Answer
\[\frac{\pi }{2}\]
Work Step by Step
\[\begin{align}
& \int_{0}^{1}{\int_{-\sqrt{1-{{y}^{2}}}}^{\sqrt{1-{{y}^{2}}}}{dx}dy} \\
& \text{Find the integral} \\
& =\int_{0}^{1}{\left[ x \right]_{-\sqrt{1-{{y}^{2}}}}^{\sqrt{1-{{y}^{2}}}}dy} \\
& =\int_{0}^{1}{2\sqrt{1-{{y}^{2}}}dy} \\
& \text{By tables }\int{\sqrt{{{a}^{2}}-{{u}^{2}}}du}=\frac{1}{2}\left( u\sqrt{{{a}^{2}}-{{u}^{2}}}+{{a}^{2}}{{\sin }^{-1}}\left( \frac{u}{a} \right) \right)+C \\
& =2\left[ {{\sin }^{-1}}y \right]_{0}^{1} \\
& =\left[ y\sqrt{1-{{y}^{2}}}+{{\sin }^{-1}}\left( y \right) \right] \\
& =\sqrt{1-{{1}^{2}}}+{{\sin }^{-1}}\left( 1 \right) \\
& =\frac{1}{2}\pi \\
& \text{Using the graph to switch the order of integration} \\
& \int_{-1}^{1}{\int_{0}^{\sqrt{1-{{x}^{2}}}}{dy}dx} \\
& \text{Find the integral} \\
& =\int_{-1}^{1}{\left[ y \right]_{0}^{\sqrt{1-{{x}^{2}}}}dx} \\
& =\int_{-1}^{1}{\sqrt{1-{{x}^{2}}}dx} \\
& =\frac{1}{2}\left( x\sqrt{1-{{x}^{2}}}+{{\sin }^{-1}}\left( \frac{x}{a} \right) \right)+C \\
& =\frac{1}{2}\left[ \sqrt{1-{{1}^{2}}}+{{\sin }^{-1}}\left( 1 \right) \right]-\frac{1}{2}\left[ -1\sqrt{1-{{\left( -1 \right)}^{2}}}+{{\sin }^{-1}}\left( -1 \right) \right] \\
& =\frac{\pi }{4}-\left( -\frac{\pi }{4} \right) \\
& =\frac{\pi }{2} \\
\end{align}\]
