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