Answer
\[\int_{0}^{1}{\int_{0}^{\sqrt{1-{{x}^{2}}}}{f\left( x,y \right)}}dydx\]
Work Step by Step
\[\begin{align}
& {{x}^{2}}+{{y}^{2}}=1 \\
& y=\sqrt{1-{{x}^{2}}},\text{ In the first quadrant} \\
& \text{The region }R\text{ is represented in the graph shown below} \\
& R=\left\{ \left( x,y \right):0\le y\le \sqrt{1-{{x}^{2}}},\text{ }0\le x\le 1 \right\} \\
& \text{Then,} \\
& \iint_{R}{f\left( x,y \right)}dA=\int_{0}^{1}{\int_{0}^{\sqrt{1-{{x}^{2}}}}{f\left( x,y \right)}}dydx \\
\end{align}\]
