Answer
Please see the figure attached.
The iterated integral in the opposite order is
$\mathop \smallint \limits_2^3 \mathop \smallint \limits_{{x^2}}^9 f\left( {x,y} \right){\rm{d}}y{\rm{d}}x$
Work Step by Step
We have $\mathop \smallint \limits_4^9 \mathop \smallint \limits_2^{\sqrt y } f\left( {x,y} \right){\rm{d}}x{\rm{d}}y$. From the limits of the integrals we obtain the domain description:
${\cal D} = \left\{ {\left( {x,y} \right)|4 \le y \le 9,2 \le x \le \sqrt y } \right\}$
Notice that this is a horizontally simple region.
Using this description we sketch the the domain of integration. Please see the figure attached.
We change the order of integration such that the domain becomes a vertically simple region. In this case, the left and right boundaries are $x=2$ and $x=3$, respectively. Whereas, the upper boundary is $y=9$. To find the lower boundary, we use $x = \sqrt y $ and get $y = {x^2}$. Thus, the new domain description:
${\cal D} = \left\{ {\left( {x,y} \right)|2 \le x \le 3,{x^2} \le y \le 9} \right\}$
So, the iterated integral in the opposite order is
$\mathop \smallint \limits_2^3 \mathop \smallint \limits_{{x^2}}^9 f\left( {x,y} \right){\rm{d}}y{\rm{d}}x$
