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