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