Answer
$$\int_{0}^{9} \int_{0}^{y / 3} \int_{0}^{\sqrt{y^{2}-9 x^{2}}} z d z d x d y=\frac{729}{4}$$
Work Step by Step
Given $$\int_{0}^{9} \int_{0}^{y / 3} \int_{0}^{\sqrt{y^{2}-9 x^{2}}} z d z d x d y$$
So, we get
\begin{align}
I&=\int_{0}^{9} \int_{0}^{y / 3} \int_{0}^{\sqrt{y^{2}-9 x^{2}}} z d z d x d y\\
&=\int_{0}^{9} \int_{0}^{y / 3} \left[\frac{z^2}{2}\right]_{0}^{\sqrt{y^{2}-9 x^{2}}} d z d x d y\\
&=\frac{1}{2} \int_{0}^{9} \int_{0}^{y / 3}\left(y^{2}-9 x^{2}\right) d x d y\\
&=\frac{1}{2} \int_{0}^{9}\left[x y^{2}-3 x^{3}\right]_{0}^{y / 3} d y\\
&=\frac{1}{2} \int_{0}^{9}\left[ \frac{ y^{3}}{3}- \frac{3y^{3}}{27}-0\right] d y\\
&=\frac{1}{2} \int_{0}^{9} \left[ \frac{2y^{3}}{9} \right] d y\\
&=\frac{1}{9} \int_{0}^{9} y^{3} d y\\
&=\left[\frac{1}{36} y^{4}\right]_{0}^{9}\\
&= \frac{1}{36} 9^{4}\\
&=\frac{729}{4}
\end{align}
