Answer
\[\int_{0}^{6}{\int_{0}^{-y/2+3}{f\left( x,y \right)}dx}dy\]
Work Step by Step
\[\begin{align}
& \int_{0}^{3}{\int_{0}^{6-2x}{f\left( x,y \right)}dy}dx \\
& y=6-2x\to x=-y/2+3 \\
& \text{Using the graph to switch the order of integration} \\
& R=\left\{ \left( x,y \right):0\le x\le -y/2+3,\text{ 0}\le y\le 6\text{ } \right\} \\
& \text{Then} \\
& \int_{0}^{3}{\int_{0}^{6-2x}{f\left( x,y \right)}dy}dx=\int_{0}^{6}{\int_{0}^{-y/2+3}{f\left( x,y \right)}dx}dy \\
\end{align}\]
