1. Problem statement: Change the order of integration of the given double integral. 2. The integral to be converted is $$\int_0^3 \int_{2-\sqrt{4-y}}^y f(x,y)\,dx\,dy + \int_3^4 \int_{2-\sqrt{4-y}}^{2+\sqrt{4-y}} f(x,y)\,dx\,dy$$ 3. Formula and rule: To change the order integrate with respect to $y$ first for each fixed $x$ by finding, for each $x$, the minimum and maximum $y$ such that $(x,y)$ lies in the region $R$. $$\iint_R f(x,y)\,dA=\int_{x=a}^{b}\int_{y=g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx$$ 4. Identify the curve boundaries. The curved boundaries satisfy $$ (x-2)^2=4-y \quad\Rightarrow\quad y=4-(x-2)^2 $$ and one straight boundary is the line $y=x$. 5. Find intersection points of $y=x$ and the parabola to determine the $x$-span of the region. $$ x=4-(x-2)^2 $$ $$ \Rightarrow (x-2)^2 + x - 4 = 0 $$ $$ \Rightarrow x^2 -4x +4 + x -4 = 0 $$ $$ \Rightarrow x^2 -3x = 0 $$ $$ \Rightarrow x(x-3)=0 \Rightarrow x=0,\;x=3 $$ So the overall $x$-range is $0\le x\le 3$. 6. For a fixed $x$ the left-bound condition $x\ge 2-\sqrt{4-y}$ is equivalent to $$ \sqrt{4-y} \ge 2-x \quad\Rightarrow\quad 4-y \ge (2-x)^2 \quad\Rightarrow\quad y \le 4-(x-2)^2 $$ and the right-bound condition $x\le y$ becomes $y\ge x$. 7. Combine the bounds: for every $x$ in $[0,3]$ the $y$ values in the region satisfy $$ x \le y \le 4-(x-2)^2 $$ 8. Final answer (reversed order of integration): $$ \int_{x=0}^3 \int_{y=x}^{4-(x-2)^2} f(x,y)\,dy\,dx $$ This iterated integral equals the original double integral and is ready for evaluation with $dy$ inner and $dx$ outer.