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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.

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A sketch of the region R

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 $\int_{0}^3\int_{-\sqrt{9-z^2}}^{\sqrt{9-z^2}}\int_0^{3-x} (xz+3z)\;dy \;dx \; dz$

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    @NathanWilson: The outer integral is the full range in whichever coordinate. Here, if the outer one is $z$ it ranges from $0$, because we must be above the $xy$ plane, to $3$ because of the cylinder. As you go in, each integral is at a particular value of the outer variables. So if the next one in is $x$, we ask how much can $x$ vary, given the value of $z?$. To stay within the cylinder, we get the range shown. Finally, for $y$ we ask what is the range of $y$, given that $x,z$ are fixed? At $x=3, y$ is required to be $0$. At $x=0, y$ can range from $0$ to $3$2012-12-02