Identifying inequalities for 3D region.

I am having some difficulty identifying the bounds of a three-dimensional region.

I am asked to evaluate $\iiint_R (xz+3z)dV$, where $R$ is the region bounded by the cylinder $x^2 + z^2 = 9$ and the planes $x+y=3$, $z=0$, and $y=0$, above the $xy$-plane.

So now that I've a sketch of the region $R$, I am trying to find the bounds of $x$, $y$, and $z$ but I'm always confused when it comes to identifying the correct bounds.

For example, I don't know which of the following for $x$ are correct:

$-3 \le x \le 3$

$-3 \le x \le 3-y$

(are both of them wrong?)

Could someone please give me some tips on how I should go about constructing the inequalities for this 3D region?

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First you need to decide what order you will integrate in. As the integrand has no $y$ in it, that integral is easy to do and I would do it first. In that case, you can take $x$ and $z$ to be fixed (they are supplied by the outer integrals) and you need to find the range in $y$. $y$ can't be less than $0$ as that plane is one of you boundaries and can't be greater than $3-x$. So the inner integral is $\int_0^{3-x}dy$. Then if we do $x$ next, we have that $x$ ranges from $-\sqrt{9-z^2}$ to $\sqrt{9-z^2}$ and finally $z$ ranges from $0$ to $3$ because of the $xy$ plane restriction. So our final integral becomes $$\int_{0}^3\int_{-\sqrt{9-z^2}}^{\sqrt{9-z^2}}\int_0^{3-x} (xz+3z)\;dy \;dx \; dz$$
I think $z$ need only range from $0$ to $3$ since the region is bounded below by the $xy$ plane. – icurays1 Dec 1 '12 at 16:45
Hmm - I really can't see how $x$ is as you have defined. From what I can see, shouldn't it be $-3 \le x \le 3$? The thing is, I'm not sure when to use which equation. – JTJM Dec 1 '12 at 16:55
@NathanWilson: Say you are at $z=1$. In that case, $x$ ranges from $-\sqrt 8$ to $+\sqrt 8$. As we are inside the $z$ integral, we choose $z$ first, then figure out the range of $x$ – Ross Millikan Dec 1 '12 at 17:36
Right. So if, assuming it's easier to start integrating with $z$ or $x$ first, will the boundaries still be the same? (I don't think so). – JTJM Dec 2 '12 at 13:28