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I have this problem: $$\iiint_D xyz \, dx \, dy \, dz,$$ with $D=\{(x,y,z) : 0 \le x \le y \le z \le 1 \}$.

I tried solving it the same way, only that I did

  • $y$ from $x$ to $z$, and not $0$ to $z$.
  • $z$ from $y$ to $1$, and not $0$ to $1$.

obviously, I simplified the problem too much. But where do I go wrong in my reasoning? I want to learn from this mistake.

(Original picture here)

jacob
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2 Answers2

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The integral should be $$ \int_0^1 \int_x^1 \int_y^1 xyz\ dz\ dy\ dx \\= \int_0^1 \int_x^1 \int_x^z xyz\ dy\ dz\ dx \\= \int_0^1 \int_0^z \int_0^y xyz\ dx\ dy\ dz=\frac{1}{48}$$ You have 6 choices for this integral limits and the above are three of them. For this example choose and remove. For example choose $x$ then $z$ then $y$ as follows $$0 \le \color{red}x \le y \le z \le 1 \implies 0 \le x \le y \\ 0 \le y \le \color{red}z \le 1 \implies y \le z \le 1 \\ \implies 0 \le \color{red}y \le 1$$

Semsem
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  • The "choose and remove" is suggestive, and works great this time. Does it always work that smooth, I guess not? Also, I don't really understand the logic behind it. – jacob Apr 13 '14 at 10:36
  • It hardly works for all problem, but you can motivate it. Geometrically, suppose you have a volume element and you want to enlarge it to cover the whole area. Choose any of your axes, determine its limits and then remove it by solution of different equations or inequalities given, and then go to the next axis – Semsem Apr 13 '14 at 11:05
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I believe the problem is that when you integrate $y$ from $x$ to $z$, you haven't done anything to guarantee that $x \leq z$. (I'm assuming you took the outer most integral to be $x$ from $0$ to $1$). You could have instead written the integral in sort of the same spirit you had, by integrating $z$ from $y$ to $1$ and then integrating $y$ from $x$ to $1$ and then integrating $x$ from $0$ to $1$

user2566092
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