Ok. So you have the triple integral:
$$\begin{align}
\int_{-3}^3\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}\int_1^{5-y} \;dz\;dx\;dy
&= \int_{-3}^3\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}4-y\;dx\;dy \\
&=\int_{-3}^34x-xy\Bigg|_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}\;dy \\
&=\int_{-3}^38\sqrt{9-y^2}-2y\sqrt{9-y^2}\;dy \\
&= 8\int_{-3}^33\sqrt{1-\left(\frac{y}{3}\right)^2}\;dy-2\int_{-3}^3y\sqrt{9-y^2}\;dy
\end{align}$$
Now, I'm going to break this up. For the left-hand integral, we must use trig-substitution. Let $\cos(t) = \frac{y}{3}$. This implies that $dy = -3\sin(t)\;dt$. The limits of integration change as well, to $t=\arccos\left(\frac{-3}{3}\right) = \pi$ to $t = \arccos\left(\frac{3}{3}\right) = 0$.
So, the integral becomes:
$$\begin{align}
24\int_\pi^0\sqrt{1-\cos^2(t)}(-3\sin t)\;dt &= -72\int_\pi^0\sin^2(t)\;dt\\
&=72\int_0^\pi\frac{1}{2}-\frac{\cos(2t)}{2}\;dt\\
&=36\int_0^\pi1-\cos(2t)\;dt\\
&=36\left(t - \frac{\sin(2t)}{2}\right)\Bigg|_0^\pi\\
&=\boxed{36\pi}
\end{align}$$
Now, for the left-hand integral, we apply $u$-substitution. If we set $u = 9-y^2$, then $du = -2y\;dy$. The limits are transformed to $u = 9-(-3)^2 = 0$ to $u = 9-(3)^2 = 0$
So, the integral becomes:
$$\begin{align}
-2\int_{-3}^3y\sqrt{9-y^2}\;dy &= \int_0^0\sqrt{u}\;du\\
&= \boxed{0}
\end{align}$$
Well, that wasn't exciting. :)
So, putting it all together, we end up with:
$$V = 36\pi + 0 = \boxed{36\pi}$$
If you have a volume $A$, and want to express $\int_A f(x,y,z)\,d(x,y,z)$ as an iterated integral, i.e. as $$
\int_A f(x,y,z) \,d(x,y,z) = \int_{A_x} \int_{A_y(x)} \int_{A_z(x,y)} f(x,y,z) \,dz\,dy\,dx
$$
you have to find $$\begin{eqnarray}
A_x &\subset& \mathbb{R} \\
A_y &\,:\,& A_x \to \mathcal{P}({\mathbb{R}}) \\
A_z &\,:\,& \{(x,y) \,:\, x \in A_x, y \in A_y(x)\} \to \mathcal{P}(\mathbb{R})
\end{eqnarray}$$
such that $$
A = \left\{(x,y,z) \,:\, x \in A_x,\, y \in A_y(x),\, z \in A_z(x,y,z) \right\}.
$$
Integration order $dz\,dx\,dy$
Let's look at the case of a pyramid with base $[-2,2]\times [-2,2]$ in the $x,y$-plane, tip at $(0,0,5)$, and integration order $dz\,dx\,dy$. We have $$
A = \left\{(x,y,z) \,:\, x \in [-2,2], y \in [-2,2], 0 \leq z \leq \tfrac{5}{2}\min \{2-|x|,2-|y|\}\right\} \text{.}
$$
So $A_y$ doesn't actually depend on $x$, and $A_z(x,y) = [0, \tfrac{5}{2}\min \{2-|x|,2-|y|\}]$. Therefore, $$
\int_A 1 \,d(x,y,z) = \int_{-2}^2 \int_{-2}^2 \int_0^{\tfrac{5}{2}\min \{2-|x|,2-|y|\}} 1 \,dz\,dy\,dx \text{.}
$$
Integration order $dx\,dy\,dz$
For integration order $dx\,dy\,dz$ we have to find $$\begin{eqnarray}
A_z &\subset& \mathbb{R} \\
A_y &\,:\,& A_z \to \mathcal{P}({\mathbb{R}}) \\
A_x &\,:\,& \{(z,y) \,:\, z \in A_z, y \in A_y(z)\} \to \mathcal{P}(\mathbb{R})
\end{eqnarray}$$
such that $$
A = \left\{(x,y,z) \,:\, x \in A_x(z,y),\, y \in A_y(z),\, z \in A_z \right\}.
$$
Obviously, $A_z = [0,5]$. For a fixed $z \in A_z$, $(x,y,z) \in A$ exactly if $$\begin{eqnarray}
z &\leq& \tfrac{5}{2}(2 - |x|) &\Leftrightarrow& |x| \leq \tfrac{2}{5}z - 2 &\text{ and } \\
z &\leq& \tfrac{5}{2}(2 - |y|) &\Leftrightarrow& |y| \leq \tfrac{2}{5}z - 2 \text{,}\\
\end{eqnarray}$$
which means $$
\int_A 1 \,d(x,y,z) = \int_0^5 \int_{-\frac{2}{5}z + 2}^{\frac{2}{5}z - 2} \int_{-\frac{2}{5}z + 2}^{\frac{2}{5}z - 2} 1 \,dx\,dy\,dz \text{.}
$$
Best Answer
I'm going to try to expand on @SangchulLee's answer a little bit.
You think through the original problem by realizing that our 3-dimensional region projects down to the region in the $xy$-plane between the $y$-axis and $x=y^3$ [or, equivalently, $y=x^{1/3}$]. In $\Bbb R^3$, $x=y^3$ is a vertical cylinder over this plane curve. For each point in that plane region, a vertical line enters the region at $z=0$ and exits at $z=y^2$. So this surface is a parabolic cylinder, with lines parallel to the $x$-axis. Here's a sketch with the assistance of Mathematica:
Indeed, the region lies over the rectangle $R=[0,8]\times [0,4]$ in the $xz$-plane, and the two cylinders intersect along the crease $y=z^{1/2}=x^{1/3}$, which projects to the curve $C$ given by $x^2=z^3$ in the $xz$-plane. If we take a point in $R$, we want to know when a line parallel to the $y$-axis enters and exits our region. If we're below $C$ in $R$, then $z^3<x^2$ and the line enters at $y=x^{1/3}$ and exits at $y=2$. (As a check, if $y\ge x^{1/3}$, then $y\ge z^{1/2}$ as well in this range, so we're "inside" both cylinders.) If we're above $C$ in $R$, then $z^3>x^2$ and the line enters at $y=z^{1/2}$ and exits at $y=2$.
Thus, we end up with the following iterated integral in the order $dy\,dz\,dx$: $$\int_0^8\int_0^{x^{2/3}}\int_{x^{1/3}}^2 dy\,dz\,dx + \int_0^8\int_{x^{2/3}}^4\int_{z^{1/2}}^2 dy\,dz\,dx.$$
(By the way, you should be able to set up the limits in the orders $dx\,dz\,dy$ or $dx\,dy\,dz$ much more easily.)