Answer
$\mathop \smallint \limits_{y = 0}^9 \mathop \smallint \limits_{x = - \sqrt {9 - y} }^{\sqrt {9 - y} } f\left( {x,y} \right){\rm{d}}x{\rm{d}}y$
Work Step by Step
From the double integral $\mathop \smallint \limits_{x = - 3}^3 \mathop \smallint \limits_{y = 0}^{9 - {x^2}} f\left( {x,y} \right){\rm{d}}y{\rm{d}}x$ we obtain the domain description:
${\cal D} = \left\{ {\left( {x,y} \right)| - 3 \le x \le 3,0 \le y \le 9 - {x^2}} \right\}$
Notice that in this description, ${\cal D}$ is a vertically simple region, upper bounded by the parabola $y = 9 - {x^2}$ and lower bounded by $y=0$.
From $y = 9 - {x^2}$, we obtain $x = \pm \sqrt {9 - y} $.
So, we can also describe ${\cal D}$ as a horizontally simple region, bounded left by $x = - \sqrt {9 - y} $ and bounded right by $x = \sqrt {9 - y} $:
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le y \le 9, - \sqrt {9 - y} \le x \le \sqrt {9 - y} } \right\}$
Thus, the double integral as an iterated integral in the order $dxdy$:
$\mathop \smallint \limits_{y = 0}^9 \mathop \smallint \limits_{x = - \sqrt {9 - y} }^{\sqrt {9 - y} } f\left( {x,y} \right){\rm{d}}x{\rm{d}}y$
