Answer
$$\int_{0}^{3} \int_{-\sqrt{9-y^{2}}}^{\sqrt{9-y^{2}}} \int_{0}^{y^{2}} y\ d z \ d x \ d y=\frac{324}{5}$$
Work Step by Step
Given$$\int_{0}^{3} \int_{-\sqrt{9-y^{2}}}^{\sqrt{9-y^{2}}} \int_{0}^{y^{2}} y d z d x d y $$
So, we have
\begin{align}
I&=\int_{0}^{3} \int_{-\sqrt{9-y^{2}}}^{\sqrt{9-y^{2}}} \int_{0}^{y^{2}} y \ \ d z \ d x \ d y\\
&= \int_{0}^{3} \int_{-\sqrt{9-y^{2}}}^{\sqrt{9-y^{2}}}[y z]_{0}^{y^{2}} dx \ dy\\
&= \int_{0}^{3} \int_{-\sqrt{9-y^{2}}}^{\sqrt{9-y^{2}}}\left(y\left(y^{2}\right)-0\right) dx \ dy\\
&=\int_{0}^{3} \int_{-\sqrt{9-y^{2}}}^{\sqrt{9-y^{2}}}\left(y^{3}\right) dx \ dy
\\
&=\int_{0}^{3} y^{3}\left[ x\right]_{-\sqrt {9-y^{2}}}^{\sqrt {9-y^{2}}} dy\\
&=\int_{0}^{3} y^{3} \left[ \sqrt{9-y^{2}}-(- \sqrt{9-y^{2}} )\right] \mathrm{dy}\\
&=\int_{0}^{3}\left[ 2y^{3} \sqrt{ 9-y^{2} } \mathrm{dy}\right]\\
&=2\int_{0}^{3}\left[ y^{2} \sqrt{ 9-y^{2} }\ \ \ y \ dy \right]\\
\end{align}
let $ 9-y^{2}=u^2$ this implies $ y^{2}=9-u^2$, $\ \ -2y \ dy =2u \ du$ and $\ \ y \ dy =-u \ du$
at $y=0 \Rightarrow u=3$,
$y=3 \Rightarrow u=0$
So, we get
\begin{align}
I &=2\int_{3}^{0}\left[( 9-u^{2}) \sqrt{ u^{2} }\ \ \ ( -u \ du) \right]\\
&=-2\int_{3}^{0} u^2 ( 9-u^{2}) du \\
&=2\int_{0}^{3} (-u^4+ 9u^{2}) du \\
&=2 (\frac{-u^5}{5}+ 9 \frac{u^{3}}{3})_{0}^{3}\\
&=2 (-\frac{3^5}{5}+ 9 \frac{3^{3}}{3}-0)\\
&=2 (-\frac{3^5}{5}+ 3^{4})\\
&=\frac{324}{5}
\end{align}
