Expressing a triple integral as an iterated integral in 6 different ways How can I express the triple integral (for volume) as an iterated integral in six different ways, where the solid I would like the triple integral for is bounded by the following surfaces?
$$z=0,\; x=0,\; y=2,\; z=y-2x$$
Additional advice appreciated.
 A: You want to integrate over:
$z\le y-2x$
$0\le z$
$0\le x$
$y\le 2$
If you sketch a picture, you'll see that this is a tetrahedron with vertices $(0,0,0)$, $(0,2,0)$, $(1,0,2)$, $(1,2,0)$.
Obviously, you have 6 possible permutations of variables $x$, $y$, $z$.
Let's have a look at the range of these variables. We have $y\ge 2x+z \ge 0$, hence
$$0\le y \le 2.$$
We also have $x\le \frac{y-z}2 \le \frac{2-0}2=1$, thus $$0\le x\le 1.$$
From $z\le y-2x\le 2-0 = 2$ we have $0\le z\le 2$.

All you have to do now is to choose some order. E.g. we can start with $x$. Then express possible range for $y$ using $x$. And then find in what range $z$ will be, if $x$ and $y$ are given. In this way we get:
$$\begin{align}
0 &\le x \le 1\\
2x &\le y \le 1\\
0 &\le z \le y-2x
\end{align}$$
For example, we have used $y\ge 2x+z \ge 2x$ in the second line. (Since $z\ge 0$.)
If we choose different ordering $x$, $z$, $y$ we get $z\le y-2x \le 2-2x$, i.e.
$$\begin{align}
0 &\le x \le 1\\
0 &\le z \le 2-2x\\
2x+z &\le y \le 2
\end{align}$$
We can try to start with $y$:
$$\begin{align}
0 &\le y \le 2\\
0 &\le z \le y\\
0 &\le x \le \frac{y-z}2
\end{align}$$
I guess you'll be able to do the remaining three possibilities yourself.
