I want to express $\iiint_E f(x,y,z) \ dV$ as an iterated integral.
The solid is bounded by the surfaces $y^2+z^2=9$, $x=-2$, $x=2$.
Looking at the projection on the plane x=0, the trace is a circle with radius 3 and center $(y,z)=(0,0)$. So $[(y,z)|-3\leq y\leq 3, -\sqrt{9-y^2} \leq z \leq \sqrt{9-y^2}]$
This would give the iterated integral:
$$\int_{-3}^{3}\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}\int_{-2}^{2} \ dx\ dz\ dy$$
In a similar way, I came up with the following variations of the integral:
$$\int_{-3}^{3}\int_{-\sqrt{9-z^2}}^{\sqrt{9-z^2}}\int_{-2}^{2} \ dx\ dy\ dz$$
$$\int_{-3}^{3}\int_{-2}^{2}\int_{-\sqrt{9-z^2}}^{\sqrt{9-z^2}} \ dy\ dx\ dz$$
$$\int_{-2}^{2}\int_{-3}^{3}\int_{-\sqrt{9-z^2}}^{\sqrt{9-z^2}} \ dy\ dz\ dx$$
$$\int_{-3}^{3}\int_{-2}^{2}\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}} \ dz\ dx\ dy$$
$$\int_{-2}^{2}\int_{-3}^{3}\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}} \ dz\ dy\ dx$$
Have I created the bounds on the integrals correctly, and are these all the ways to form the iterated integral?