Answer
\[\int_{0}^{1}{\int_{0}^{\sqrt{1-{{x}^{2}}}}{f\left( x,y \right)}dy}dx\]
Work Step by Step
\[\begin{align}
& \int_{0}^{1}{\int_{0}^{\sqrt{1-{{y}^{2}}}}{f\left( x,y \right)}dx}dy \\
& x=\sqrt{1-{{y}^{2}}}\to y=-\sqrt{1-{{x}^{2}}},\text{ }y=\sqrt{1-{{x}^{2}}} \\
& \text{Using the graph to switch the order of integration} \\
& R=\left\{ \left( x,y \right):0\le y\le \sqrt{1-{{x}^{2}}},\text{ }0\le x\le 1\text{ } \right\} \\
& \text{Then} \\
& \int_{0}^{1}{\int_{0}^{\sqrt{1-{{y}^{2}}}}{f\left( x,y \right)}dx}dy=\int_{0}^{1}{\int_{0}^{\sqrt{1-{{x}^{2}}}}{f\left( x,y \right)}dy}dx \\
\end{align}\]
