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I'm trying to figure out how to switch the order of integration for this problem. I'm given the region (I don't know how to use latex, maybe someone can clean this up for me?)

$z$ from $0$ to $x+y$

$y$ from $0$ to $1-x$

$x$ from $0$ to $1$

That was for the integral $dz\;dy\;dx$ and I have to change it to $dy\;dx\;dz$. I think maybe I drew out the region wrong and that's why I can't figure this out. The one I drew looks like a right triangle. In the first octant The $x$ axis from 0 to 1 and the hypotenuse of the triangle is the line $1-x$ then it goes out in $y$ from 0 to 1 (I can upload the picture if it helps). I don't think I'm taking the $x+y$ part into account anywhere? The new integral I came up with is

$y$ from $0$ to $1$

$x$ from $0$ to $1-y$

$z$ from $0$ to $1$

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    I don't think that $x$ should depend on $y$, if you go in this order.2012-01-21

1 Answers 1

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Write your limits of integration as inequalities:

$0\le z \le x+y$ $0\le y \le 1-x$ $0 \le x \le 1$

and you should be able to see $0 \le y \le 1-x \le 1$ and $0 \le z \le x+y \le x+1-x =1$. You also have $z-x \le y \le 1-x$ so you could have

$\max(0, z-x) \le y \le 1-x$ $0 \le x \le 1$ $0\le z \le 1$

making your formulation

$\int_{x=0}^1 \int_{y=0}^{1-x} \int_{z=0}^{x+y} dz \, dy \, dx = \int_{z=0}^{1} \int_{x=0}^1 \int_{y=\max(z-x,0)}^{1-x} dy \, dx \, dz $

but it might be simpler if you split this depending on whether $x$ or $z$ is bigger so

$z-x \le y \le 1-x$ $0 \le x \le z$ $0\le z \le 1$

and

$0 \le y \le 1-x$ $z \le x \le 1$ $0\le z \le 1$

making your formulation

$\int_{x=0}^1 \int_{y=0}^{1-x} \int_{z=0}^{x+y} dz \, dy \, dx = \int_{z=0}^{1} \int_{x=0}^z \int_{y=z-x}^{1-x} dy \, dx \, dz + \int_{z=0}^{1} \int_{x=z}^1 \int_{y=0}^{1-x} dy \, dx \, dz $

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    I got it now, thanks!2012-01-22