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