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Would you solve this problem?

$$ \int\int\int_D\ (x+y+z)\ dx\,dy\,dz\qquad D= \left\{ 0\leq x\leq y\leq z\leq1 \right \} $$

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Yes, I would, but what have you tried yourself? –  Anon Jan 30 '13 at 22:41
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Hint: First do the double-integral version of this, i.e. $\int\int_D (x+y)dxdy,\,D=\{0\leq x\leq y \leq 1\}$ if the triple one looks complicated. –  Anon Jan 30 '13 at 22:43
    
By permuting $x,y,z$, convince yourself that this is just $(1/2)\int\int\int x dx dy dz$, where the integral is over the unit cube. –  mjqxxxx Jan 30 '13 at 23:37
    
I think that 2 should be a 3, @mjqxxxx ? –  DonAntonio Jan 31 '13 at 3:13
    
I vote for a 1/6 instead of a 1/2 or 1/3 in mjqxxx comment, since there are 6 permutations of x,y,z. –  coffeemath Mar 24 '13 at 20:24

2 Answers 2

Hint: the integrand is symmetric in $x,y$, and $z$. So the integral over the portion of $[0,1]^3$ where $x \leq y \leq z$ is the same as the integral over the portion where $y \leq z \leq x$ which is the same as the integral any of the other permutations. So the overall integral can be gotten in terms of the overall integral over $[0,1]^3$, which is easier to compute.

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I would first get a glimpse of this $D$ region. Then I'd see if symmetry is my friend. Not always is, but is a good thing to check that.

Which are the limits of the integrals? Imagine there is no third dimension, no $z$. So we are looking at the $XY$ plane. $D$ would be $\{0 \leq x\leq y \leq 1\}$...and that's the region over and including the $x=y$ line but stopping right at $y=1$.

That's easy to see because, everything north of the $y=x$ line are those points with $y$ coordinate bigger or equal than their $x$ coordinate.

You can now see what are the limits for the base of $D$: every $x$ goes from $x=0$ to $x=1$, but every $y$ goes from the line, namely from $y=x$ to $y=1$. Or, if you like, every $y$ goes from $y=0$ to $y=1$, but every $x$ goes from the line $x=0$ and crashes into the line $x=y$. Any description is fine.

Now you do the same in three dimensions, just add another letter. And there you have you integral limits, which is the most challenging thing from this exercise. The rest is up to you.

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