SOLUTION
$$\begin{aligned} I &=\int_\sqrt{x}^{\pi}\int_0^{\pi^2} \sin(y^3) \,\mathrm{d}y \,\mathrm{d}x \\ &=\int_\sqrt{x}^{\pi}\int_0^{\pi^2} \sin(y^3) \,\mathrm{d}x \,\mathrm{d}y \\ &=\int_\sqrt{x}^{\pi} \pi^2 \sin(y^3) \,\mathrm{d}y \end{aligned}$$
So you want to integrate the function $f(x,y)=\sin (y^3)$ with bounds $\int_{x=0}^{x=\pi^2}\int_{y=\sqrt x}^{y=\pi}$.
This can be represented as the set
\begin{align} \ S & =\{ (x,y) \in \Bbb R^2:0≤x≤\pi^2 \;,\; \sqrt x≤y≤\pi \} \\ \ & =\{ (x,y) \in \Bbb R^2:0≤x≤\pi^2 \;,\; x≤y^2≤\pi^2 \} \\ \ & =\{ (x,y) \in \Bbb R^2:0≤x≤y^2≤\pi^2 \} \\ \ & = \{ (x,y) \in \Bbb R^2:0≤y^2≤\pi^2 \;,\; 0≤x≤y^2\} \\ \ & = \{ (x,y) \in \Bbb R^2:0≤y≤\pi \;,\; 0≤x≤y^2\} \end{align}
Basically this is our region:
Thus, another way to parameterize this region is to have $\int_{y=0}^{y=\pi}\int_{x=0}^{x=y^2}$ instead. Hence
\begin{align} \ \int_{x=0}^{x=\pi^2}\int_{y=\sqrt x}^{y=\pi}\sin (y^3)dydx & = \int_{y=0}^{y=\pi}\int_{x=0}^{x=y^2}\sin (y^3)dxdy\\ \ & = \int_{y=0}^{y=\pi} y^2\sin (y^3)dy \\ \ & = \Bigl[ -\frac 13 \cos(y^3) \Bigr]_{y=0}^{y=\pi} \\ \ & = \frac 13 \Bigl (1-\cos (\pi ^3) \Bigr ) \end{align}