# Rewriting triple integrals

I'm having trouble rewriting a triple integral.

The question is rewrite the following integral in five different ways:

$\int_0^1\int_y^1\int_0^z f(x,y,z) dx dz dy$

I am having trouble with visually seeing the projections onto the different planes because x runs from 0 to z, but z runs from y to 1, so it's hard for me to picture what the bounded regions look like.

the region of integration is simply $$\{(x,y,z) : 0\le x\le z, 0\le y\le z, 0\le z\le 1 \}$$

Now you can rewrite the integrals in the following ways: $$\int_{x}\int_y\int_z\\ \int_{x}\int_z\int_y\\ \int_{y}\int_x\int_z\\ \int_{y}\int_z\int_x\\ \int_{z}\int_y\int_x\\ \int_{z}\int_x\int_y\\$$

the step is everytime the same: for the most outer integral, take the smallest possible lower bound and the greatest possible upper bound. Then for the next integral, take the smallest possible lower bound given the value of the outer variable and go on.

For example:

the integral given is $$\int_{y=0}^1\int_{z=y}^1\int_{x=0}^z\\$$

Now if you want to change this to $$\int_{x}\int_y\int_z$$then $x$ varies between $0$ and $\max z = 1$. So your first integral is $$\int_{x=0}^1$$

Then you go on until $$\int_{x=0}^1 \int_{y=0}^1 \int_{z=\max(x,y)}^1$$

• Do you mean z = min(x,y)? I still don't get how you find the max and min given the outer variables – bkmoney Oct 21 '14 at 22:05
• the max is just the maximum possible value: here, as in the first integral the max for $x$ is $z$ whose upper bound is 1, when $x$ becomes the outer variable its upper bound becomes 1. – mookid Oct 21 '14 at 22:14
• So if I'm following you correctly, the answer in the order dzdydx should be $\int_{x=0}^1\int_{y=0}^1\int_{z=0}^1$ ?? – bkmoney Oct 21 '14 at 22:16
• no. because for the inner integrals, the bounds depends on the values of the outer variables. – mookid Oct 21 '14 at 22:45
• see my message: "Then for the next integral, take the smallest possible lower bound given the value of the outer variable" – mookid Oct 21 '14 at 22:46