^{2}}}du}=\frac{1}{2}\left( u\sqrt{{{a}^{2}}-{{u}^{2}}}+{{a}^{2}}{{\sin }^{-1}}\left( \frac{u}{a} \right) \right)+C \\
& =\left[ x\sqrt{4-{{x}^{2}}}+4{{\sin }^{-1}}\left( \frac{x}{2} \right) \right]_{-2}^{2} \\
& =\left[ 2\sqrt{4-{{2}^{2}}}+4{{\sin }^{-1}}\left( \frac{2}{2} \right) \right]-\left[ -2\sqrt{4-{{\left( -2 \right)}^{2}}}+4{{\sin }^{-1}}\left( \frac{-2}{2} \right) \right] \\
& =4{{\sin }^{-1}}\left( \frac{2}{2} \right)-4{{\sin }^{-1}}\left( \frac{-2}{2} \right) \\
& =4\pi \\
& \text{Using the graph to switch the order of integration} \\
& \int_{-2}^{2}{\int_{-\sqrt{4-{{y}^{2}}}}^{\sqrt{4-{{y}^{2}}}}{dy}dx} \\
& \text{Find the integral} \\
& =\int_{-2}^{2}{\left[ x \right]_{-\sqrt{4-{{y}^{2}}}}^{\sqrt{4-{{y}^{2}}}}dx} \\
& =\int_{-2}^{2}{2\sqrt{4-{{y}^{2}}}dx} \\
& \text{By symmetry with the integral }\int_{-2}^{2}{2\sqrt{4-{{x}^{2}}}dx} \\
& \int_{-2}^{2}{2\sqrt{4-{{y}^{2}}}dx}=4\pi \\
\end{align}\]
