Changing order of iterated integral

Question

I am given an integral $$\int_0^1dz \int_0^{1-z}dy\int_0^1f(x,y,z)dx$$ and I want to re-iterate it to have integration with respect to $z$ on the inside and integration with respect to $x$ on the outside.

So, the region is $0 \leq x \leq 1, 0\leq y \leq z-1$, and $0\leq z \leq 1$. Moving these around one can come up with a number of different combinations of bounds, but how do I know which set of bounds is the correct one to get the integral set up the way I want it?

If I plot this region in 3D, it sort of looks like a door wedge I believe, but I do not know if this will help.

Any insight into this would be appreciated!

Answer 1

The integral on the left is the equivalent integral with $z$ on the 'inside' and $x$ on the 'outside'.

$$\int_0^1\int_0^1\int_0^{1-y} f(x,y,z)\,dz\,dy\,dx=\int_0^1\int_0^{1-z}\int_0^1 f(x,y,z)\,dx\,dy\,dz$$

The diagram illustrates the order of integration. The red dot is free to move in the $x$ direction between values of $x=0$ and $x=1$.

The orange dot is free to move in the $y$-direction between the values of $y=0$ and $y=1$.

But the green interval parallel to the $z$-axis is constrained to have $z$ values between $0$ and $1-y$.

Answer 2

We have, that $$~~~~~~~~~~~~~~~0\le z\le 1,~~~ \text{and} ~~~0\le y\le 1-z)\\~~~~~~~\Longleftrightarrow (0\le z\le 1,~~~ \text{and} ~~~0\le y\le 1-z\le 1)\\~~~~~~\Longleftrightarrow (0\le y\le 1,~~~ \text{and} ~~~0\le z\le 1-y\le1)\\\Longleftrightarrow (0\le y\le 1,~~~ \text{and} ~~~0\le z\le 1-y)$$

Hence, $$~~~~~~(0\le x\le 1,~~~~0\le z\le 1,~~~ \text{and} ~~~0\le y\le 1-z)\Longleftrightarrow (0\le x\le 1,~~~~0\le y\le 1,~~~ \text{and} ~~~0\le z\le 1-y))$$

Therefore, By Fubuni, it follows that, $$\int_0^1dz\int_0^{1-z}dy\int_0^1 f(x,y,z)\,dx = \int_0^1dy\int_0^{1-y}dz\int_0^1 f(x,y,z)\,dx$$