What is this triple intergration's lower and upper bound? How does one find the upper bound for the question below. I know how to do the intergration; I just can’t start the question as I am unsure of how to find these values. I know that all the lowers are $0$.
The question is:

$$\textrm{Let } M = \iiint_B x\,\mathrm{d}z \,\mathrm{d}y \,\mathrm{d}x  \\ \textrm{ where } B \textrm{ is the region inside } x \ge 0, y \ge 0, z \ge 0 \textrm{ and } x + 2y + 3z \le 6.
\\ \textrm{Sketch region } B \textrm{ in the } (x,y,z) \textrm{ space and show that } M = 9.$$

 A: Remember that the limits of integration for the first (outermost) variable encompass all the values in the universe that that variable could possibly take; then the limits of integration for the next variable encompass the possible values given a particular value for the first variable; and so on.
In this case, the outermost variable is $x$, then $y$, then $z$. (Other choices would be equally valid, but the $dz\, dy\, dx$ at the end shows that here $x$ is the outermost and $z$ is the innermost variable.) So here are the questions, in order, we need to answer:


*

*What are all the possible values of $x$? Obviously $x\ge0$; and also $x$ can't be larger than $6$, because if it were then $x+2y+3z$ would never be $\le6$ (since $y$ and $z$ are nonnegative). So the limits of integration for $x$ will be $\int_0^6$.

*Given a particular choice of $x$, what are all the possible values of $y$? Obviously $y\ge0$ still. But notice that $x+2y+3z\le6$ implies (since $z$ is nonnegative) that $2y \le 2y+3z \le 6-x$, or $y\le(6-x)/2$. And any such value of $y$ is possible (take $z=0$ for example). Therefore the limits of integration for $y$ will be $\int_0^{(6-x)/2}$.

*Finally, given particular choices of $x$ and $y$, what are all the possible values of $z$? Here the two inequalities give us exactly what we need: $z\ge0$, and $x+2y+3z\le6$ or equivalently $3z\le6-x-2y$ or equivalently $z\le(6-x-2y)/3$. Therefore the limits of integration for $z$ will be $\int_0^{(6-x-2y)/3}$.


[Note in particular that the limits of integration for a variable can depend on earlier variables (variables that are "more outer"), but they can't depend on later variables (variables that are "more inner").]
All in all, the triple integral becomes
$$
\int_0^6 \int_0^{(6-x)/2} \int_0^{(6-x-2y)/3} x\,dz\,dy\,dx.
$$
