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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.

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The volume isn't uniquely determined by those bounding surfaces -- it could have $x\lt0$ or $x\gt0$, or $y\lt2$ or $y\gt2$. –  joriki Apr 7 '12 at 8:48
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1 Answer

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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.

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I believe there is a small mistake here, but thank you for the key ideas! I reached an answer. –  dmonopoly Apr 25 '12 at 4:09
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